Energy conversionlecture notes and study-groupproblems

O.SydorukImperial College London

List of corrections

Corrected the last integral in the formula sheet

Added a figure of C-core for problem 2.6

p. 10: corrected the sign in front of the integral QFR

ldl

p. 13, 3rd paragraph from the top, line 2 from its bottom should read:ˆ D

Hsphere D dS D D

Hsphere dS D D � .area of the surface/

p. 14, 2nd paragraph from the bottom: replaced ‘The normal to the sidewill always point vertically (either upwards or downwards), but the fieldis horizontal’ with ‘The normal to the side will always be parallel to thecharged plane, but the field is perpendicular to it’.

p. 19, 3rd paragraph from the bottom: corrected the sign of potentials'.M/ and '.B/ to C1 V.

p. 21, 2nd equation from the bottom: corrected the first term on the right-hand side to q1q2=.4�"0r12/

Formula sheetIt will be available at the exam

Maxwell’s equations in integral form Il

.E � dl/ D �ddt

ZS

.B � dS/Il

�H � dl

�D

ZS

.J � dS/ Cddt

ZS

.D � dS/IS

.D � dS/ D

ZV

�dVIS

.B � dS/ D 0

Gauss’s law for electric fields in differential form, Cartesian coordinates

"0

�@Ex.x; y; z/

@xC

@Ey.x; y; z/

@yC

@Ez.x; y; z/

@z

�D �.x; y; z/

Gauss’s law for electric fields in differential form, centrosymmetric distributions, spherical coordinates1

r2

ddr

.r2E.r// D�.r/

"0

Electric flux density and field strength: D D "0"dE. Magnetic flux density and field strength: B D �0�rH.

Coulomb’s lawF D

q1q2

4�"0"dr3r

The Biot-Savart lawdB D

�0

4�I

Œdl � r�

r3

Voltage, potentialUAB D '.A/ � '.B/ D

Z B

A.E � dl/

Electrostatic energyW D

1

8�"0

Xi¤j

qi qj

rij

Capacitance: C D q=U . Inductance: L D ˆ=I . Force on a charge in electric field: F D qE; in magnetic field:F D qŒv � B�.

Rotating machines. Torque, definition (force perpendicular to arm): T D Fa. Torque for a motor with N coils:T D KˆI , where K D 2N=�. Back-emf: e D Kˆ!. Equation of motion: T D Id!=dt .

Useful integralsZ dxp

x2 C a2D ln.x C

pa2 C x2/

Z dx

.x2 C a2/3=2D

x

a2p

x2 C a2

Zxdx

.x2 C a2/3=2D �

1p

x2 C a2

f ormul a shee t

f ormul a shee t

Problems 1: Electrostatics

1. Four point charges are placed in the corners of a square with side a, asshown in the figure. Find the forces acting on all charges. Also, find theelectric fields in the middle of each side.

Cq

�q

�q

Cq

a

Cq

�q

d

2?. An arrangement of two opposite charges, Cq and �q, placed at a dis-tance d from each other is called the electric dipole. Find the field of thedipole at distances much larger than d .

problems marked by ?

are more complicated

3. Find the electric field on the axis of a uniformly charged ring. Theradius of the ring is a; the total charge is q.

4. An infinite plane is charged with the surface charge density � . Find theelectric field of the plane by integration. Compare the result with the onegiven by Gauss’s law. Hint: split the plane into rings and use the resultof the previous problem

5. Two parallel planes have surface charge densities � and �� . Find theelectric field.

6. An infinitely long thin line is charged with the linear density �. Findthe electric field.

7?. A solid sphere of radius R charged with the volume density � hasan uncharged spherical hollow of radius Rh. The centre of the hollow isdisplaced from the centre of the sphere by a (a CRh < R). Find the electricfield inside the hollow.

R

Rh

8. Find the work done by a spatially homogeneous electric field E to movea charge q along a straight line and along a semicircle (see figure).

E

Lq

9. Find the electric potential of the dipole from problem 2.

10. Find the electric potential on the axis of the ring from problem 3, first,by direct integration and, second, by integrating the field.

11. Find the electric potential of the charged plane from problem 4.

12. Find the electric potential of the charged line from problem 6.

13. Find the electrostatic energy of the system of charges from problem 1.

14. Find the capacitance of a spherical capacitor. The capacitor consistsof two concentric spheres of radii R1 and R2.

R2

R1

probl ems 1 : e l e ct ro stat i c s

probl ems 1 : e l e ct ro stat i c s

Problems 2: Magnetic fieldsand electromagnetic induction1?. Find the magnetic flux density in the plane of a ring of a radius R

carrying a current I . Show that far away from the ring the flux densitydeclines as 1=a3, where a is the distance from the ring centre.

2. A long straight hollow conductor carries a current I , which is dis-tributed uniformly. Find the magnetic field strength. The inner radiusis R1, the outer is R2.

R1

R2

3?. An infinitely long solenoid of circular cross-section has n turns per unitlength and carries the current I . I

(a) Show that the magnetic field of the solenoid is directed along its axis.(b) Assuming that the magnetic field of a current carrying loop is propor-tional to the current and decays at large distances a as 1=a3, show that themagnetic field of the solenoid is zero infinitely far away from it.(c) Using the two above results and Ampere’s law, determine the magneticfield inside and outside the solenoid.

4. Using the result of the previous problem, find the inductance of a longsolenoid of circular cross-section with radius R. The solenoid length is l;the total number of turns is N .

5. A ferromagnetic toroid has a core of rectangular cross-section of widthw D 1 cm. Calculate the minimum mean radius r for which the variationof B from the inner to the outer radius is 2% or less.

NI gB

l

6. An iron C-core as shown has a gap length of 1 mm, a total path lengthin the iron of 50 cm, and the coil provides 1000 A�turns. If the iron has�r D 10000, calculate the flux density in the gap, stating any assumptionsmade.

7. For the following applications, state whether high or low values of Bs

and Hc are required and why: (a) strong electromagnet; (b) strong perma-nent magnet; (c) hard drive data storage; (d) loudspeaker; (e) transformercore; (f ) compass needle

a

l

B

8. A square frame with side a enters with a velocity v a region of constantmagnetic field B that has a length l. Calculate and plot the time depen-dencies of the magnetic flux through the frame and the induced emf.

ac

b9. Find the mutual inductance between a straight infinitely long wire anda rectangular frame lying in the plane of the wire.

l10. Find the mutual inductance between two long coaxial solenoids, onewrapped tightly around the other. The numbers of turns are N1 and N2,the area of the cross-sections is S . The length of both solenoids is l.

probl ems 2 : mag ne t i c f i e ld s and em ind uct ion

11. Open and closed circuit tests are carried out on a transformer, withthe input and output voltages, and the input current and power, measuredfor each test. The results are

Xt Rt

XmRiU1 ZL U2

ItImIin

U1 U2 Iin Pin

open circuit test 1000 V 240 V 0.732 A 435 Wclosed circuit test 90 V 0 V 16.1 A 1089 W

Calculate the turns ratio, and the values of the circuit elements.

U RL

LL M

C CRR

12. Two identical metal coils have a capacitance C , an inductance L, anda resistance R. They are coupled magnetically by a mutual inductance M .One of the coils is powered at the resonant frequency 1=

pLC , the other

has a load RL. Find the power efficiency. Analyse the effects of the mutualinductance and the resistance.

13. Find an expression for the group velocity of magneto-inductive wavesand show that the direction of the velocity depends on the sign of themutual inductance.

14. A dc motor has an armature winding of 20 turns and 10 � resistance,and the field windings generate a total flux of 50 mWb. A voltage of 10 V isapplied to the armature and the steady state speed is found to be 100 rpm.Find the torque.

15. The motor of the previous question is now connected to a differentload having the characteristics ! D 10T 2

L . If 10 V is applied to the armature,what will be the equilibrium speed in rpm?

16. The dc motor loaded as in the previous question has 10 mH of arma-ture inductance. Find the initial acceleration d!=dt if the armature voltageis suddenly changed from 10 V to 5 V.

pr obl ems 2 : mag ne t i c f i e ld s and em ind uct ion

Problems 3: Maxwell’s equations

1. The potential of a point charge q is ' D q=.4�"0r/. Using the expressionE D �

�@'

@xex C

@'

@yey C

@'

@zez

�calculate the electric field E.

2. Show that the field inside and outside a uniformly charged solid sphereobeys Gauss’s law in the differential form

@Ex

@xC

@Ey

@yC

@Ez

@zD

�

"0

where � is the volume charge density.

3. At the surface of a hollow sphere of a radius R charged with a surfacedensity � , the relationship between the fields inside and outside the sphereis Eout.R/ � Ein.R/ D � . Find the electric field using Gauss’s law in thedifferential form for centrosymmetric distributions

1

r2

ddr

.r2E.r// D�.r/

"0

probl ems 3 : max w el l ' s e q uat ion s

probl ems 3 : max w el l ' s e q uat ion s

Electric charges and fields

As you know, electromagnetic phenomena are caused by electric charges.We will first study how charges behave when fixed in their positions. Suchcharges are called static, and the theory describing them, electrostatics.

Charges can be of two types: positive and negative. The unit of charge iscoulomb [C]. An electron, for example, has a negative charge of �1:69 �

10�19 C. The charge of a proton is positive.

.xi ; yi ; zi /

qi

Charges can occupy space. The simplest is a point charge that has no size.To describe it, we need to know only its value q and its position. Theposition can be given, for example, by the Cartesian coordinates .x; y; z/.In diagrams, we will draw a point charge as a small circle. We can take N

point charges of different values and place them in different points. Weconstruct so a discrete charge distribution. We can describe it by a set ofcharge values qi and a set of coordinates .xi ; yi ; zi /, where i D 1 : : : N .

x1 x2

�.x/

Charges can also form continuous distributions. These can be of threetypes: charges can be distributed along a line, on a surface, and in a vol-ume. The simplest is a distribution along a straight line. Just like for adiscrete charge distribution above, we can define the position of the lineand its total charge. But the line can be charged inhomogeneously, so thatdifferent parts of the line will have different charges. So, it is not enoughto know the total charge; we must also know the charge density.

For a straight line lying along the x-axis, the charge density � is a functionof the coordinate, � D �.x/. The total charge is then

q D

Z x2

x1

�.x/dx

In this course, we will most often meet homogeneous charge distributions,whose charge density is constant. The total charge of a homogeneouslycharged line segment of length l is simply q D �l, and that of a ring ofradius R is q D 2�R�.

The charge density of a surface charge distribution (we will denote it by�) depends on two coordinates, �.x; y/ in a Cartesian system. A volumecharge density � depends on three coordinates, �.x; y; z/. For homogeneousdistributions, � and � are constant, and the total charges are q D �S andq D �V , where S is the area and V is the volume occupied by the charge.In this course, you will meet surface charges distributed on planes, hollowspheres, and hollow cylinders. Examples of volume charge distributionsare solid spheres and cylinders. It is useful to remember the expressionsfor their surface areas and volumes.

charg e s and f i e ld s 1

Force between point charges

Charges exert forces on each other. The simplest interaction is betweentwo static point charges. The force is then given by Coulomb’s law. Fortwo charges q1 and q2 placed in vacuum at the distance r12 from each other,the magnitude of the force is

F D1

4�"0

jq1q2j

r212

F12

F21

r12

q1

q2

The force also has a direction. To express it mathematically, we can relyon vectors. Vector r12 on the diagram points from q1 to q2. However,it also has a length. To avoid dealing with it, we define the unit vectorOr12 D r12=jr12j. Its length is equal to unity jOr12j D jr12j=jr12j D 1. (It iscommon to denote vectors using a bold typeface and scalars using an italictypeface, so the length of a vector can be written both as jrj and as r.) Wecan then write the force on charge q2 as

F12 D1

4�"0

q1q2

r312

r12

If the charges are of the same sign, the product q1q2 is positive, and forceF12 will have the same direction as vector r12. So, the force is repulsive. Ifthe charges are of the opposite sign, the product is negative and the forceis attractive. We can obtain the force on charge q1 by interchanging thesubscripts 1 and 2. And because r12 D �r21, we have F12 D �F21, as thethird Newton’s law demands.

The factor 1=.4�"0/ is a constant. We call "0 the permittivity of free space(also the vacuum permittivity). It is equal to about 8:85�10�12 F/m (faradsper meter).

The diagram on this page shows type of vectors: vectors connecting thecharges (r12) and forces (F12). The former are vectors in real space; theystart in one point of the real space and end in another. Their length ismeasured in metres and denotes distance between the charges. The forcevectors do not really start in one point of space and end in another, al-though it looks as if they start on a charge and end at a different point. Itis common to draw force (also velocity and field) vectors in the real space,but you have to know the limitations of doing so.

In an arrangement of charges, the force acting on one of them is the vec-torial sum of the forces due to each of the other charges. This propertyis called the superposition principle. For three charges, q1, q2 and q3, theforce on q1 will be F1 D F21 C F31 D q1=.4�"0/

�q2r21=r3

21 C q3r31=r331

�. Andfor N charges, the force acting on charge j is

Fj Dqj

4�"0

NXiD1i¤j

qi

r3ij

rij

The statement i ¤ j below the sign of the sum means that the charge doesnot exert a force on itself.

2 char g e s and f i e ld s

Electric field

Coulomb’s law works perfectly well in electrostatics, where we assume thatcharges are fixed and remain so forever. But if we assume that chargescan move, it will not be able to describe the interaction correctly. Weknow, for example, that it takes time for an electromagnetic signal to travelsome distance. For example, the signals from the Voyager 1 space probereach the Earth in about twenty hours. When the probe sends a signal, itgenerates electric charges. After the signal is sent, the charges in the probetransmitter disappear. Only in twenty hours will this signal create othercharges in a receiver on the Earth. Between these events, no charges exist,but we say that the energy is contained in the electromagnetic field.

A fixed electric charge will create an electric field. The field will exist every-where and regardless of whether other charges are present. The measure ofthe electric field is the electric field strength; its unit is [V/m]. People oftencall electric field strength just ‘electric field’. The electric field of a pointcharge at an arbitrary point A is equal to the force this charge would exerton a unit charge (q D 1 C) placed at this point. From Coulomb’s law then

E.r/ D1

4�"0

q

r3r

Here r is the vector from the position of the charge to point A. Whenpoint A changes, so will r and, hence, the electric field. We can say thatthe electric field is a function of point, E.r/.

In the three-dimensional space, any point has three coordinates, say, x,y, and z in a Cartesian system. The field vector E, on the other hand,has three components, Ex, Ey , and Ez. So, formally, the notation E.r/ isa shorthand for three functions of three variables: Ex.x; y; z/, Ey.x; y; z/,and Ez.x; y; z/. In other words, we assign to every point of space threevalues that characterise the electric field at this point. If the origin of thecoordinate system coincides with the charge, then the x-component of thefield is of the form

Ex.x; y; z/ Dq

4�"0

x

.x2 C y2 C z2/3=2

Mathematically, introducing the field would not be much useful if all threefield components Ex, Ey and Ez were independent of each other. How-ever, they are not; and a major task of this course is to reveal relationshipsbetween the fields and the charges. Those of you familiar with partialdifferentiation can see, for example, that the field components are inter-connected by proving for the field of a point charge that

@Ex

@xC

@Ey

@yC

@Ez

@zD 0

at any point except where the charge is. Those unfamiliar with partialdifferentiation should return to the problem at the end of the term.

charg e s and f i e ld s 3

Because the electric field strength in defined as a force on a unit charge, theelectric field of a charge distribution should obey the superposition princi-ple. The total electric field of a distribution (either discrete or continuous)is the vector sum of the fields created by all its parts (either point chargesor parts of the continuous distribution).

Once we know the electric field, E, we can easily calculate the force on acharge Q placed at any point as F D QE. We have thus modified our initialpicture of ‘charges act on charges’ into ‘charges create electric fields’ and‘electric fields act on charges’. The new picture leads to two different typesof problems. The first is to find the electric field of a charge distribution.The second is to find the force on a charge placed in an electric field. Forexample, we might look at a charge placed in a homogeneous electric fieldEx D E, Ey D 0, Ez D 0. Later, we will also look at how materials behavewhen placed in a field. It is a very convenient approach, as we then neednot worry about how to create such a field.

E E

denser lines D stronger field

It is often easier to analyse an electric field if we can visualise it. We havealready encountered some problems with visualising force vectors in realspace. Visualising fields can become even more confusing, because fieldscan exist everywhere in space. We will need, therefore, to work out aconvention to visualise fields, always keeping in mind its limitations. Thecommon way is to use field lines. These are constructed as follows. First,at every point, the field vector is tangential to a field line. Second, thedirection of a field line coincides with the direction of the field. Third, thelines start at positive and end at negative charges. These rules determinethe direction of the field, but we also have to show the field strength. Theclassic method is by line density: the denser the lines the higher the fieldsin this region. With the advance of computer graphics, people have startedto use a different method: to superimpose the field lines on a colour-mapplot of field strength.

Below are examples of field lines for one and two point charges.

4 char g e s and f i e ld s

Mathematics for electromagnetic theory

The laws for electric and magnetic fields can be formulated math-ematically, using a branch of mathematics called vector calculus. You willnot study it until the second year. In this course, you will not need to knowmuch about the techniques of vector calculus, but you still will need to un-derstand several concepts. Fortunately, it can be done based on what youalready know. We will introduce two such concepts here: line and surfaceintegrals. We start, however, by discussing how to find electric fields ofcontinuous charge distributions.

Electric field and integral

Coulomb’s law allows us to find the electric field of a discrete charge distri-bution (an arrangement of point charges). But the most practically impor-tant are continuous distributions. Here, we will use an example to showhow to calculate the fields of continuous distributions. A straight line ofcharge has the length 2a and the linear charge density �. We need to findthe electric field at the distance b from the centre of the line, on the axisgoing through the line.

�a 0 a b x

First, we need to define the geometry of the line mathematically. Forthis, we have to choose a coordinate system. A proper choice will oftersimplify the mathematics considerably, especially if the geometry has asymmetry. Our problem is one-dimensional: it is convenient to choose anaxis (and use x for the coordinate) along the the line, and put the originat the centre of the line. In this system, the electric field must point alongthe axis E D Eex. Here, ex is the unit vector of the x-axis. The directionof the field being clear, we now have to find its magnitude.

E1 E2

E1 E2 E3 E4

Although we cannot yet say how large the field is, we can use the superpo-sition principle. It states that the field of a whole distribution is the sumof the field produced by its parts. So, the whole line produces the samefield as its two halves together, E D E1 C E2. The same as true for the fieldproduced by all quarters, E D E1 C E2 C E3 C E4. We can keep splitting theline into smaller segments, so that for the total N of them

E D

NXiD1

Ei

The length of each segment will be �x D 2a=N ; and its charge, �q D �x�.

We have now presented the unknown field of the whole line by the sumof the unknown field on the segments. However, we notice that the moresegments in the line, the smaller the size of an individual segment, and sothe closer its field will be to the field of an equivalent point charge.

mat hemat ic s 5

If we denote the field of the i-th point charge by �Ei , we can write

E �

NXiD1

�Ei

This expression contains two important steps. First, we replaced the strictequality ‘D’ by an approximate one ‘�’. We know, however, that the ex-pression will become more and more exact as we increase the number ofsegments. Putting it mathematically jE �

P�Ei j ! 0 for N ! 1. We are,

of course, on the way to an integral.

Second, we replaced the unknown fields of the segments with the fields ofpoint charges, whose expressions we know. Assuming that a point chargeis in the middle of a segment, the coordinate of the leftmost point chargeshould be x1 D �a C a=N . The next point charge to the right will be atx2 D �a Ca=N C2a=N . The i-th charge is then at xi D �a Ca.2i �1/=N . Wecompute the field at a distance b from the origin, and hence at the distanceri D b � xi from the i-th charge. The field due to the i-th charge can bewritten then as

�Ei D1

4�"0

�q

r2i

D�

4�"0

�x

r2i

D1

2�"0

�a

N .b C a � a.2i � 1/=N /2

The field of the whole line is then

E ��a

2N�"0

NXiD1

1

.b C a � a.2i � 1/=N /2

As discussed above, the exact expression for the field can be written as

E D limN !1

�a

2N�"0

NXiD1

1

.b C a � a.2i � 1/=N /2

Evaluating this expression directly is not easy. However, to say N ! 1

is equivalent to saying �x ! 0, so the expression for the field can also bewritten as

E D lim�x!0

�

4�"0

NXiD1

�x

.b � xi /2

which is, of course, the integral

E D�

4�"0

Z a

�a

dx

.b � x/2D

�

4�"0

1

b � x

ˇ̌̌̌a�a

D1

2�"0

�a

b2 � a2

We always can find the field of a continuous distribution by splitting itinto small parts, approximating each by a point charge, and summing theirfields. But this sum can always be written as an integral, and integratingis much easier than calculating the sum and finding its limit. And so, weneed not write any sums at all; we could just formulate the integral.

x b � x

0 dx dETo do so, we choose an arbitrary small segment on the line. The segmenthas the coordinate x and the length dx. The field it creates is that of a pointcharge, dE D �dx=.4�"0.b � x/2/. We obtain the total field by integratingE D

R a

�adE. Using the differentials dx and dE instead of �x and �En saves

time and effort.

6 mat hemat ic s

It is often useful to check the validity of a result by looking at limitingcases. If b � a then E � �a=.2�"0b2/, which is the field of a point charge.The result confirms one’s expectations: far away from the line, its geometryshould not matter, and it should ‘look like’ a point charge.

We can now to formulate a procedure for calculating the electric fields ofcontinuous distributions.

1. Choose a coordinate system. It means choosing the type of the system(Cartesian, polar etc.), the direction of the axes, and the place for the ori-gin. Often, a symmetry will suggest a choice. For example, if a rotationaxis exists, a cylindrical or polar system might be suitable. It is also oftenconvenient to put the origin at a centre of symmetry.

2. Split the charge distribution into equal small segments whose fieldsyou can approximate analytically. Almost always, it means approximat-ing by point charges. Exactly how you do it depends on the problem. Wesplit the straight line into straight segments of length dx. If we had a ring,on the other hand, it would be convenient to split it into segments eachoccupying the angle d' when viewed from the centre.

3. Take a small segment with an arbitrary position and find the expres-sion for its field. This stage involves geometric, trigonometric, and vectormanipulations. An appropriate choice of the system will pay off.

4. Write the total field as an integral. Usually, it would merely meandefining the integration limits. The problem is now solved from the pointof view of physics. You might be able to evaluate the integral analytically(always possible for problems in this course but rarely in real life). If not,you can evaluate the integral numerically or, perhaps, try to find an ana-lytical approximation.

5. Analyse the result. Is the dimension of your expression correct? Itshould be [charge/("0�distance2)], and mistakes in integration often leadto a wrong power of the distance. How should the field behave far away,and does your expression comply?

We formulated the field of a line charge as an integral. We arrived to theintegral by considering a sum of the fields of point charges. Herein liesthe meaning of the integral. How we evaluate it (using antiderivatives) isa different matter. We might not know how to evaluate an integral but wewill still be able to understand what it means.

Using the same approach, we will introduce below the concepts of line andsurface integrals.

mat hemat ic s 7

Work and line integral

The concept of work is important in many fields. We will see later, forexample, that electrostatic work defines voltage. We start to discuss workusing a mechanical example.

A B

F Imagine a person is rolling a ball on a flat surface. There is no gravity andno friction, but there is a wind blowing horizontally and with a constantforce against the motion of the ball. What is the work the person has todo to move the ball to a distance L, from point A to point B?

It is one of the rare cases when the physical and the colloquial meaningscoincide. It is obvious that the work is the larger the stronger the windand the longer the path. Denoting the force of the wind by F , the work is

WA!B D FL

The wind is constant, so the work depends only on L, but not on the po-sitions of points A and B.

F

L˛

A B

Let us consider the wind at an angle to the horizon. The wind createsless resistance to the movement, and so the work should be smaller. As alimiting case, the wind blowing from the top will not oppose the move-ment at all. In mathematical terms, the person does the work against theprojection of the force on the horizontal line

WA!B D FL cos˛

We can write the same differently. The force is a vector, and we can intro-duce the path vector L as starting at A and ending at B. The angle betweenthe two vectors is ��˛, so the scalar (dot) product is .F �L/ D FL cos.��˛/ D

�FL cos˛. The work should have the opposite signWA!B D �.F � L/

What is then the work the person has to do to bring the ball in the oppositedirection, from B to A? The path vector is now �L, so

WB!A D .F � L/ D �WA!B

The work becomes negative, which is no longer intuitive. The reason lies inhow we formulated the question. We wanted to know the work the personhas to do. But the person need not apply any effort to bring the ball from Bto A. The wind will do it. In fact, the person might use the kinetic energythat the moving ball will acquire. It is reflected by the negative of sign ofthe work the person has to do.

The matter becomes more confusing when we consider the work that thewind does, not the person. The magnitude will be the same but the signwill reverse. This work is positive if the ball moves from from B to A andnegative when it moves in the opposite direction.

xA xB

F.x/ The next generalisation is to take the wind whose strength and directionvary from point to point. To describe the force, we introduce the hori-zontal x-axis, so that the force is a vector that depends on x, F.x/. Thecoordinates of points A and B become xA and xB, and the path vector be-comes L D .xB �xA/ex. The force is varying, so the above equations for thework are invalid.

8 mat hemat ic s

But we have already dealt with the same situation when finding the electricfield of a line charge (see p. 6). The work is additive: the work done movingthe ball from A to B is the sum of the works needed to move it from A toC and then from C to B. Therefore, we can split the path into small equalsegments of the length �x. The work done to move the ball along the i-thsegment is �Wi . The total work is WA!B D

P�Wi .

The shorter the individual segment, the less will the force vary along it. Wecan assume then that the force does not change along a segment and besure that the error we make decreases as we take more and more segments.Then �Wi � �.F.xi /�ex/�x, where xi is the position of the segment. Takingthe limit �x ! 0 gives

WA!B D �

Z xB

xA

.F.x/ � ex/ dx D �

Z xB

xA

Fx.x/dx

Here Fx.x/ is the projection of the force on the x-axis. Because the forcenow varies, the work depends not on the total length of the path but alsoon the starting and finishing positions. If the force is constant, Fx can betaken out of the integral sign. Then R xB

xAdx D xB � xA D L, and we arrive

at the above result WA!B D FxL.

A

B

We have so far stayed within the boundaries of the familiar calculus, butwill now take a further step and assume that the path from A to B is nolonger a straight but a curved line. First, we need to describe the linemathematically. We will assume two dimensions (three are treated in thesame way) and introduce a Cartesian coordinate system. We can thendescribe the path as a function, say, of the form y D l.x/. We also specifypoints A and B and so determine the direction of movement. Second, weneed to know the force on this path, F.x; y/.

�li

F.xi ; yi /To find the work we proceed as before. We split the path into many seg-ments of equal length. The more segments we have the less the forcechanges along a segment, so we can assume that the force is constant alonga segment. We can also assume that the more segments we have, the bet-ter each of them can be approximated by a straight line. Then, for eachsegment, we can approximate the work assuming a straight path and aconstant force, so that for the i-th segment

�Wi � �.F.xi ; yi / � �l.xi ; yi //

The vectors �l.xi ; yi / vary from segment to segment, because their orien-tation changes. The total work is then

WA!B D � lim�l!0

Xi

.F.xi ; yi / � �l.xi ; yi //

This limit is called line integral and is denoted as

WA!B D �

Zl

.F � dl/

Here, we do not write the dependencies .x; y/ for brevity, but also becausewe may use a different (say, polar) coordinate system.

mat hemat ic s 9

You should understand that the line integral is not the usual integral you arefamiliar with. Here, l is not the dummy variable; the force is not definedas F.l/, and the limits are not some l1 and l2. You should see behind thenotation of the line integral the above procedure of splitting the path intosegments, approximating each by a straight line, assuming a constant force,and finding the work along this segment depending on the force and theorientation of the segment.

Of course, techniques exist to evaluate line integrals. The equation of theline is known, so the values of y are determined for any x. It should thenbe possible to find expressions for dl and F that depend only on x. Afterthat, the line integral can be transformed into a usual one. We do notconsider these techniques in any detail in this course; in our problems,evaluating line integrals will be simple. One such simple situation is whenthe projection of F.x; y/ is constant on l. (It does not imply that F has aconstant magnitude or direction.) Denoting this projection by QF , we canwrite

WA!B D QF

Zl

dl D QF L

where L is the total length of the path. Another obvious situation, whichwe have already discussed, is when the path is a straight line.

Line integrals appear in many applications and describe not only work.We will see later, for example, that line integrals of magnetic fields describeelectric currents.

Often, we will consider a line integral along a closed path: we start atone point, move along the path, and return to the same point. We callsuch integrals closed path integrals and denote them by the symbol H . Theclosed path integral H

l.F � dl/ is called circulation of F around l.

Note that we use the same symbol l to denote the path, the path length,and an element of the path. Likewise, the symbol S can denote a surface,its area, and an element of the surface.

10 mat hemat ic s

Flux and surface integral

When we were discussing work, we extended the concept of integral fromfunctions that vary along a straight axis (i.e. depend on a single variable) tofunctions that vary along an arbitrary line. The next step would be to con-struct integrals of functions that vary along a surface. Such integrals willallow us to calculate the flux of a vector field. As before, we will introducesuch surface integrals using a mechanical example.

L D vt S

v

Imagine that water flows along a pipe of a square cross section. The watervelocity is v, the cross-sectional area is S . What is the volume V of waterthat will pass through the pipe in time t? The answer is: all water thatreaches the cross section in this time. The distance that the water covers isL D vt , so V D SL D Svt . We will assume that the velocity is constant withtime, so time is of no interest to us. We take t D 1. The resulting volumeper unit time is called flux (its dimensions will be [m3=s]). We will denoteis as ˆ, so that ˆ D Sv.

S S 0 v

n

˛˛

We will now find the flux through a different surface, S 0, that rests onthe cross-section S as shown in the figure. The flux should not change,because the same amount of water goes through S 0 and S . In terms ofS 0, ˆ D vS 0 cos˛, where ˛ is the angle between S and S 0. This expressionlooks like a dot product. Dot product is defined between two vectors. Theproblem is that velocity is a vector but area isn’t. To create a vector thatis related to the surface, we can introduce a normal to the surface, n. Wedefine it as a dimensionless vector whose direction is perpendicular to thesurface n?S 0 and whose length is unity jnj D 1. Then we can write

ˆ D .v � n/S 0

Introducing the normal allows us to extract further information from theexpression for the flux. If we reverse the direction of the water flow, thenthe velocity vector will change its sign vrev D �v . The flux is then

ˆrev D .vrev � n/S D �.v � n/S D �ˆ:

ˆinto > 0 ˆout of < 0

n

Snegative positive side

So, when the velocity reverses its direction, the flux changes its sign. Andchoosing the normal n allows us to define a ‘positive’ and a ‘negative’ sidesof the surface and so to distinguish between water flowing ‘into’ and ‘outof ’ the surface. Of course, what will be ‘positive’ and ‘negative’ sides de-pends on our choice. We could equally well choose the normal that wouldpoint in the opposite direction, thus swapping positive and negative sides.But no matter the choice, opposite signs of the flux will mean oppositedirections of the velocity.

mat hemat ic s 1 1

S

v

n

�S

We can now generalise the concept of flux to surfaces of complex shapesand to varying velocities. The way to do it is the same as for work. We firstnote that flux is additive: the flux through the whole surface is the sum offluxes through all its parts. And so, we split the now-complex surface S

into smaller surfaces of equal areas. We then approximate each of them bya rectangular patch, noticing that the smaller the area, the smaller error wemust be making. Each rectangular patch will have the area �S and its ownposition, defined by three coordinates, say, .x; y; z/. For each patch we candefine a normal. We also make sure that the direction of the normal doesnot change from patch to patch. In other words, if we define a positiveside of the whole surface S , all patches must comply. The flux through thei-th patch is then �ˆi D .v.xi ; yi ; zi / � n.xi ; yi ; zi /�S/. We can find the totalflux as

ˆ D lim�S!0

X.v.xi ; yi ; zi / � n.xi ; yi ; zi //�S

We call this limit surface integral and write

ˆ D

ZS

.v � n/dS

Here, we dropped the coordinates .x; y; z/, to allow for non-Cartesian sys-tems. Some books use alternative notations. First, we can define forbrevity a new vector dS D ndS , so that ˆ D

RS

.v � dS/. Second, to dis-tinguish between surface and line integrals, the double integral symbolcan be used, ˆ D

’S

.v � dS/. Also, closed surfaces are very common, forwhich the symbols H and – are used, similar to line integrals.

n

direction of the closed pathWe said above that the choice of positive and negative side of an open sur-face is arbitrary. However, the border of an open surface is a closed path.The direction of a path is important for calculating line integrals. A con-vention exists: once we choose the direction of the normal, the direction ofthe path is determined automatically by the right-hand rule. To apply therule, point the right-hand thumb along the normal and curl the remainingfingers. These fingers will point into the direction of the border path.

closed surface

outward-pointing normalsClosed surfaces are different. They enclose volumes and have natural ‘in-side’ and ‘outside’ regions. The directions of their inward- and outward-pointing normals will be strictly defined. We will always be choosingoutward-pointing normals.

We have assumed that the vector v is a velocity and defined ˆ as volumeper unit time. But we can now take any vector field u and define for it fluxthrough a surface S as ˆ D

RS

.u � dS/. The flux so defined will no longerhave the meaning of volume per unit time. We might not be able to easilyinterpret it at all. Nevertheless, fluxes of electric and magnetic fields playan important role in electromagnetics, as we shall see next.

12 mat hemat ic s

Gauss’s law

Having introduced the concepts of surface and line integral, we cannow return to the study of electric fields. One of the laws for electric fieldsis Gauss’s law.

Earlier, we defined electric field strength, E, as a force acting on a unit(test) charge. We still consider charges in vacuum and define now anothervector, the electric flux density, as

D D "0E

R

nD

For a point charge q in vacuum, D D qr=.4�r3/. Let us find the flux ofD through an imaginary sphere of a radius R with the point charge at itscentre. Such imaginary surfaces are also called Gauss surfaces. By def-inition ˆ D

Hsphere.D � dS/. The sphere is a closed surface so we use the

symbol H . Also, dS D ndS , where n is a normal to the sphere. Remem-ber that we always take the outward-pointing normal. The normal to thesphere at any point will be pointing along the radius. But so will alsothe electric flux density D. Hence, D k n and .D � dS/ D DdS . Then, theflux is ˆ D

Hsphere D dS . The magnitude of the flux density is not con-

stant in space. But it is constant everywhere on the sphere S . On thesphere with the radius R, D D q=.4�R2/ D const. Thus, we can writeˆ D

Hsphere D dS D D

Hsphere dS D D � .area of the surface/. The area of

the surface of the sphere is 4�R2, so we finally have Hsphere.D � dS/ D q.

Electric flux through a closed surface is equalto the charge this surface encloses

We will now claim that this result is general. For any charge distributionand any closed surface, the total electric flux through the surface is equalto the charge enclosed by this surface. Mathematically, we write it asI

any surface.D � dS/ D qenclosed

Q

S1 S2

Let us see how Gauss’s law works. We take some charge distribution withan arbitrary shape and density; its total charge is Q. We surround thisdistributions by two surfaces. One of them, S1, is closer to the charge; theother, S2, is farther from it. Nevertheless (and regardless of the shapes ofS1 and S2), we have H

S1.D � dS1/ D

HS2

.D � dS2/ D Q.

S3The surface S3 does not enclose the charge (the charge is not inside thesurface). Hence, the total flux is zero H

S3.D � dS3/ D 0. But do not confuse

flux with electric field. Although the flux through S3 is zero, the fieldanywhere on and inside S3 might be non-zero.

S4The surface S4 encloses only a part of the charge distribution, and so onlythis part will contribute to the flux through S4. The greyed part will not.

Gaus s ' s l aw 13

Using Gauss’s law to calculate electric fields

We already know a way to calculate electric fields of continuous chargedistributions: split the distributions into small parts, approximate each ofthem as a point charge, and integrate. For some distributions, Gauss’s lawgives an alternative, much simpler way.

Assume that we have some charge distribution whose field we want tofind. Assume we can surround the charge we an imaginary surface withthe following properties. First, everywhere on this surface D k n. Andsecond, everywhere on this surface D D const. Then Gauss’s law is simplyDS D q, where S is the area of the surface and q is the charge enclosed.But D D "0E, so that

BAC

E

E Dq

"0S

Let us look at several examples. The first is an infinite vertical planecharged with the constant surface charge density � . We take a point Ato the right from the plane. Because the plane is infinite, this point isequivalent to any other point B at the same distance from the plane. So,E.A/ D E.B/. If we flip the plane along the horizontal axis going throughA, the plane will remain the same (its top and bottom are the same, soflipping should play no role). Because the geometry does not change, thefield should not change as well. The only possibility is that E.A/ is ori-ented horizontally and points either towards or away from the plane. Thedirection will depend on the sign of the charge. If the plane is chargedpositively, the field will point away from the plane. It also means that atpoint C to the left of the plane, the field will be in the opposite direction.If points A and C are equidistant from the plane then E.A/ D �E.C/.

D

n

n

A

C

We will now build an imaginary Gauss cylinder around a part of the planeas follows. The bases of the cylinder have the area S . One base is centredat point A and the other at point C. So, the plane cuts the cylinder in half.The charge enclosed by the cylinder is q D �S . The total flux through it isthe sum of the fluxes trough each of the bases and through the side. Thenormal to the side will always be parallel to the charged plane, but the fieldis perpendicular to it. We have .D � n/ D 0, so that the flux though the sideis zero. The outward-pointing normal to the right base points to the right.So does the field. On the other hand, the field is constant everywhere onthe base. Hence, .D � n/ D "0.E � n/ D "0E. The flux through the left baseis the same, so that the total flux through the cylinder is equal to the fluxthrough the two bases, 2"0ES . It also must be equal to the charge enclosedso that 2"0ES D �S or

E D�

2"0

Deriving the same result by integration requires more effort. The salientfeature of this result is that the field is constant no matter how far awayfrom the plane.

14 Gau s s ' s l aw

Our next example is a solid sphere of radius R with the charge q. Thevolume charge density is then � D 3q=.4�R3/. Like the plane, the spherehas a symmetry: we can rotate and flip it around the centre, and it will notchange. Therefore, the electric field, both inside and outside the sphere,should be oriented along the radius-vector from the centre, E k r.

R

r

Dn

R

r

Gauss surface

q

outside inside

We will first find the field outside the sphere, at a distance r from its cen-tre. The choice of the Gauss surface, as suggested by the symmetry, is thesphere of radius r with the same centre. The normal to this surface clearlyis along the radius, n k r. We again have the situation when D k n andD D const on S . Because the surface encloses the whole of the charge, wehave 4�r2"0E.r/ D q so that

Eoutside.r/ Dq

4�"0r2

So, the field outside a charged sphere is the same as the field of a pointcharge q placed in the centre of the sphere. In particular, the field strengthdecays as the square of the distance from the sphere.

We will now find the field inside the charged sphere. The Gauss surfaceshould again be a sphere, flux through which is 4�r2"0E.r/. However, theGauss sphere encloses not the whole charge but the charge q 0 D 4�r3=3� D

qr3=R3. Equating the charge and the flux, we obtain

Einside.r/ Dq

4�"0R3r

The field is proportional to the distance from the centre. Also, at the sur-face of the charged sphere Einside.R/ D Eoutside.R/.

Gaus s ' s l aw 15

16 Gau s s ' s l aw

Voltage, potential andelectrostatic energyWe previously considered work required to move an object on which aforce acts. If the object moves from point A to point B along a path l, thework done by the force can be expressed by a line integral WA!B D

R BA .F �dl/.

(Remember: it is a line integral where l is not the dummy variable; A andB are not its limits. This notation means integrating from point A to pointB along the path l.)

If the object is a point charge q placed into an electric field E then F D qE

and WA!B D qR BA .E �dl/. We now assume that q D 1. We define the voltage

between point A and point B as the work that has to be done (by the forcesof the field) to move a unit charge from A to B, so that

UAB D

Z B

A.E � dl/

From this definition, the dimension of voltage is ŒU � D Œfield strength �

distance� D ŒV=m � m� D ŒV�, as it should be.

q

BA

PO

N

We found before that the value of a line integral and of the work will, gen-erally, depend both on the end-points and on the path. But we have nowdefined voltage referring only to the end-points and ignoring the path.It turns out that the path does not matter. As a demonstration, we willconsider the voltage due to a point charge. The point charge q creates theelectric field E D qr=.4�"0r3/. We will calculate the voltage between pointsA and B that lie on the same radius-vector. We will take two paths. Thefirst path is along the straight line connecting A and B (solid line). Thesecond path (dashed line) is ANOPB, with pairs NO and PB lying on thesame circles.

For the straight path AB, we have dl D dr, so that .E � dl/ D qdr=.4�"0r2/.The voltage between A and B is then

UAB D

Z B

A.E � dl/ D

Z rB

rA

qdr

4�"0r2D

q

4�"0

�1

rA�

1

rB

�Here rA and rB are the distances from the points A and B to the charge q.

We now consider the more complicated path ANOPB. The line integralalong it can be split into the sum of integrals along AN, NO, OP, and PB.Then, the voltage between A and B calculated along this path is

UAB D

Z N

A.E � dl/ C

Z O

N.E � dl/ C

Z P

O.E � dl/ C

Z B

P.E � dl/

The values of E and dl will be different for each of these integrals. Thepaths from A to N and from O to P are along radii, so we haveZ N

A.E � dl/ D

q

4�"0

�1

rA�

1

rN

�and

Z P

O.E � dl/ D

q

4�"0

�1

rO�

1

rP

�

voltag e 17

The paths NO and PB are along circles, where the movement is orthogonalto the electric field. Hence, E � dl D 0, and the integrals are zero.

q

BA

So, the voltage between A and B found along ANOPB is the sum of inte-grals along AN and OP. Because points N and O and points B and P areat the same distance from the charge, rN D rO and rB D rP. Finally:

UAB Dq

4�"0

�1

rA�

1

rN

�C

q

4�"0

�1

rO�

1

rP

�D

q

4�"0

�1

rA�

1

rB

�The result is the same as when moving straight from A to B.

We now claim that the result is true for any electrostatic field, not nec-essarily that of a point charge: for arbitrary electrostatic field, the voltagebetween two points does not depend on the path chosen between them.

The voltage between two pointsdoes not depend on the path

UAB D

Zany path between A and B

.E � dl/

A practical side of this result is that we try to choose the simplest pathwhen finding the voltage between two points. For the above example ofthe point charge, finding the voltage along the straight path connectingpoints A and B was simpler than along the path ANOPB.

E l1

l2

AB

The result of voltage being independent of path can be written in a differentway. Assume we have an arbitrary spatially-varying field E. The voltagebetween points A and B calculated along a path l1 is UAB D

Rl1

.E � dl/.The voltage between points B and A calculated along a different path l2 isUBA D

Rl2

.E � dl/. But UAB D �UBA, so that UAB C UBA DR

l1.E � dl/ C

Rl2

.E �

dl/ D 0. On the other hand, going from A to B along l1 and then from Bto A along l2 means going along the closed path built by combining l1 andl2. Because l1 and l2 are arbitrary, so is the closed path. The circulation ofelectric field around any closed path is zero. We can writeI

any path.E � dl/ D 0

Fields obeying these property are called conservative. Electrostatic fieldsare conservative, and so are gravitational ones. However, as we will seelater, time-varying electric fields are not conservative.

The two equations, H .E�dl/ D 0 and Gauss’s law H.D�dS/ D qenclosed; describe

electrostatic fields completely. They are constituents of a general system ofMaxwell’s equations, which describe electric and magnetic fields.

18 voltag e

Potential

q

BA

M

Let us return to the problem of the point charge q. As we found, thevoltage between point A and point B is

UAB Dq

4�"0

�1

rA�

1

rB

�We now introduce a third point, M, infinitely far away from the charge.Because the voltage between point A and B is independent of the path, itcan be written as UAB D UAM C UMB D UAM � UBM. We changed the signin the second term because we swapped the points. The voltage UAM is

UAM Dq

4�"0

�1

rA�

1

rM

�but point M is at infinity, where the electric field is zero and rM D 1, so

UAM Dq

4�"0

1

rAand, analogously, UBM D

q

4�"0

1

rB

We now introduce electrostatic potential for points A and B as '.A/ D

q=.4�"0rA/ and '.B/ D q=.4�"0rB/. Then, the voltage UAB is the differencebetween the potentials, UAB D '.A/ � '.B/. Voltage is also known as ‘po-tential difference’. For an arbitrary point at a distance r away from thecharge, the potential is '.r/ D q=.4�"0r/.

Let us recap the logic. Voltage is found between two points, and dependson the position of both. However, we can choose one point and assign toit a (zero) value of potential. Then, any other point can be characterisedby its own potential. The voltage equals the difference of the potentials.

Choosing point M at infinity and assigning zero potential to it is conve-nient and physically sound. However, we could choose any other pointand assign a different value of potential to it. Let us, for example, agreethat '.B/ D 1V. Then, the potentials at other points will change. The po-tential at infinity is �.M/ D 1V�q=.4�"0rB/, and the potential at point A is'.A/ D 1V C q=4�"0.1=rA � 1=rB/. The voltages, however, remain the same,regardless of the convention for the potentials.

The same arguments apply to all electrostatic fields, not only those of pointcharges. If we know the potential of point M, then for point A we have

'.A/ D

Z M

A.E � dl/ C '.M/

equipotential linefield line

If we know the potential, we can calculate the electric field strength (howto do it, we will discuss later). As a result, we can equally well describe afield by its potential or by its strength. But potential is a scalar, whereaselectric field strength is a vector. So, potential is simpler to deal with. Inparticular, it is easier to plot. Plotting potential is done by equipotentiallines. These are lines along which the potential remains constant. In otherwords, the voltage between any two points on an equipotential line is zero.The voltage is proportional to .E � dl/, and it can be zero either when E D 0

or when E ? dl. Therefore, equipotential lines are orthogonal to field lines.

voltag e 19

Calculating potential

Our results show that we can calculate the potential if we know the electricfield. We will now discuss how to find the potential of a given chargedistribution directly.

We will rely on the superposition principle. The potential, by definition,is work, and work is additive. So, if we have two point charges q1 and q2,the potential they will create at an arbitrary point A will be the sum of thepotentials of the individual charges, so that

'.A/ D '1 C '2 Dq1

4�"0r1

Cq2

4�"0r2

Here r1 and r2 are the distances from point A to the charges. Extendingthis expression to a discrete distribution of N point charges, we can write

'.A/ D

NXiD1

qi

4�"0ri

For continuous charge distributions, we can apply the same logic as wedid when finding electric fields. We split the distribution into small parts.Due to the superposition principle, the potential of the whole distributionwill equal the sum of the potentials created by each part. By making theparts smaller (and so increasing their number), we can approximate each ofthem by a point charge. We can then write the correct correct expressionfor the potential as an integral.

x b � x

0 dx d' We will apply this method to a line charged homogeneously will a density�. The length of the line is 2a. Similar to our earlier example of the electricfield (see p. 6), we will find the potential on the axis defined by the line, atthe distance b from its centre. The potential d' of an element dx is

d' D�

4�"0

dx

b � x

The potential of the whole line is then

' D�

4�"0

Z a

�a

dx

b � xD �

�

4�"0

ln.b � x/ja�a D

�

4�"0

lnb C a

b � a

We saw earlier that the potential is given by an integral of the field. Wecould therefore find the field by differentiating the potential. For our ex-ample, the field on the axis of the line is given by

E D �d'

dbD

1

2�"0

�a

b2 � a2

which is the same result as we found before.

Generally, if we know the potential '.x; y; z/ of a charge distribution, wecan find the electric field as

E D �

�@'

@xex C

@'

@yey C

@'

@zez

�Because potential is a scalar and electric field is a vector, it is often easierto find the potential first and then to differentiate it to find the field.

20 voltag e

Electrostatic energy

q1

r12

q2

A single point charge q1 creates an electric field, but no fields act on thischarge. That is why no work needs be done to move this charge around.Assume now that we keep the charge q1 fixed at its position and moveanother charge, q2, from infinity close to q1. At infinity, the charges donot interact. But when we start bringing them closer to each other, chargeq1 will exert a force on charge q2. If the charges are of equal sign, thencharge q1 will oppose the movement of charge q2. To bring the charges ata distance r12 from each other, we have to do a work equal to

W12 D q2'1 Dq1q2

4�"0r12

where '1 is the potential due to q1. We say that this work goes into chang-ing the potential energy of the system of two charges. If the charges arereleased, the repulsion forces will move them apart. Assuming there is nofriction, the charges will accelerate until the distance between them be-comes infinite. If the charges have equal masses m, the maximum kineticenergy they will get will be equal to the potential energy they had in thebeginning. Denoting the terminal velocity by v, we can write

2mv2

2D W12 D

q1q2

4�"0r12

Potential energy has been converted into kinetic energy of charge move-ment (and, hence, electric current).

On the other hand, if charges q1 and q2 are of the opposite signs, they willbe attracted to each other. The attraction forces (not the external force)can now will bring the charges together. It will be reflected in the negativesign of the potential energy W12.

q1 q2

q3

r12

r13 r23

Suppose now we have the two charges, q1 and q2, separated by the distancer12 and we want to bring a third charge q3 from infinity close to them. Bothcharges q1 and q2 will exert forces on q3, so the work that has to be doneto move q3 is

W13 C W23 D q3'1 C q3'2 Dq1q3

4�"0r13

Cq2q3

4�"0r23

where r13 is the distance between charges q1 and q3, and r23 is the distancebetween q2 and q3. Moving charge q3 will change the potential energy ofthe system. The total potential energy of the system of two charges will beequal to

W D W12 C W13 C W23 Dq1q2

4�"0r12

Cq1q3

4�"0r13

Cq2q3

4�"0r23

We can now see a pattern: the total potential energy of a system of pointcharges is the sum of the energies of all charge pairs. Mathematically

W D1

2

Xi¤j

qi qj

4�"0rij

D1

8�"0

Xi¤j

qi qj

rij

The factor 1=2 stands in front of the first sum because the sum takes eachpair twice. For example, the sum calculates the energy between charge q4

and q8 first as q4q8=r48 and then as q8q4=r84.

voltag e 21

22 voltag e

Conductors

We have so far discussed charges placed in vacuum. We will now con-sider how materials react to and modify electrostatic fields. Electrostaticsconsiders two types of materials: conductors and dielectrics. Here, we willdiscuss conductors.

In conductors, charges are free to move anywhere inside a material with-out resistance. We will also assume that the number of free charges ina conductor is infinitely large. Such conductors are called perfect, andmetals with high conductivity are the closest to being perfect conductors.The conductivity of metals is due to free electrons, so we will refer to freecharges as electrons. The total charge of electrons is compensated by thetotal charge of immobile ions that constitute the material. So that the totalcharge of a conductor is zero unless it is charged externally.

E E

Ei

metalvacuum vacuumAssume now we place a conducting plate in a constant homogeneous elec-tric field E. The field is horizontal, and the plate is infinite in the verticaldirection. If the field penetrates inside the conductor, it will act upon thefree electrons. The electrons will move to the left, leaving behind positivelycharged ions. So, the left part of the conductor will start to charge neg-atively and the right one, positively. These positive and negative chargeswill create their own, internal, field Ei directed from the right to the left—opposing the external field. The total field inside the conductor will beE C Ei and will thus decrease. Because the number of free electrons isinfinite, a nonzero field inside the material will lead to further electronsmoving to the left. The internal field will increase, and the total field willdecrease. The process will take place until the total field inside the con-ductor vanishes, and electrons stop moving. We study electrostatics, wherefields do not change with time. It means that we consider situations whenthe charge redistribution inside the conductor has already happened, nocharges move, and the field inside the conductor is zero.

Electric field inside a conductor is zero

Non-zero charge can exist only onconductor surface

If the field everywhere in the conductor is zero, no matter what the closedimaginary surface we choose inside the conductor, the electric flux throughthis surface is also zero, ˆ D "0

R.E � dS/ D "0

R.0 � dS/ D 0. It can only be

when the charge inside the conductor is zero. So, all the electrons thatmoved to the left and all the uncompensated positive ions can only bepositioned on the surfaces.

These results are true not only for the conducting plate but for an arbitrarilyshaped conductor. Regardless of the shape, the field inside is zero, and thecharges can be only on the surfaces. Because the conductor is electricallyneutral, the total charge on the surfaces should be zero.

cond uctor s 2 3

q Similar processes happen when conductors are charged externally insteadof being placed into a field. The charges brought inside a conductor repeleach other, and because they are free, they move to the conductor’s surface.The charge and the field inside the conductor are, again, absent.

We are interested, for now, in electrostatics, which means equilibrium.So, the electrostatic charges on conductor surfaces do not move. There-fore, there is no electric field that could move the charges on the surface onconductors. In other words, the tangential component of the electric field(directed along a surface) is zero. Electric field can only be directed per-pendicular to a conductor surface. As follows, the integral R .E � dl/ alongany path on the surface is zero. All points of the surface are at the samepotential, and the surface is an equipotential one.

q

Gauss surface We will now apply these results to a conducting shell with a positive chargeq in the centre. The shell is a sphere that has a concentric spherical hollowof a smaller radius.

The field inside the shell must be zero. So, if we consider a sphericalGauss surface inside the shell, the flux through it is also zero. Accord-ing to Gauss’s law, the total charge enclosed by the Gauss surface mustalso be zero. The surface encloses the charge q. For the total charge to bezero, a negative charge �q must appear on the inner shell surface.

On the other hand, the shell itself is neutral, so its total charge must bezero. Once a �q charge is induced on the inner shell surface, a compensat-ing Cq charge must appear on the outer surface. Due to the symmetry, thecharges on both surfaces will be distributed homogeneously. The electricfields both inside and outside the shell will be radial.

24 cond uctor s

Capacitance and capacitors

qWhen we discussed charging a conductor, we assumed that chargesare brought to it from some reservoir at infinity. A different situation iscommon in practice: one conductor is charged by transfer of charge froma different conductor. One conductor becomes positively charged and theother, equally but negatively charged, so that their total charge is zero.Two separate conductors constitute a capacitor.

�q

Cq

An electric field will appear between the positively and negatively chargedconductors. There will also be a voltage between them. As we learned be-fore, voltage is work that the field does to move a unit charge, and, clearly,work has to be done to move charges between the conductors. Because allpoints of a charged conductor are at the same potential, the voltage be-tween the conductors does not depend on the points we choose on them.

Increasing the charge of the conductor will increase the field and, hence,the voltage. (The capacitor charge is that of the positively charged conduc-tor.) Relying on the superposition principle, we can say that the voltageU must be proportional to the charge q. Therefore, the ratio of the chargeand voltage does not depend on any of them, but will depend only on thegeometry of the conductors. We thus define the capacitance as

C Dq

U

Capacitance is measured in farads [F].

Cq �q

d

E D Eleft C Eright

The simplest capacitor is parallel-plate one. We will now calculate its ca-pacitance. To do so, we assume that the left plate has a charge Cq and theright one, �q. The distance between the plates is d , their area is S . If theheight of the plates is much larger than the separation between them, thenfor most points inside the capacitor the edges will be ‘far away’. In otherwords, we can ignore the effects of the edges and find the field betweenthe plates assuming that the plates are infinitely high. We have alreadyfound the field of a charged plane when discussing Gauss’s theorem. Thesurface charge density of the left, positively charged, plate will be � D q=S .It will create a constant horizontal field equal to Eleft D �=.2"0/. The fieldwill point away from the plate. On the other hand, the right plate will becharged with an equal but negative charge. It will also create a horizontalfield Eright D �=.2"0/ that will point towards the plate. Between the platesboth field will point in the same direction, so that the total field is

E D 2�

2"0

Dq

"0S

It will point from the left to the right plate.

c apac i tanc e and c apac i tor s 2 5

The voltage between the plates is, by definition, U DR rightleft .E � dl/. In prin-

ciple, we could choose any path for the integral, but the most convenientone is a straight horizontal line connecting the plates. Then dl k Etotal and

Cq �q

dl

E

U D

Z d

0

Edl D E

Z d

0

dl D Ed Dqd

"0S

Finally, the capacitance isC D

q

UD

"0S

d

Because the dimension of the vacuum permittivity is [F/m], the dimensionof C is here, as it should be, [F].

Cq �qU Cq1 �q1U

Cq2 �q2U

We will now derive the known equations for parallel and series connectionsof two capacitors. If we take a parallel capacitor and split it horizontally,we can move the two parts apart. We can then connect the plates of thenewly formed capacitors in parallel. Then their total charge and the voltagebetween the plates remain the same as in the original capacitor. Hence, thetotal capacitance of the two new capacitors must be equal to the originalone. Assuming that one of the capacitors has the charge q1 and the otherthe charge q2, we have C1 D q1=U and C2 D q2=U . For the original capacitorwe have C D q=U . But q D q1 C q2, so that

C Dq

UD

q1 C q2

UD

q1

UC

q2

UD C1 C C2

Cq �qU

Cq �q

U1

Cq �q

U2

The vertical planes between the plates of a capacitor are equipotential sur-faces. On the other hand, surfaces of conductors are also always equipo-tential. That is why we can insert a vertical conductor between the plateswithout changing the overall voltage, charge, and capacitance. We canthen use the conductor to form two capacitors in series. The two middleplates are at the same potential. They have charges of opposite signs in-duced on them, so that their total charge is zero. It means that the chargesof the each of the new capacitors will be the same as the charge of the orig-inal one. The voltages, however, will change, so that the original voltage U

is the sum of the voltages across the new capacitors U1 and U2. To see thereason for it, recall that voltage is, by definition, work. The work the fielddoes to move a charge from the left-most to the right-most plate (U ) isthe sum of the works needed to move the charge from the left-most plateto the middle ones (U1) and then from the middle ones to the right-mostone (U2). No work needs be done to move the charge between the middleplates, because they are at the same potential.

Then, starting with the original capacitance, we can writeC D

q

UD

q

U1 C U2

D1

U1=q C U2=qD

1

1=C1 C 1=C2

or, in a more familiar form,1

CD

1

C1

C1

C2

26 c apac i tanc e and c apac i tor s

Our next example is a cylindrical capacitor. It consists of two conductingcylinders of height l sharing the same axis. The inner cylinder has theradius a, and the outer one, the radius b. We find the field between thecylinders using, again, Gauss’s law. We surround the inner conductor by aGauss cylinder of radius r. Due to the symmetry, the electric field pointsradially away from the conductor and is the same on the side of the Gausscylinder. The flux is thus ˆ D "0ES D "0E2�rl. The charge enclosed is q, sothat the field is E D q=.4�"0rl/. The simplest path to calculate the voltageis along the radius so that

side view top view

la

r

b

U D

Z b

a

Edr Dq

2�"0l

Z b

a

dr

rD

q

2�"0lln

b

a

and the capacitance isC D

q

UD

2�"0l

ln.b=a/

Often, people are interested in the capacitance per unit length, which isequal to C=l.

dq

We now return to the general capacitor and discuss how it can store en-ergy. If the capacitor is not charged, it stores no energy. To charge thecapacitor, we remove a small positive charge dq from one of the conductorsare move it to the other. Once we remove the negative charge, the con-ductor (initially neutral) becomes charged negatively. This negative chargewill create a field and attract the positive charge we’ve just removed, thusopposing our attempt to move the charge. So, to place the charge dq onthe second conductor, we have to do work equal to dW D Udq, where U isthe voltage between the conductors.

Once we place the charge dq, both conductors are charged, one positivelyand one negatively. If we want to bring a further charge dq the same way,we will have to counteract both the attraction from the negative and re-pulsion from the positive charges. Again, we will have to do work. It isstill dW D Udq, but is larger than for the previous charge because U hasincreased. The total work to charge the capacitor to a charge q is

W D

Z q

0

Udq D1

C

Z q

0

qdq Dq2

2CD

CU 2

2

By doing the work, we have increased the potential energy of the capaci-tor. The energy stored in the capacitor equals W . If we now connect theconductors through a resistor, a current will flow.

c apac i tanc e and c apac i tor s 2 7

28 c apac i tanc e and c apac i tor s

Dielectrics

Unlike conductors, dielectrics have no free charges. Nevertheless,a dielectric placed into an electric field will react to it. We will explain whathappens in a dielectric using a simple picture. In this picture, dielectrics aremade of atoms consisting of a positive nucleus surrounded by a sphericalcloud of negative electronst. The atom is electrically neutral, so the positivecharge of the nucleus is equal to the negative charge of the electron cloud.The charge in the cloud is distributed homogeneously, and we know fromGauss’s law that the electric field such a sphere produces is equal to thefield of an equivalent point charge placed in the centre of the sphere. Thisnegative charge coincides with the nucleus and compensates its field. Thetotal field the atom produces is zero.

E

d

The nucleus is immobile. The electrons are attracted strongly to the nu-cleus and cannot overcome its attraction. But they could move slightly tothe side of the nucleus if placed into an electric field. Then, the centreof the electron cloud no longer coincides with the nucleus. The equiva-lent negative charge of the electrons and the positive charge of the nucleuswill be separated by a small distance, thus forming a dipole. Dipoles, asyou know, produce electric fields. We say that a dielectric placed into anelectric field becomes polarised. The charges that appear in the dielectricare called ‘polarisation’ or ‘bound’ charges. The latter emphasises that thecharges cannot move inside the dielectric but are bound to the nucleus.

Cq�q

dp D qdWe can describe the dipoles induced in the dielectric by their dipole mo-

ments, p. By definition, p D qd , where q is the charge, and d is the sep-aration. The value of q will be determined entirely by the material (thenumber of electrons in an atom). The value of d will depend both on thematerial (how easy it is to separate the electrons and the nucleus) and onthe electric field strength, E. The higher the field the further it can sepa-rate the charges, so we assume that d � E. The dipole moment then will beproportional to the field, and we can write p D qd D ˛E. The constant ˛ iscalled polarisability. A bulk material will contain many such dipoles. De-noting the number of the dipoles per unit volume by N , we can introducethe polarisation density as P D Np D qNd .

no netcharge

��bound �bound

d d

Let us now consider what happens if we put a dielectric between the platesof a charged capacitor. An electric field exists between the plates and it willpolarise the dielectric. If the left plate is charged positively, the negativecharge of the dipoles will shift to the left by a distance d . We can see thateverywhere inside the material the positive charge of a dipole is compen-sated by the negative charge of its neighbour, so that the dielectric has novolume charge. However, the charges at the dielectric boundaries are notcompensated. The left boundary contains negative surface charge with thedensity ��bound, and the right boundary contains an equal positive charge.

di e l e ct r i c s 2 9

The charge due to a single dipole is q. The volume density of dipoles is N .However, the surface dipoles are contained in a thin layer of thickness d ,so that the surface density of dipoles is Nd . The surface charge density isthen �bound D qNd . But qNd is also equal to the polarisation density, sothat

Gauss surface

�free ��bound�bound D P

Both the plates and the surfaces of the dielectric are charged. The chargesof the dielectric create a field opposing the field of the capacitor, so thatthe total field inside the dielectric will reduce. We can now apply Gauss’slaw to a surface that goes through a positively charged capacitor plate andthe the negatively charged surface of the dielectric. Denoting the surfacecharge density on the capacitor as �free, we can write

"0E D �free � �bound

Let us discuss the difference between the free charges on the conductor andthe bound charges on the dielectric. If we remove the dielectric from thecapacitor, the capacitor will remain charged and the charges on plates willremain there. But if we remove the free charges from the capacitor, thebound charges on the dielectric will disappear. The bound charges existonly in the presence of the field. In fact, as we derived above �bound D

P D Nqd D ˛NE. Substituting this expression into Gauss’s law we have"0E D �free � ˛NE. Rearranging, "0.1 C ˛N="0/E D �free. Introducing therelative permittivity of the dielectric as "d D 1 C ˛N="0, we can rewriteGauss’s law in the form

"0"dE D �free

By including the effects of the dielectric into the permittivity we avoiddealing with two different kinds of charges (free and bound), and insteadhave to consider only free charges.

We assumed that the dielectric is homogeneous, so that N is constant. Itis not always so. Inside inhomogeneous dielectrics the polarisation densitycan vary. It would result in non-constant relative permittivity "d. Then,writing the electric flux density as

D D "0"dE

we rewrite Gauss’s law for dielectrics in the formI.D � dS/ D qenclosed

And if "d D const, Gauss’s law can be written in the form

"0"d

I.E � dS/ D qenclosed

In general, the equations that contain "0 for vacuum are changed for di-electrics by replacing "0 with "0"d. For example, Coulomb’s law becomes

F Dq1q2

4�"0"dr3r

Because "d > 1, the force inside a dielectric is smaller than the force be-tween the same charges in vacuum. Inside dielectrics, the forces and fieldsare weaker due to the compensating effect of the bound charges.

30 d i e l e ct r i c s

Summary

We have finished the study of static electric fields and will nowsummarise the results. They can be put into three sentences:

1. charges create electric fields;2. electric fields act on charges;3. electric fields, regardless of what created them, have certain properties.

Charges create electric fields. Gauss’s law tells us how. It says that theelectric flux through a closed surface is equal to the charge that this surfaceencloses. Mathematically, we write it asI

.D � dS/ D qenclosed

where D D "0"dE. Here, E is the electric field strength; D is the electricflux density; "d is the relative permittivity of the dielectric; qenclosed is thecharge; and dS is an element of the surface. The sign H shows that thesurface is a closed one.

We can use Gauss’s law in this form to find the electric fields of symmetriccharge distributions. We considered spheres, planes, lines, and cylinders.We can also find the fields of continuous distributions in a different way.To do so, we split the charge distribution into small parts, approximateeach of them as a point charge, find the field due to the point charges, andintegrate over the whole distribution. Mathematically, we can denote thismethod as E D

R dE.

Electric fields act on charges. We talked about a point charge q placedinto an electric field E. Then, the force acting on the charge is F D qE.

Electric fields, regardless ofwhat created them, have certain properties.Every static electric field is conservative. This sentence can be translatedinto two equivalent ones. First, the work that the field does to move acharge between two points depends only on the position of the points butnot on the path between them. Second, the work needed to move a chargealong a closed path is zero. Voltage is the work that the field does to movea unit charge. Mathematically, we writeI

any path.E � dl/ D 0

where dl is an element of the path.

s ummary 31

32 summary

Magnetic fields

Previously, we saw that the behaviour of charges and constant elec-tric fields can be described mathematically. Our task now will be to studythe laws governing dc currents and constant magnetic fields. Just like forelectric fields, we will rely on three claims:

1. dc currents create magnetic fields;2. magnetic fields act on dc currents;3. magnetic fields, regardless of their source, have certain properties.

You already know how to describe dc currents, so we will concentrate hereon magnetic fields. The measure of a magnetic field is magnetic flux den-sity, denoted as B. It is a vector, so at any point of space is has a lengthand a direction. To say it differently a magnetic field at any point is char-acterised by three numbers. In a Cartesian coordinate system, these arethe components Bx.x; y; z/, By.x; y; z/, and Bz.x; y; z/. The unit of the mag-netic flux density is Tesla [T]. To keep our discussion brief, we will talkabout magnetic field meaning vector B (as in ‘find the magnetic field B’ or‘current placed into the magnetic field B’).

Magnetic field can change from point to point, and we can introduce mag-netic flux through a surface S as the surface integral ˆ D

RS

.B � dS/. Theunit of magnetic flux is Weber [Wb]. We also can consider magnetic fluxthrough a closed surface. Like in electrostatics, the magnetic flux througha closed surface has a special meaning. As turns out, we will find that forany magnetic field and any closed surfaceI

any surface.B � dS/ D 0

This equation is known as Gauss’s law for magnetic fields. The left-handside is similar to Gauss’s law for electric fields. The right-hand side ofGauss’s law for electric fields contained the electric charge enclosed by thesurface. But we see here that for magnetic fields the right-hand side isalways zero. We can, therefore, interpret Gauss’s law by saying that nomagnetic charges exist.

B

N

S

Just like electric fields, we can visualise magnetic fields by field lines. Theseare defined in such a way that the vector B is tangential to a line at anypoint. Also, the direction of the lines coincides with the direction of thefield. As we know, the electric field lines start and end on charges. Butno magnetic charges exist, and so the magnetic field lines cannot start andend anywhere. They must be closed lines. (In a magnet, the magneticfield lines appear to start at the north pole and end at the south pole onlybecause we do not usually draw the lines inside the magnet. If we did, wewould connect the south and north pole and, so, close the field lines.) Thefield that go through infinity are plotted as non-closed lines; we assumethat they close at infinity.

mag ne t i c f i e ld s 3 3

Moving charges in magnetic fields

Magnetic fields exert forces on moving charges. The force on a charge q

moving with a velocity v is

q v

B�

F

F D qŒv � B�

The term Œv �B� is the cross (vector) product. The magnitude of the force isF D qvB sin � , where � is the angle between the velocity and the magneticfield. So, is the charge does not move, v D 0, the magnetic force is zero.It will also be zero if v k B. The force will be largest when v ? B. Thedirection of the force is given by the right-hand rule. Extend your thumband curl the remaining fingers from v to B. The thumb will point in thedirection of the force.

B

dl

S

I

We will now consider the force on a linear (but not straight) conductor ofa cross-section S . The conductor carries a current I and is placed into amagnetic field B. We take along a conductor a small element of volumedV D Sdl. All charges in this element can be seen as moving in the samedirection with the velocity v . The force acting on one charge is qŒv � B�.But there are dN D ndV charges in the volume element, where n is thedensity. The total force on the element will be the sum of forces on allelements: dF D qŒv � B�dN D qnSŒv � B�dl. The displacement vector dl hasthe same direction as the velocity, so that vdl D vdl. Then, the force isdF D qnvSŒdl � B�. But qnvS D I is the total current flowing through theconductor, so that

dF D I Œdl � B�

The total force on a conductor defined by the path l is then F D IR

lŒdl�B�.

Biot-Savart law

B

dl

I

r

By now, we have considered the general properties of magnetic fields andthe magnetic forces on currents. The remaining question is how currentscreate magnetic fields. The answer, for linear currents, is given by the Biot-Savart law. A linear current I , whose orientation and direction at any pointare given by a vector dl, creates a magnetic field dB equal to

dB D�0

4�I

Œdl � r�

r3

Here, r is the radius-vector connecting the current and the point wherewe calculate the field. The symbol �0 denotes the free-space permeability,which is equal to 4� � 10�7 [H/m] (Henry per metre).

The Biot-Savart law plays the role similar to that of Coulomb’s law inelectrostatics. Coulomb’s law deals with point charges; the Biot-Savartlaw with small current elements. We can use Coulomb’s law to find theelectric field of a charge distribution by splitting the distribution into smallparts and integrating. Likewise, we can find the total magnetic field of acurrent flowing along a line l by the line integral

B D�0

4�I

Zl

Œdl � r�

r3

34 mag ne t i c f i e ld s

We will now look at two examples of how to find magnetic fields using theBiot-Savart law. Our first example is a current I flowing from left to rightalong a horizontal line of length 2l. We need to find B above its centre.

�l l x

y

dx x

R'

rdB

We have to go through the same steps as when we calculated electric fields.First, we need a coordinate system. A Cartesian one is the best, withthe x-axis along the line, y-axis vertical, and the origin at the centre ofthe line. Then, we choose a small element dl D dxex with an arbitrarycoordinate x. The radius-vector r from this element to a point on the y-axis lies entirely in the xy-plane. Applying the right-hand rule, the cross-product Œdl � r� must be along the z-axis, pointing out of the plane of thepaper. The direction of dB known, we have to find its absolute value. It is,from the Biot-Savart law,

dB D�0

4�I

rdx sin'

r3D

�0

4�Isin'dx

r2

where ' is the angle between dl and r. We denote the distance from theline to the point where we calculate the field by R. Then, from the geom-etry, r2 D x2 C R2 and sin' D R=

px2 C R2 so that

dB D�0

4�I

Rdx

.x2 C R2/3=2

We find the total field, due to the whole line, by integratingB D

�0IR

4�

Z l

�l

dx

.x2 C R2/3=2D

�0I

2�R

lp

l2 C R2

I

B

If the line is infinitely long, l � R and l=p

l2 C R2 D 1, so that the field isB D �0I=.2�R/. The magnetic field lines are circles centred on the line.

dB z

r

' a

dl I

dB z

r

' a

'

side view

dBopp

Our next example is a current I flowing along a circular loop of radius a.We have to find the field on the axis of the loop. A cylindrical coordinatesystem whose z-axis coincides with the axis of the loop is the best for thisproblem. We again choose a small element dl at an arbitrary position onthe loop and plot a vector r connecting the element and the point z on theaxis, where we calculate the field. We denote the angle between dl and r

by '. The field dB is given by a cross-product, and so must be orthogonalto r. From geometry, then, ' is also the angle between dB and the z-axis.

The vector dB is oriented in such a way that it has projections both on thez-axis and on the plane of the loop. But we can also consider the elementlying on the opposite side of the diameter. This element will have thecurrent flowing in the opposite direction, and so will create a magneticfield dBopp that is a reflection of dB along the z-axis. The sum of these twofield will have only a z-component. So, the total magnetic field on the axisof the loop will be directed along the axis. For the element dl, we havedBz D dB cos'. So, we have from the Biot-Savart law.

dBz D�0

4�Idl cos'

r2

From geometry, r2 D a2 C z2 and cos' D a=r. And because the loop iscircular, dl D ad� .

mag ne t i c f i e ld s 3 5

The total field is thenBz D

�0

4�I

Z 2�

0

a2d�

.a2 C z2/3=2D

�0I

2

a2

.a2 C z2/3=2

And if z � a, Bz � �0Ia2=.2z3/, i.e. decays at a cube of the distance. Smallcurrent-carrying loops are called magnetic dipoles.

The magnetic field lines of a loop current outside the axis are closed paths.

Ampere’s law

Just as we found for electric fields that Coulomb’s law is a special case ofthe fundamental Gauss’s law, the Biot-Savart law for linear currents is aspecial case of Ampere’s law. We begin our discussion of Ampere’s law bylooking at an infinitely long straight current I . We found earlier that themagnetic field lines are circles surrounding the current, and B D �0I=.2�R/.

I

l

S

n

Let us define a path as follows. The path l is a circle of radius R. It’s centrecoincides with the current, and its plane is perpendicular to the current. Inother words, the path coincides with a magnetic field line. This path alsodefines a surface S—a filled circle. The current I then flows through thissurface. Of two possible normals to the surface we choose the one whosedirection coincides with the direction of the current. As we learned whendiscussing line and surface integrals, choosing a normal fixes the directionof the path. With this choice, the direction of the path coincides with thedirection of the field lines.

We will now find the path integral C DH

l.B �dl/. As we know, such closed

path integral is called the circulation of B along l. Everywhere on l, B D

const and B k dl. So C D BH

ldl D B � 2�R D �0I .

Circulation of B along a path is proportionalto the current this path encloses

This result, known as Ampere’s law, turns out to be general. The circu-lation of B along any closed path is proportional to the total current thispath encloses regardless of the current’s shape, direction, and distribution.Mathematically, we write I

l

.B � dl/ D �0Ienclosed

Once we determine the path, the direction and the normal to the surface,a current can be flowing either into or out of the surface. We take thecurrents flowing into the surface (along the normal) with the positive sign,and the currents flowing in the opposite direction, with a negative sign.

I1

I2

l l

C D I1 C I2C D I1 C I2 C D I1 � I2If the current is distributed with the surface density J , then Ienclosed DR

S.J � dS/, and Ampere’s law takes the formI

l

.B � dl/ D �0

ZS

.J � dS/

36 mag ne t i c f i e ld s

Inductance

According to the biot-savart law, the field created by a homogenousdc current is proportional to the current strength, B � I . If the currentflows along a closed loop, we can also calculate the magnetic flux throughthe surface of the loop as ˆ D

RS

.B �dS/. So, the flux ˆ will be proportionalto B and, hence, to the current, ˆ � I . The ratio of flux and dc current iscalled inductance

B

I

L Dˆ

I

We measure inductance in Henrys [H].

If we have N identical loops, each carrying the same current I , then theflux through all of them is Nˆ. The inductance is then

L DNˆ

I

S R

r

B

As an example, we will calculate the inductance of a toroid. It has a circularcross-section of an area S and a radius R. A total of N turns are woundaround the toroid. To find the flux ˆ, we assume a current I . The currentflowing through each loop will create a magnetic field. We can find it usingAmpere’s law. To apply the law, we select a circular path of radius r insidethe toroid. Due to the symmetry, the field is constant and tangential tothe path, so that H .B �dl/ D 2�rB. On the other hand, the total current thispath encloses is NI . From Ampere’s law, then, 2�rB D �0NI , so that

B D�0NI

2�r

We now have to find the flux ˆ through a turn. Because B changes fromone point to another on the cross-section, we would need to find a surfaceintegral R BdS . But we can simplify the calculation if we assume that theradius of the toroid R is much larger than the radius of the cross-section.Then, the magnetic field changes little, and we can approximately take itas constant, B � m0NI=.2�R/. The flux then is ˆ D BS , so that

L DNˆ

ID

�0N 2S

2�R

We notice that the inductance grows as a square of the number of turns,so increasing their number is a good way to increase the inductance. Also,the approximate expression for the inductance will work for toroids witharbitrary shape of the cross section, provided the radius is large enough.

i nd uctanc e 37

38 ind uctanc e

Magnetic materials

When discussing dielectrics, we looked at a simple model of an atomwith a nucleus surrounded by electrons (see p. 29). The electrons movearound the nucleus. But movement of electric charges is an electric cur-rent, so we can see the atom as a small current-carrying loop. As you know,such loops are magnetic dipoles, and they create magnetic fields.

I

Although quantum mechanics tells us that this model of atoms is too sim-plistic, the main conclusion holds even then: atoms in a material can cre-ate magnetic fields. Without external fields, these fields are oriented ran-domly, and they cancel each other within the whole material.

But if we put the material into a magnetic field, the dipoles will re-orientthemselves so that their own magnetic fields will align with the externalone. The total field in the material becomes the sum of the external andinternal magnetic fields. Our aim is to describe the effect mathematically.

Im

Im

Assume that the material has a cylindrical shape, and that the externalfield is along its axis. The aligned dipoles inside the material will havetheir currents lying perpendicular to the external field. They will create atotal magnetisation current Im. According to Ampere’s law, the circulationof B along a closed path is proportional to the current. We have so farconsidered only conduction currents, but these will now be supplementedwith the magnetisation currents. Ampere’s law will take the formI

.B � dl/ D �0.I C Im/

We are now dealing with a situation similar to dielectrics. In dielectrics,we had two types of charges, free and bound. In magnetic materials, wehave to types of currents. In dielectrics, the bound charges disappear ifthe external electric field disappears. The same is in magnetic materials:the magnetisation current Im exists because there is an external magneticfield, and the current is zero without the field. Therefore, just as we did fordielectrics, we would like to include the effects of Im into the left-hand sideof Ampere’s law and keep only the conduction current in the right-handside. To do so, we introduce a new vector, magnetic field strength as

H D1

�0

B � M

The units of the magnetic field strength are Amperes per metre [A/m].The vector M is called the magnetisation vector, and it characterises thedensity and the orientation of the magnetic dipoles inside the material(similar to the polarisation vector P in dielectrics). Naturally, M and Im

are connected, and it can be shown thatIm D

I.M � dl/

mag ne t i c mat er i a l s 3 9

Substituting this expression into Ampere’s law givesI.B � dl/ D �0

�I C

I.M � dl/

�or I ��

1

�0

B � M

�� dl

�D I

and, finally, I �H � dl

�D I

Let us return to the definition of the magnetic field strength, H D B=�0�M.We can assume that the stronger the field, the more dipoles will align alongit, so that M � H. Then B � H, and we can write

B D �0�rH

where we introduced yet another quantity, relative magnetic permeability,�r. The relative permeability characterises the magnetic response of a ma-terial to magnetic fields, so that we need not consider magnetisation cur-rents. In vacuum �r D 1. If the relative permeability is constant, Ampere’slaw takes the formI �

H � dl�

D I orI �

B � dl�

D �0�rI

However, Gauss’s law for magnetic fields remains unchanged in magneticmaterials, H .B � dS/ D 0.

Most materials have only weak magnetic properties. If they slightly am-plify the external field, they are called paramagnetic, and their relative per-meabilities is larger than unity. On the other hand, materials that slightlyresist external fields are called diamagnetic, and their relative permeabilitiesare smaller than unity.

B

H

vacuum, B D �0H

diamagnet, �r < 1

paramagnet, �r > 1

40 mag ne t i c mat er i a l s

Ferromagnetic materials

Electrical engineers find most uses for ferromagnetic materials that havevery strong (and complicated) magnetic response. The most famous ferro-magnetic materials are iron and nickel. As often happens, only quantumtheory can fully describe ferromagnetism. We can, however, establish aqualitative explanation based on our picture of magnetic dipoles.

When a material is placed into a magnetic field, its magnetic dipoles alignalong the field. In para- and diamagnets, these dipoles are independent ofeach other. But in a ferromagnetic material, a dipole will experience notonly the external magnetic field but also the field from the neighbouringdipoles. If the neighbours are already aligned along the field, they will‘encourage’ the dipole to align too. In other words, a small external fieldwill cause, due to the dipole cooperation, a strong magnetic response.

B

HO

ABs

Let us consider in detail how ferromagnets magnetise. We start with aferromagnet whose dipoles are oriented randomly. The the magnetic fieldstrength is zero, H D 0, and so is magnetic flux density, B D 0. We nowmagnetise the material, which means increasing H . As we saw above, B

will grow strongly. This is shown by the beginning of curve OA on the B-H

diagram. As we increase H further, more dipoles find themselves alignedwith the field, and less dipoles remain non-aligned. The process begins toslow down, and B will no longer grow strongly. At some value of H , alldipoles are aligned with the field. The magnetisation saturates and can nolonger increase, M D const. From now on, B D �0H C const. The slope ofthe B-H curve is equal to �0. It is much smaller than the initial slope, andso the curve will be much more horizontal. The saturation point is A onthe diagram, and the corresponding value of the magnetic flux density isBs. B

H

R

ABrLet us now take the material at point A and decrease H . Smaller H means

that the dipoles in the material are less compelled to align along the field.Thermal fluctuations will break the saturation alignment of point A, so B

will decrease. However, in ferromagnetic materials, a dipole experiencethe field of the neighbours. Even if H decreases, the alignment of point Ais still felt by the dipoles. It will try to keep the dipoles aligned. As a result,as we decrease H , the B-H curve will follow not the original path AO back,but the less steep path AR. At H D 0 some of the dipoles remain aligned(residual magnetisation) creating the flux density Br (called remanence).It explains why permanent magnets exist. B

H�Hc

A

R

C

If we want to demagnetise the material completely, we must apply H inthe direction opposite to the residual magnetisation. In the B-H diagramit means taking negative H . As we increase H (make it more negative),more and more dipoles lose their orientation, until at some value Hc (calledcoercitivity) B reaches zero.

mag ne t i c mat er i a l s 4 1

If we continue toward more negative values of H , the material will startmagnetising the material again, but in the direction opposite to that of OAcurve. It means that B will start to grow towards negative values. The B-H

will be in the third quarter. The process continues until the magnetisationsaturates. The curve reaches pout D that mirrors point A.

B

H

A

R

C

D

If we now want to reach point A from point D, the material will haveto follow a curve whose form is identical to curve ARCD but that goesthrough the fourth quarter. The result will be a hysteresis loop. The termmeans that the value of B depends not only on the value of H but also onthe prior history of the material. As the loop shows, two values of B aregenerally possible at a single value of H .

B

H

Br

�Br

Bs

�Bs

�Hc Hc

The dependence of B on H is nonlinear, and so if we still want to definethe relative permeability from B D �0�rH , we will find that �r D �r.H/.In iron, it changes between about 100 and 100,000.

42 mag ne t i c mat er i a l s

Magnetic circuits

We previously considered a toroid—a torus with wire windings. Similarto it is a solenoid, but its windings are placed along a line. We can see eachwinding as a circular loop. The current flowing through the solenoid willbe in the same direction for each loop, so we can understand the total fieldof the solenoid as a sum of the fields from each loop.

IIf the solenoid is in air, the magnetic field lines outside of it will divergequickly (as they do for a single loop). But if the winding are around aferromagnetic core, the magnetic field lines generated inside the core will‘prefer’ to stay inside. They will follow the core as long as the core extends.In other words, the ferromagnetic material will guide the magnetic flux ofthe solenoid like metal wires guide electric currents. It occurs because themagnetic permeability of the core is much larger than that of the air.

NI gB

l

This analogy between magnetic flux and current allows us to consider mag-netic circuits. As an example, we will look at a C-shaped core. On oneside, the core is inside a solenoid consisting of N windings and carrying acurrent I . On the other side, the core has a gap of a thickness g.

The magnetic flux of the solenoid is bound to the core. The magnetic fieldlines will follow the core from the left to the right side until they reachthe gap. The lines then exit the core and go through the gap, re-enter thecore and follow it back to the solenoid. The result is a closed field line.We will assume, as we did for the toroid, that the magnitude of B is thesame through the core. The magnetic flux then is ˆ D BS . Because themagnetic field lines do not leave the core, ˆ D const. We will also assumethat the field lines in the gap are strictly vertical. It is similar to having nofringing fields in a capacitor. Then the flux in the gap also equals BS .

We now take a closed imaginary path that goes through the solenoid, thecore, and the gap (dashed line). According to Ampere’s law, H

l.H�dl/ D NI .

Assuming that the field is tangential to the path, we can rewrite Ampere’slaw as

Hcorel C Hgapg D NI

where l is the length of the path inside the core. On the other hand, forthe ferromagnetic core and the air-filled gap, we can write B D �0�rHcore

and B D �0Hgap. Also B D ˆ=S . Thenˆ

�l

�0�rSC

g

�0S

�D NI

NI

ˆ

Rcore

Rgap

We have already said that the flux ˆ is analogous to current. The termFM D NI , called magneto-motive force, is analogous to voltage. The twoterms in the bracket are called reluctance, and are analogous to resistance.The definition of the reluctance is R D FM=ˆ. We obtained for the core

Rcore Dl

�0�rS

In the C-shaped circuit, the reluctance of the core is connected in serieswith that of the gap, so that we can plot an equivalent magnetic circuit.

mag ne t i c mat er i a l s 4 3

44 mag ne t i c mat er i a l s

Summary

We have finished the study of dc magnetic fields, and will now sum-marise the results.

DC currents create magnetic fields. Mathematically, the field created bya current is described by Ampere’s lawI

l

�H � dl

�D I

where H is the magnetic field strength, and I is the current enclosed by thepath l. The magnetic field strength is linked to the magnetic flux densityB by the expression B D �0�rH, where �0 is the permeability of the freespace, and �r is the relative permeability of a material.

Magnetic fields act on dc currents. The force on a charge q moving in amagnetic field is F D qŒv � B�. The force acting on a linear current element(which comprises many moving charges) is

dF D I Œdl � B�

Magnetic fields, regardless of their source, have certain properties. Allmagnetic fields obey Gauss’ law of the formI

S

.B � dS/ D 0

for any closed surface S . A consequence of this law is that all magneticfield lines are either closed paths or begin and end at infinity. Because theright-hand side is zero, we also say that no magnetic charges exist.

s ummary 45

46 summary

Electromagnetic induction

So far, we have considered static electric and magnetic fields and dc cur-rents. The related quantities, such as magnetic and electric flux, were alsostatic. We will now look at time-varying phenomena.

We will first discuss what happens if the magnetic flux varies with time,ˆ D ˆ.t/. The magnetic flux, by definition, is ˆ D

RS

.B � dS/. To be able touse the results we obtained previously, we would like to keep the magneticfield constant. To vary the flux, we will vary the surface S .

l

x

Bbar

frameTo create a varying surface, we will take a vertical conducting bar that canslide, without friction but with electrical contact, on a conducting frame.The length of the bar is l. A constant magnetic field B is perpendicular tothe frame, pointing out of the paper plane. The bar and the left part of theframe form a surface S . Denoting the length of the frame to the left of thebar by x, we have S D lx. On the other hand, the magnetic flux through S

is ˆ D BS D Blx.

B vF

We now start sliding the bar to the right with a constant velocity v. Movingthe bar means that free charges inside it also move to the right. In themagnetic field, the moving charges experience the force F D qŒv � B�. Theforce on positive charges is directed downwards. It will redistribute thecharges, so that the upper end of the bar will charge negatively and thelower end, positively.

E

Because the top and bottom of the bar become electrically charged, anelectric field E appears in the bar. The field is directed from the bottomto the top, against the force F . Because E and F are counter-directed, theforce F must do work to move the charges from the top to the bottom ofthe bar. Per unit charge, this work is W D F l D vBl. The energy must beconserved, so this work goes into increasing the voltage between the topand bottom of the bar. This voltage, called electromotive force (or emf ) isnegative and equals E D �vBl. We use the term emf to emphasise that thework is done by non-electric forces.

Let us now look at how the magnetic flux through the frame changes.The flux is ˆ D Blx. On the other hand x D vt , where t is the time. So,dˆ=dt D Blv. Comparing this expression with the expression for the emf,we can write

E D �dˆ

dt

This process, by which an emf is induced due to a time-varying magneticflux, is called electromagnetic induction. The above expression linking theemf and the time derivative of the flux is called Faraday’s law (of electro-magnetic induction).

e l e ct romag ne t i c i nd uct ion 47

Because the frame is conducting, the emf induced in the bar will create acurrent Iind flowing through the frame and the bar in the clock-wise di-rection. The induced current will create a magnetic field Bind directed, likethe external field, perpendicular to the frame but in the opposite direction.

Iind

Iind

Bind

IindFind

B

On the other hand, the external magnetic field will exert a force on thecurrent. We are interested in the force on the moving bar. The inducedcurrent in the bar flows downwards, and the force induced on the bar,Find D IindŒdl � B�, is directed to the left. Therefore, this force opposes themovement of the bar.

The force has appeared due to the current; the current due to the emf; andthe emf due to the movement of the bar. We can thus say that the emf isinduced in such a way that it counteracts its source. This result is calledLenz’s rule, and mathematically it is expressed by the negative sign in theexpression for Faraday’s law.

If we moved the bar to the left with the same velocity, the area of thesurface made by the frame and the bar would decrease, so that dˆ=dt < 0.Then, the emf induced in the bar would create a force directed to the right,and thus counteracting the decrease of the flux.

This behaviour of the induced emf is what we would expect from theenergy-conservation law. To create a current in the frame, we must dowork, and hence to move the bar against a force. If the force were directedto the right, we would not need any effort to maintain the velocity v of thebar. Instead, the bar would accelerate on its own.

We do positive work, so the work that the induced force does is negative.When the bar moves at a distance dx, the induced force does the workdW D �Finddx. However, Find D IindlB and ldx D dS , so that

dW D �Iinddˆ

The configuration of a frame with a moving bar is significant because itshows how the energy of a mechanical movement can be converted intovoltage. And indeed, electromagnetic induction is the principle behindelectric motors and generators, which we will consider later.

S

l

n

B.t/ However, Faraday’s law turns out to be fundamental (and one of fourMaxwell’s equations). It holds if the magnetic flux varies is time not onlydue to time-varying surface S but also due to time-varying magnetic fieldB.t/. It is also true if the surface S is an imaginary one, and not made, aswere the frame and the bar, of any material. To formulate Faraday’s law,we choose an arbitrary open surface S in a time-varying magnetic field.The border of the surface also defines a closed path l. If we choose onthe surface a positive direction for the normal, we automatically define thedirection for the path l. The flux through the surface is ˆ D

RS

.B.t/ � dS/.The emf created along the path is, by definition, E D

Hl.E.t/ � dl/. Faraday’s

law then takes the formIl

.E.t/ � dl/ D �ddt

ZS

.B.t/ � dS/

48 e l e ct romag ne t i c i nd uct ion

Mutual inductance

As Faraday’s law says, a time-varying magnetic flux will create (or, in gen-eral, modify) currents in closed conductors. Assume we have two loops,one carrying a dc current I1 and the other I2. Each loop creates a magneticfield that can reach the other loop and, therefore, create a flux through itssurface. We will denote the magnetic flux through loop 1 created by loop 2by ˆ12. The flux through loop 2 due to loop 1 is ˆ21.

I1 I2

ˆ12 ˆ21If the loops are infinitely far from each other, ˆ12 D ˆ21 D 0. We then fixloop 2 in its position and take loop 1 and start bringing it close to loop 2.As we move loop 2, ˆ21 increases. Because ˆ21 changes with time, it willinduced an emf in loop 2. According to Lenz’s rule, the effect of this emfis to counteract the change of the flux. In other words, the emf creates aforce on loop 2 that will oppose the movement. Therefore, we have to dowork to bring loop 2 closer to loop 1. We derived above the expression forwork as dW D Idˆ (we take here positive sign, because we do the work,not the field). The total work to move loop 1 from infinity to a distance r

from loop 2 will be

W1 D

Z r

1

I1dˆ12 D I1Œˆ12.r/ � ˆ12.1/� D I1ˆ12.r/

If we interchange the loops (fix loop 1 and move loop 2), the work wehave to do to move loop 2 will be W2 D I2ˆ12.r/. No matter which loop wemove, the initial and final positions of the loops are the same, so the worksmust also be the same W1 D W2 leading to I1ˆ12 D I2ˆ12. We introducemutual inductance between the loops as

M Dˆ12

I2

Dˆ21

I1

There are two sources of magnetic flux through loop 1. The first sourceis the current I2 in loop 2, which creates the flux ˆ12. The second is thecurrent I1 in loop 1 itself, which creates a flux equal to ˆ1 D L1I1, whereL1 is the inductance of loop 1. The total flux will be ˆ1 tot D L1I1 C MI2.Analogously, the total flux through loop 2 will be ˆ2 tot D L2I2 C MI1.

I

I

Consider now two loops fed by the same current I that flows through theloops in the same direction. If no mutual inductance between the loopsexists, then the total flux through both loops is ˆtot D ˆ1Cˆ2 D L1ICL2I D

.L1CL2/I . The total inductance is then Ltot D ˆtot=I D L1CL2, which is, ofcourse, the well-known to you rule for a series connection of two inductors.However, if the loops are mutually coupled, then the flux through eachof the loop increases by MI so that ˆtot D L1I C MI C L2I C MI andLtot D L1 C L2 C 2M . But we can arrange the loops in such a way that thecurrent flows through them in the opposite directions. Then ˆ1 D L1I �MI

and ˆ2 D L2I � MI , so that Ltot D L1 C L2 � 2M .

L L

M

Ltot D L1 C L2 C 2M

L L

M

Ltot D L1 C L2 � 2M

To distinguish between the two situations in circuit notations, we use the‘dot convention’. We place a dot at a side of an inductor, and if the bothcurrents flow either into or out of the dots, we take the positive sign forthe mutual inductance term. Otherwise, the sign is negative.

e l e ct romag ne t i c i nd uct ion 49

We will now look at an example of how to calculate mutual inductance.The example is two loops of radius a1 and a2 whose axes coincide. Wewill assume that a1 � a2. The distance between the loops is d . Accordingto the definition, we can find the mutual inductance by considering theflux of either loop. However, it is much easier to consider the flux thatthe larger loop 1 creates through the smaller loop 2. Because loop 2 issmall, we can assume that the magnetic field through it due to loop 1 isapproximately constant and equal to the field on the axis. We previouslyfound the field on the axis of a circular loop as

a1

a2

d

B1 D�0I1

2

a21

.a21 C d 2/3=2

Because B1 is approximately constant across loop 2, the flux ˆ21 � B1�a22.

The mutual inductance is thenM D

ˆ21

I1

��0�a2

1a22

2.a21 C d 2/3=2

Energy stored in an inductor. Voltage

We previously saw that the energy stored in a capacitor equals the workneeded to charge it (see p. 27). Work has to be done because to charge thecapacitor, we need to overcome the field it creates. We consider the energyof an inductor in a similar way.

i

u12

1 2

We start with a loop of an inductance L and assume that no current flowsthrough it. We now want to increase the current from zero to some valueI . As we start to increase the current, it will create a time-varying fieldin the loop. The time-varying field creates a time-varying flux, which inturn induces an emf in the loop. The self-induced emf will oppose theincrease of the current, so to continue the increase, we have to do work.The flux is ˆ.t/ D Li.t/, where i is the instantaneous value of the current.So, dˆ D Ldi and the work dW D idˆ D Lidi . Hence, the total work doneto increase the current is

W D

Z I

0

dW D L

Z I

0

idI DLI 2

2

This work equals the energy stored in the inductor.

i

u12

We can look at the situation from a different angle. A current flowingalong the loop means transferring certain amount of charge from point 1to point 2. At the same time, the self-induced emf creates an electric fieldthat opposes the charge movement, i.e. a field directed from point 2 to 1.If we are able to sustain the current i in the loop, we must have createdan opposing field pointing from 1 to 2. This field will create a voltage u12.If the resistance of the loop is negligible, the two fields cancel each other,and u12 C E D 0. As follows

u12 D Ldi

dt

50 e l e ct romag ne t i c i nd uct ion

Energy density

Above, we have obtained an expression for the energy stored in an inductoras W D LI 2=2. This expression is in terms of the current flowing throughthe inductor. But the current creates a magnetic field, so we can also saythat the energy is stored in the field. The latter statement is more general:we can have a time-varying magnetic field but no current, and still someenergy would be stored in the field. (Recall the example from the begin-ning of this course when charges or currents existed in neither receiver northe transmitter but the energy was carried between them by the field.)

S R

r

B

Because the field exists in space and is not localised, we describe the energyby energy density w equal to the energy per volume. We will now derive theexpression for the magnetic energy density of a toroid. We have derivedthe expression for its inductance as (see p. 37)

L DNˆ

ID

�0�rN2S

2�R

The relative permeability �r describes the material of the toroid core. Themagnetic field exists solely inside the toroid. The volume of the toroid isV D 2�RS , so that the magnetic energy density is

w DW

VD

�0�rN2S

2�R

I 2

4�RSD

�0�r

2

N 2I 2

.2�R/2

On the other hand, we have obtained for the field inside the toroid, usingAmpere’s law,

H DNI

2�R

which leads to w D .�0�rH 2/=2. However, B D �0�rH , so we finally canpresent the magnetic energy density in the form

w DBH

2

Although all expressions for w of the toroid are equivalent and can bederived from each other, the latest one is the most general. It turns out tobe true not only for the toroid, but for an arbitrary magnetic field.

e l e ct romag ne t i c i nd uct ion 51

Transformers

You are already familiar with transformers from the course in circuit analy-sis, and we will look here at them in more detail. A transformer consists oftwo coils with a common flux that links them. The coils are called primaryand secondary, and their windings have N1 and N2 turns, respectively. Theflux can be transferred by a ferromagnetic core.

U1 N1

I1

U2N2

I2

If a time-varying current I1 flows through the primary coil, it will createa time-varying magnetic field. The field will induce a back-emf in thecoil. For every turn of the winding, the back emf is related to the flux byFaraday’s law E D �dˆ=dt . For many turns, the total emf will be

E1 D �N1

dˆ

dt

Because a current is flowing through the coil, this emf is counteracted bya voltage U1 across the coil equal to

U1

N1 W N2

I1

U2

I2

U1 D N1

dˆ

dt

We will now assume (as we did for magnetic circuits, see p. 43) that theentire flux ˆ is guided by the ferromagnetic core to the secondary coil. Thistime-varying flux will create an emf in the secondary coil. If the secondarycoil is closed, the emf will excite a current in it. The direction of the currentwill depend on how the secondary coil is wound around the core relativeto the primary coil. As we have already discussed, the polarity is indicatedusing dots next to coils in circuits. We will assume that the currents flowinto the dotted terminal, resulting in the circuits shown here.

alternative notation

N1 W N2 The voltage U2 across the secondary coil is then U2 D N2dˆ=dt leading toU1

U2

DN1

N2

The ratio N1 W N2 is shown in the circuit.

We will now find the relationship between the currents I1 and I2. UsingAmpere’s law (exactly as we did for magnetic circuits), we obtain Hl D

N1I1 C N2I2, where l is the circumference of the core. We will assume, fornow, that the permeability of the core is very large �r � 1. But B D �0�rH ,and the flux ˆ is proportional to B. The flux is finite, and so must be B.Because �r is very large, we can have finite B only if H is very small, H � 0.Then Ampere’s law takes the form N1I1 D �N2I2, so that I1=I2 D �N2=N1.

U1

N1 W N2

I1 I2 IL

ZL

U1

I1

Z 0L

We will now connect a load ZL to the secondary coil. For the voltagedirection chosen, the current flowing through the load is opposite to thecurrent through the coil, IL D �I2. On the other hand Ohm’s law statesU2 D ZLIL. From here, and using the relationships between U1 and U2 andbetween I1 and IL, we obtain

Z 0L D

U1

I1

D

�N1

N2

�2

ZL

The load Z 0L is called referred load and shows how the transformer together

with the actual load ZL ‘look like’ from the input terminal.

52 e l e ct romag ne t i c i nd uct ion

Above, we considered an ideal transformer. We assumed no losses andcomplete flux linkage between the coils. Both assumptions will fail tosome extent in real transformers. We now consider real transformers anddiscuss how to describe them using circuit elements.

X1 R1 X2R2

The first obvious effect is ohmic loss in the coils. We can take it intoaccount by two resistors, R1 and R2, in series with the coils. The secondeffect is flux leakage. In real transformers, not all the flux generated by thecurrent in one coil flows through the other coil. In principle, this effectcould be taken into account by the mutual inductance between the coils.However, a simpler picture exists. Flux leakage means that not all voltageU1 applied to the primary coil will create an emf in the secondary one.The reduction of the effective voltage can be modelled by connecting theprimary coil, in series, to some leakage reactance jX1. The same applies tothe secondary coil, giving a leakage reactance jX2.

Xt RtJust as we did for the load impedance earlier, we can refer R2 and jX2 fromthe output to the input side and combine the referred impedances with theimpedances of the primary coil into two series components Rt and Xt.

Xt Rt

Xm

We will now consider imperfections due to the core. First, realistic rela-tive permeabilities of the core are finite, which means that the magneticfield strength in the core is non-zero. Consequently, Ampere’s law isHl D N1I1 C N2I2. We introduce a magnetisation current as Im D Hl=N1.Then, I2 D N1.I1 � Im/=N2. Compared to the ideal transformer, the effec-tive current flowing through the primary coil has reduced. We can modelthe reduction of current by a component connected in parallel to the pri-mary coil. Because the magnetisation current isn’t a dissipative effect, thiselement must be a reactance jXm.

B

H

The final mechanism is the loss in the core (iron loss). It has two sources:hysteresis of the core material and eddy-current loss. Earlier, we derivedthe expression for the magnetic energy density as w D BH=2. It is correctwhen the relative permeability �r is constant. In ferromagnets, however,�r D �r.H/, and the correct expression is R HdB. Geometrically, it is anarea defined by the B-H curve and the B-axis. As we know, the B-H curvein ferromagnets will depend on whether we magnetise or demagnetise thematerial. Consequently, the magnetic energy stored in the core will bedifferent. The difference between the areas during the magnetisation anddemagnetisation cycles will dissipate as heat. We can see it as a sort of‘friction’ that occurs when the magnetic field reorients the dipoles insidethe ferromagnet. This loss can be minimised by cores with low hysteresis.

B

eddy currents

B

no eddy currents

The eddy-current loss exists because the core (say, iron) is a conductor.A time-varying magnetic field creates an emf inside the conductor; theemf creates an electric current. Due to the finite conductivity, the electriccurrent leads to loss. This type of loss can be reduced in laminated ferro-magnets that consists of thin layers of a ferromagnetic material separatedby thin insulating films.

e l e ct romag ne t i c i nd uct ion 53

The core losses can be combined into a resistor Ri. Like the magnetisationcurrent, this resistor reduces the value of I2, and so we put it in parallelwith the magnetisation reactance jXm.

Xt Rt

XmRiU1 ZL

ItImIin An equivalent circuit has emerged with an ideal transformer and four ad-ditional components, two resistive and two reactive, which represent im-perfections. This circuit is an approximation; for example, the values of allfour additional parameters may change with frequency. That is why theeasiest way to determine the values of the components is to measure them.The problem is that we cannot directly measure the currents through andthe voltages across the four components. We can, however, measure thevoltage U1, the input current Iin and the input power Pin.

I(open)in

XmRiU(open)1

Im We first do the open-circuit test by leaving the secondary coil open. ThenI2 D 0, and because I2 D �N1I1=N2, we also have I1 D 0. No current flowsthrough Rt, and the input power P

(open)in is dissipated entirely through Ri.

By measuring P(open)in and U

(open)1 , we determine

Ri DU

(open)1

2

P(open)in

Analysing the circuit, we then determine

Im D

vuutI(open)in

2

�

U

(open)1

Ri

!2

and Xm D U(open)1 =Im.

Xt Rt

U(short)1

I(short)in We then do the short-circuit test by shortening the secondary coil. Cur-

rents will now flow in all four components, but usually the values of theseries components are much smaller than that of the parallel ones. As fol-lows, the current through the parallel components is small and they canbe neglected. The power is dissipated only through Rt, so that

Rt DP

(short)in

I(short)in

2

Analysing this circuit, we obtain

U(short)1

2

D I(short)in

2 �R2t C X2

t

�from which we can find Xt.

For real transformers, we can introduce two additional characteristic pa-rameters. The first one is the efficiency � defined as the ratio of the outputto input power, � D Pout=Pin. The difference between the two powers isthe dissipated power Pout D Pin � Ploss, so that � D 1 � Ploss=Pin. On theother hand, Ploss D I 2

t RtCU 21 =Ri and Pin D U1Iin cos�, where � is the phase

angle between the input voltage and current.

The second parameter is the regulation, which shows the difference be-tween the output voltage U2 for the open circuit and for a given load

Reg D 1 �U2

U(open)2

54 e l e ct romag ne t i c i nd uct ion

Inductive powering and communication

We continue discussing applications of electromagnetic induction andwill consider inductive powering and communication. Most electrical de-vices is powered either by wires or by batteries. Electrical appliances, forexample, are plugged into electric sockets, and the sockets are connectedto power plants by wires. Mobile devices, on the other hand, are usuallypowered by batteries. But in some situations, neither wires nor batteriesare a good powering option. One of them is powering human-implantabledevices, such as cardiac and neural stimulators. Wires are out of question.Batteries are a better solution and are, indeed, widely used. But batteriesdo not last forever and must be replaced.

Mutual inductance between two coils offers an alternative. As we haveseen before, a time-varying current in one coil can induce a current in theother, without physical connection between them. The secondary coil canbe connected to an implantable device and be placed inside a human body.The primary coil remains outside. Beside being contactless, inductive linksare insensitive to the dielectric properties of the surrounding media—afurther advantage for biomedical applications.

The same principle can be applied for contactless sensing. As a simpleexample, the secondary coil can be connected to a temperature-sensitiveresistor. As we saw for transformers, the current flowing through the pri-mary coil depends on the load of the secondary one. And so by measuringthe current, we can deduce the load, and, hence, temperature. A moreadvanced application is near-field communication, an example of which iscontactless card payment.

The major difference between inductive powering and transformers is thatthe coils are now loosely coupled and the mutual inductance between themis important. We will therefore begin by discussing the mutual inductancebetween two coils in more detail.

Mutual inductance between circular coils

We derived earlier an analytical expression for the mutual inductance oftwo coaxial coils, one of which was small (see p. 50). Analytical expressionsfor coils of equal size also exist but are complicated, so we will consider themutual inductance only qualitatively. We will assume that the coils lie inparallel planes, and that the currents flow in them in the same direction.The two basic configurations are the axial one, when the coils share thesame axis, and the planar one, when the coils lie in the same plane.

i nd uct i v e p ow er ing 55

The magnetic fields of two coils in the axial configuration are co-directed.The flux of one coil will reinforce the flux of the other. Therefore, themutual inductance M between the coils is positive. On the other hand, theflux of one coil will reduce the flux of the other in the planar configuration.The mutual inductance in the planar configuration is negative.

axial planar

M > 0 M < 0

If we start in the axial configuration and move one of the coils perpendic-ular to the axis then at large distances we will almost arrive at the planarconfiguration. We will start with a positive and end with a negative mu-tual inductance, and therefore a position should exists where the mutualinductance is zero.

distance

�We assume that both coils have equal self-inductances L. We can thencharacterise the amount of coupling between the coils by a dimensionlesscoefficient � D M=L. Practical values of j�j are of the order of 0:1 and lower.They are larger for the axial than for the planar configuration (because moremagnetic field lines generated by one loop can penetrate the other in theaxial configuration).

Power efficiency

U RL

LL M We consider now power transfer between two coils. The primary coil isconnected to an ac source with a voltage U and an angular frequency !.The secondary coil has a resistive load RL. The inductance of both coils isL. We will present the ac signal in the complex form. Then, the derivativesd=dt can be replaced by j!. Denoting the currents flowing through theprimary and secondary coils as Ip and Is, we can write for the primarycoil U D j!LIp C j!MIs. The first term on the right-hand side is thevoltage across the inductor, and the second term represents the voltagedue to the secondary coil and the mutual inductance. For the secondarycoil, we have 0 D .j!LCRL/IsCj!MIp. Solving the system of two coupledlinear equations, we get

Ip D �j!M

!2.M 2 � L2/ C j!LRLU

From here we can find Is as

Is D �j!M

j!L C RLIp

The power efficiency is the ratio between the power absorbed in the loadand the input power, so that

� DPabs

PinD

jIsj2RL

Re .IpU �/

where the asterisk denotes complex conjugation. Using the expressions forIp and Is and for the coupling coefficient � D M=L, we obtain

� D�3

1 CR2L

!2L2

56 ind uct i v e p ow er ing

Because the practical coupling coefficients are low, the power efficiencywill be also low. Not all the input power is usefully absorbed in the loadbecause the reactive power is large. But we can compesate the reactancesby putting capacitors C in series with the inductors. At the resonant fre-quency !0 D 1=

pLC , the total impedance of the capacitor and inductor

vanishes, j!0LC1=.j!0C / D 0. We then have for the two loops U D j!0MIs

and 0 D RLIsCj!MIp. From here, we obtain that � D 1; the power transferis 100% efficient regardless of the coupling coefficient.

U RL

LL M

C C

That is why practical systems use resonant coils. Realistic power efficien-cies are lower than unity, because some power dissipates in the resistanceof the coils. For lossy coils, the coupling coefficient � cannot be takensmall when working at the resonant frequency (see a study-group prob-lem). If the coupling is too small, most of the input power dissipate in theresistance of the primary coil. Usually, the distance between the coils is ofthe order of their size.

Magneto-inductive waves

To increase the distance between the primary and the secondary coils, wecan use relay coils. When many relay coils are put at the same distancefrom each other (periodically), a new phenomenon emerges: signals prop-agate between coils in the form of waves.

C C C C

L L L LM M M

In�1 In InC1

Let us consider a chain or resonant coils, each with an inductance L anda capacitance C . The distance between the neighbouring coils is a, andeach coil is coupled magnetically to two its nearest neighbours by a mutualinductance M . We will number the coils consecutively from left to right.Assuming no sources, we can write for the n-th coil�

j!L C1

j!C

�In C j!MInC1 C j!MIn�1 D 0

A trivial solution for this equation exists: In D 0 for any n. But there arealso non-trivial solutions. We can assume that the current in any n-thelement is of the form In D I0 exp.�jkan/. Here I0 is the amplitude, andk is called the wavenumber. This mathematical form represents waves ofcurrent propagating along the line with the wavelength 2�=k. These wavesare magneto-inductive waves. Then, we obtain�

j!L C1

j!C

�I0e�jkan

C j!MI0e�jka.nC1/C j!MI0e�jka.n�1/

D 0

Diving by I0 exp.�jkan/ and taking into account that

exp.jka/ C exp.�jka/ D 2 cos.ka/

we get �j!L C

1

j!C

�C 2j!M cos.ka/ D 0

Finally, diving by j!L, using the definition of the resonant frequency !0 D

1=p

LC and the coupling coefficient � D M=L, we get�1 �

!20

!2

�C 2� cos.ka/ D 0

i nd uct i v e p ow er ing 57

As we can see, the wavenumber and the frequency of the waves dependon each other. The relationship between ! and k is called the dispersionrelation, and it is one of the most important characteristics of a wave. Forlight or for waves on two-wire transmission lines, the dispersion relationis ! D ck, where c is the light velocity.

!

!0

��� ka

1=p

1 � 2j�j

1=p

1 C 2j�j

The dispersion curve of magneto-inductive depends on the magnitude andthe sign of the coupling coefficient. For frequencies 1=

p1 � 2j�j < !=!0 <

1=p

1 C 2j�j, the wavenumber is real. Outside this range (called the prop-agation or pass band), the wavenumber can only be imaginary. ThenIn D I0 exp.�jkjan/. The signal decays exponentially along the line, andno wave propagate.

Any point on the dispersion curve describes a wave propagatating at a par-ticular frequency ! with a particular wavenumber k. We can characterisethis wave by two different velocities. The first one called the phase velocityis defined as vph D !=k. This velocity with which the phase of the wavemoves. The second one is the group velocity defined as vgr D d!=dk. Thegroup velocity is geometrically the slope of the dispersion curve. It is thevelocity with which the signal and energy move. For positive �, we can seeon the right diagram below that both the phase and the group velocitiesare positive. The waves with vph � vgr > 0 are called forward waves. On theother hand, for negative �, we see that vph > 0 but vgr < 0. Such waveswith vph � vgr < 0 are called backward waves.

!

!0

!

!0

ka ka

� < 0 � > 0

vgr Dd!

dk

vph D!

k

Most of the waves that electrical engineers deal with are forward waves.For example, waves propagating in transmission lines, coaxial cables, rect-angular waveguides, and optical fibres are forward waves. Backward wavesare somewhat exotic, but they play a pivotal role in, for example, metama-terials. Metamaterials are made of small electromagnetic resonators thatmimic natural atoms. They can show rffects like negative refraction of rays.

58 ind uct i v e p ow er ing

Motors and generators

Faraday's law says that time-varying magnetic flux creates electricfields. Herein lies the principle of rotating machines: electric generatorsand motors. We begin the discussion with the elementary ac generator.

The generator consists of a metallic (copper) coil placed between the polesof a permanent magnet. The moving part of a generator (the coil here) iscalled the rotor or armature. The stationary part is called the stator. Asthe coil is rotating around its axis, the magnetic flux through it changes,and a voltage appears between the ends of the coil.

But the ends of the coil move, and we cannot connect a load directly tothem. We can, however, connect the ends to two slip rings. Althoughthe rings rotate together with the coil, they do so around stationary axes.We can now make a connection to the rings using stationary brushes.The brushes are carbon blocks rubbing over the rotating copper slip rings.

slip ringsbrush

N

S

We can determine the voltage across the load using Faraday’s law. Weassume that the magnetic field B is constant and uniform, and that thecoil rotates with the angular frequency !. The area of the coil is S . Thestarting position will be when the coil is perpendicular to the field lines.As the coil rotates, the flux through it is ˆ D BS cos.!t/, where !t equalthe angle from the vertical line. The voltage across the load is then

N S

!t

U D �dˆ

dtD BS! sin.!t/

Rotation of the coil generates an ac voltage. In the starting, vertical, posi-tion the voltage is zero.

voltage

time

motor s and g enerator s 5 9

Elementary dc generator

To obtain a dc voltage, we could rectify the voltage generated by an acgenerator. However, a better way exists, called commutation. It is basedon mechanical switching. The ends of the coil are connected to a singleslip ring that is split into two halves, which forms a simple commutator.(Here, and below we do not show, for simplicity, the magnets.)

voltage

time

When the coil starts rotating from the vertical position, the first half ofthe period will generate the same voltage as the ac generator. But in thesecond half of the period, the ends of the coil swap their connection tothe brushes. As the result, the generated voltage reverses the polarity andhas the form of j sin.!t/j. The resulting pulsating voltage has a non-zerocomponent, but the ac variation (ripple) is too large. A way to reduce theripple and have a higher dc voltage is to add coils.

For example, the voltages of two coils rotated by 90ı will also differ by 90ı.To make the communator, we now split the ring into four equal segments.The brushes will be connected to one of the coils for a quarter of a period,after which the commutator will switch the coils. The resulting voltagewill look like this

voltage

timeMore coils give less ripple, but the commutation is more complicated.

DC motor

U

IN

S

We can reverse the situation by energising the coil of the dc machine andplacing it into a magnetic field. We will also change the configuration ofthe magnet. It now consists of two pieces. The top one has its N-polefacing the coil; the bottom one, the S-pole.

Close to the coil conductors, the field lines of the semicircular magnets willbe directed along the line connecting the conductors. A dc current placedin a magnetic field experience a force F D I ŒdI � B�. This force will rotatethe coil. Had the current flown in the coil always in the same direction,the direction of the force would change every half of the period. However,the coil is energised through the commutator, and so the current switchesit’s direction. Referring to the side view of the motor, the top conductorwill always have the current flowing towards the viewer.

N

S

FB

The force on each of the conductor is then F D IBL, where L is the lengthof the armature coil. The forces will cooperate to create the total torque ofT D 2Fa D 2BILa, where a is the coil radius. On, the other hand, the totalflux created by the magnets can be written as ˆ D �aLB. So, the torquebecomes T D 2ˆI=�.

60 motor s and g enerator s

Instead of a single coil, we can make the armature out of N coils, each con-tributing the same amount of torque. Then, T D 2NˆI=�, or introducingK D 2N=�, T D KˆI .

As the coil rotates, the magnetic flux through it changes. It creates, ac-cording to Faraday’s law, a back-emf with the amplitude e D ˆ! in thecoil. This emf will counteract the coil movement. If the armature has N

coils, the back-emf increases, but not N times. The reason is that the back-emfs from each coil will be out of phase. The combined back-emf becomese D Kˆ!.

RI

e U

We can now present an equivalent electric circuit of the motor, denotingthe resistance of the coil by R. If the input voltage is U and the current is I ,then U D e C IR. Multiplying both sides by I , we have UI D eI C I 2R. Theleft-hand side is the input power Pin D UI . The second term of the right-hand side is the power lost to heat Ploss D I 2R. We can understand whatthe remaining term means by rearranging eI D Kˆ!I D T=I!I D T!. Bylooking at definitions and dimensions of the torque and frequency, we seethat ŒeI � D Œforce � distance=time� D Œforce � velocity� D Œmechanical work�.So, the term Pmech D eI represents the useful mechanical work. We ob-tained the equation showing the balance of power

Pin D Pmech C Ploss

In principle, a torque would produce acceleration, and hence varying an-gular frequency ! (the equation of motion is T D Id!=dt). However, weassume that the motor has some load, which it has to move. The mechan-ical resistance of the load counter-acts the torque, creating a steady-staterotation with the frequency !. We can find this frequency from the equa-tion U D e C IR. Expressing e D Kˆ! and I D T=.Kˆ/ we obtain

! DU

Kˆ�

TR

.Kˆ/2

When the torque increases, the frequency decreases. The maximum fre-quency is !max D U=.Kˆ/ for T D 0. Because there is no torque, there isno current in the coil, and the input voltage is fully compensated by theback-emf. It is, in principle, possible to increase T so that ! becomes zero,and rotation stops. It occurs for T D KˆU=R.

RI L

e U

This equilibrium machine model works when the motor rotates with aconstant speed. But if U varies, the current in the armature coil will varyas well. We then need to take the coil inductance into account and changethe circuit. The voltage across the inductor is LdI=dt and we have

dI

dtD

U � e � IR

L

The sign of dI=dt depends on the sign of U � e. So, if the armature voltageexceeds the back-emf, dI=dt > 0, the current grows, and so creates positivetorque that accelerates the motor. Conversely, if the armature voltage isbelow the back-emf, the torque of coil will decelerate the motor.

motor s and g enerator s 6 1

Polyphase ac generators

We now return to ac generators. So far, we assumed a rotating coil anda stationary magnetic field. But it is possible to reverse the situation andhave a magnet rotating inside a coil. The coil in this figure is perpendicularto the plane of the paper. Although now strictly true, the voltage generatedin coil aa0 will be sinusoidal.

N

Sa a0

N

Sa a0

b

b 0

Like for dc motors, we can add coils to this generator. If coil bb 0 is ro-tated 90ı against coil aa0, the voltage generated in bb 0 will be phase-shiftedagainst the voltage in aa0 by 90ı. But we also connect the coils in seriesby joining the outputs a0 and b. We can calculate the resulting voltagebetween a and b 0 using phasors. It will be phase-shifted by 45ı degreesagainst Uaa0 and Ubb0 but exceed their amplitude by p

2. The voltages pro-duced by a generator are called phases. For practical purposes, the phasesUaa0 and Ubb0 are identical, but they differ from the phase Uab0 . So, we willhave two-phase output from the generator with two coils. Because thecoils share a conductor, the generator has three outputs.

voltage

timeUbb0

Uaa0

Uab0 a a0 b b 0

Ubb0Uaa0

Uab0

Ubb0 Uab0

Uaa0

N

S

We can add more coils. If we have N of them, the voltage between theneighbouring ones will be shifted in phase by �=N degrees. If we connectthem in series, we can find the total voltage, again, by phasors. We can seethat the amplitude of the resulting voltage UN is less that the sum of theamplitudes of the voltages in each coil, jUN j <

Pi jUi j. It is a disadvantage

and occurs because of the big total phase shift between the voltages in thecoils. Some coils are almost out of phase. We cannot change the phaseshift, but we can make several groups of coils and connect those together.A good compromise is to have three groups of coils, and therefore, threeoutput phases. On one hand, adding a number of voltages allows us toobtain large amplitudes. On the other hand, three phases give a flexibilityto be combined into a number of outputs.

a0

a

b 0

b c

c 0

We can arrange the three phases to have equal amplitudes and be shiftedby 2�=3 relative to each other. Their vector sum is zero.voltage

time

Ucc0

Uaa0

Ubb0

Ubb0

Uaa0

Ucc0

wye

neutral

delta

Three separate phase need six leads. But we can reduce the number ofleads by sharing them between phases. The two common solutions arewye- and delta-connections. Delta connection has three ‘live’ phases; wyeconnection has three live phases and a neutral.

62 motor s and g enerator s

Maxwell’s equations

We now return to the fundamental questions of how to describe electro-magnetic phenomena mathematically. Let us first recall what we alreadyknow. We started with electrostatics and considered electric fields that donot vary in time (but can vary in space). Such fields are created by staticelectric charges. The two equations that describe electrostatics completelyare H

S.D � dS/ D q and H

l.E � dl/ D 0. Afterwards, we considered magnetic

fields that do not vary in time. We learned that such fields can be createdby dc electric currents. The two equations describing dc magnetic fieldsare H

S.B � dS/ D 0 and H

l

�H � dl

�D I . As we can see from the above equa-

tions, the static electric and magnetic phenomena are not linked. Theyexist independently of each other.

However, we then considered electromagnetic induction and Faraday’slaw. We wrote the law mathematically asI

l

.E.t/ � dl/ D �ddt

ZS

.B.t/ � dS/

We saw that if the magnetic flux varies in time, it creates a circulation ofan electric field. In other words, a time-varying magnetic field can createa time-varying electric field.

lS2

S1

I

In the previous discussion, we assumed quietly that time-varying magneticfields can be created by time-varying electric currents. But will Ampere’slaw, H

l

�H � dl

�D I , which worked for dc fields and currents, work accu-

rately for time-varying ones? The answer is negative. We can see it bylooking at the familiar situation of a capacitor in a circuit. As we know,an ac current can flow through the capacitor. It means that an ac conduc-tion current can flow through the wires. We surround a wire by a path l.According to Ampere’s law, the circulation H

l

�H � dl

� must be equal to thecurrent that passes through an open surface resting on the path l. But wecan create many surfaces like that. One of them, S1, can cross the wire.The conduction current through this surface equal IS1 D I . By a differentsurface S2 can go between the capacitor plates and cross no wire IS2 D 0.Then, for the same path l we have H

l

�H � dl

�D I and H

l

�H � dl

�D 0. Of

course, this result is meaningless.

We must change Ampere’s law. For time-varying phenomena it should berewritten as I

l

�H � dl

�D I C

ddt

ZS

.D � dS/

Now, for surface S2 we have D D "0E, so thatddt

ZS

.D � dS/ D "0SdE

dtD

"0S

d

dU

dtD C

dU

dt

On the other hand, for surface S1 we have Hl

�H � dl

�D I . Equating the

results for the two surfaces we obtain I D CdU=dt , which is the correctexpression for the current across the capacitor.

max w el l ' s e q uat ion s 6 3

We shall now generalise the equation further. So far, we have only dealtwith linear currents I . We also said that the sign of the current depends onthe current direction. In general, currents can be distributed in space. Wecan characterise these currents by their density J . The density can vary inspace. In addition, we can transform the current density into a vector byassigning it a direction. The total current becomes I D

RS

.J � dS/.

S

J

We can make a similar generalisation for the charge q enclosed by a surfaceS . If the charge is distributed in space with a volume charge density �, wehave q D

RV

�dV .

We can now summarise the results into the following four equationsIl

.E � dl/ D �ddt

ZS

.B � dS/Il

�H � dl

�D

ZS

.J � dS/ Cddt

ZS

.D � dS/IS

.D � dS/ D

ZV

�dVIS

.B � dS/ D 0

Together they are knows as Maxwell’s equations. They are fundamentaland describe all electromagnetic phenomena. We cannot derive them fromother principles. Rather, these equations summarise all facts we knowabout electromagnetic fields.

Actually, we could (and many textbooks do) postulate these equations inthe very beginning of the course and then derive from them every otherresult. For example, if nothing changes with time, d=dt D 0, then the firstand the third equations contain only the electric quantities E, D, and �, andthe second and fourths equations contain only the magnetic quantities B,H, and J . There is no link between the pairs of equations, so we see thatdc electric and magnetic fields are unrelated.

As a reminder, we discuss now the integrals in Maxwell’s equations. In thefirst two equations, S is any open surface, and l is the closed path defined byS (the ‘rim’ of S). We can define a direction to go around l. This directionis strictly linked (by the right-hand rule) to the direction of the normalswe may take on surface S . In the last two Maxwell’s equations, S is anyclosed surface, and V is the volume that S encloses. For a closed surface,we take the outward-pointing normals.

To solve Maxwell’s equations (i.e. to find, for example, the fields for agiven charge distribution), we will need to augment them with relation-ships between E, D, B, H, and J . They are called constitutive relations. Wehave considered two of them in this course: D D "0"dE for dielectric ma-terials and B D �0�rH for magnetic materials. Another one known to youfrom the course on semiconductors is J D �E, where � is the conductivity.Other materials exist, with more complicated constitutive relations.

64 max w el l ' s e q uat ion s

Maxwell’s equations in differential form

As we have written them, Maxwell’s equations are said to be in the integralform. They relate not the fields directly but rather their integral quantities,such as flux and circulation. The fields can generally vary from point topoint in space, so that for example E D E.x; y; z; t/. To find such fields,the integral form of Maxwell’s equations is inconvenient. We have alreadyseen that we can find the electric field from Gauss’s law, H

S.D �dS/ D q, only

if we know that the field is constant on S , so that the flux is proportionalto the field. But generally, the relationship between the flux and the fieldis not simple. For example, the flux through a closed surface can, as weknow, be zero although the field is non-zero everywhere.

Therefore, we need to rewrite Maxwell’s equations in a form that dependsonly on the quantities at a single point. It is known as the differential formof Maxwell’s equations. To see fully how the equations can be converted,you would need to know more about vector calculus. We will look, as ademonstration, only at a single equation.

S

Our equation will be, again, Gauss’s law HS

.D � dS/ D q. Here, q is thecharge enclosed by an imaginary surface. We choose the surface to be aparallelepiped. If it encloses the charge distribution entirely, q is equal tothe total charge of the distribution.

SWe now start shrinking the parallelepiped. When it sits inside the chargedistribution, the flux through it will be equal only to the charge it encloses.We are interested in electric fields that change smoothly and continuouslyin space. So, as the parallelepiped becomes smaller and smaller, the vari-ation of the field on it should also become smaller and smaller. We canthen assume that the field on a single side of the parallelepiped does notchange at all. You should already have recognised that it is the reasoningused to formulate limits and derivatives.

x

y

z

.x; y; z/ dx

dydz

n nWe can then introduce a Cartesian coordinate system and orient the axesalong the sides of the parallelepiped. The lengths of the sides will be dx,dy, and dz. The origin will have coordinates .x; y; z/. We first considerthe fluxes through the left and right sides (those perpendicular to the x-axis). As we said, the electric field on the left side will approximately beconstant. We can take this field equal to its value at the origin, E.x; y; z/.The area of the side is dydz. The flux through this side will be ˆleft D

�"0Ex.x; y; z/dydz. The sign is negative because the normal points in thenegative x-direction (remember that we take outward-pointing normals).We also assumed vacuum, so that D D "0E.

The same arguments apply to the right side, but the x value here willchange by dx, so that ˆright D "0Ex.x Cdx; y; z/dydz. The sign has changedbecause the normal is now pointing in the positive direction. The totalflux through both surfaces is ˆx D ˆleft C ˆright D "0ŒEx.x C dx; y; z/ �

Ex.x; y; z/�dydz.

max w el l ' s e q uat ion s 6 5

On the other hand, we can expand E.x C dx; y; z/ in the Taylor series,keeping only the first two terms. (The remaining terms disappear whenwe tend x to zero). Then

Ex.x C dx; y; z/ D Ex.x; y; z/ C@Ex.x; y; z/

@xdx

so that the flux through both surfaces is

ˆx D "0

@Ex.x; y; z/

@xdxdydz

By reasoning in the same way, we can find the fluxes through the remainingfour surfaces as

ˆy D "0

@Ey.x; y; z/

@ydxdydz ˆz D "0

@Ez.x; y; z/

@zdxdydz

The total flux through the parallelepiped is then the sum of fluxes throughall sides

ˆ D "0

�@Ex.x; y; z/

@xC

@Ey.x; y; z/

@yC

@Ez.x; y; z/

@z

�dxdydz

This flux must be equal to the charge dq enclosed by the parallelepiped. Ifthe volume charge density is �.x; y; z/ then dq D �.x; y; z/dxdydz. Equatingthe charge to the flux and cancelling out the common multiplier dxdydz,we obtain

"0

�@Ex.x; y; z/

@xC

@Ey.x; y; z/

@yC

@Ez.x; y; z/

@z

�D �.x; y; z/

This is Gauss’s law in differential form. It is equivalent to the integral formin the sense that one can be derived from the other. You are already familiarwith a simpler version of this equation from the course in semiconductors.If only the x-component of the field in non-zero and if it only varies alongthe x-coordinate, E D exE.x/, then

dE.x/

dxD

�.x/

"0

If the charge is distributed in a dielectric, we need to replace "0 with "0"d.We have written the above equations in a Cartesian system. If we changethe type of the coordinate system, the equations will change as well. Forexample, if the problem is spherically symmetric and E and � depend onlyon the distance r, then the equation for the field is

1

r2

ddr

.r2E.r// D�.r/

"0

Will will now look at an example of how to use this equation to find fields.We will take the already familiar uniformly-charged solid sphere. The totalcharge is q, the radius is R.

Rq

Ein Eout

We have two regions. Inside the sphere � D q=V D 3q=.4�R3/ D const.Outside the sphere � D 0. So, for the field inside the sphere we have

ddr

.r2Ein/ D�r2

"0

r2Ein D�r3

3"0

C A

Ein D�r

3"0

CA

r2

Here A is a constant.

66 max w el l ' s e q uat ion s

We can find the value of A by reasoning as follows. This expression mustbe true everywhere inside the sphere, including its centre. The field in thecentre cannot be infinitely large. Therefore, the term 1=r2 must vanish,and A must be zero.

Recalling that � D q=V D 3q=.4�R3/, we can rewrite the expression for thefield inside the sphere as

Ein Dq

4�"0

r

R3

which is the same as the field we obtained earlier using Gauss’s law in theintegral form.

For the field outside the sphere, we haveddr

.r2Eout/ D 0

r2Ein D C

Eout DC

r2

Here C is a constant. To find the constant, we need additional informa-tion. Such information is given by boundary conditions. The surface of thesphere is a boundary separating two domains, one with charge and onewithout. The boundary condition relates the electric fields outside and in-side the sphere. There are no surface charges on this boundary, and theboundary condition is Ein.R/ D Eout.R/. Substituting the expressions forthe fields inside and outside the sphere into the boundary conditions, weobtain

q

4�"0

R

R3D

C

R2

C Dq

4�"0

so thatEout D

q

4�"0r2

which is, again, the same result as we obtained from Gauss’s theorem inintegral form.

We have already looked further into the topic than your knowledge ofmathematics allows us, and will stop here. Boundary conditions and otheradvanced topics are discussed thoroughly in the second-year Fields course.

max w el l ' s e q uat ion s 6 7

Slow-varying phenomena and electric circuits

Maxwell’s equations describe all electromagnetic phenomena, so they mustbe true for electric circuits as well. But the link between Maxwell’s equa-tions and circuit equations is not immediately obvious. For example, a ba-sic circuit concept is voltage. We discussed that voltage can be introducedin electrostatics from the equation H

l.E �dl/ D 0. Only if this equation holds

will the work of the electric field between two points not depend on thepath, and we can define voltage. But the complete time-varying Maxwell’sequation is I

l

.E � dl/ ¤ 0 D �ddt

ZS

.B � dS/

So, the work of the electric field does depend on the path. As a result, wecannot in general define voltage if the electric field varies in time.

Another example is the problem are capacitance and inductance. We firstintroduced these concepts for static charges and dc currents. Why thencan we use them for time-varying signals?

For an ac electric field, we can write E.x; y; z; t/ D QE.x; y; z/ sin.!t/. In otherwords, we can separate the dependence on the coordinate from that ontime. Then dE=dt D ! QE.x; y; z/ cos.!t/ and

maxjdE=dt j

maxjEjD !

If the frequency ! is small, the amplitude of the electric field will be muchlarger than its time derivative. The same applies for the magnetic field. Wecan then ignore the time derivatives in Maxwell’s equations, which leadsto two pairs of equations for constant electric and magnetic fields.I

l

. QE � dl/ D 0IS

. QD � dS/ D

ZV

Q�dV

Il

�QH � dl

�D

ZS

. QJ � dS/IS

. QB � dS/ D 0

We can then solve dc electric and magnetic problems, and obtain the acsolutions by adding time variation to the static solutions. The same appliesto voltage: if we know the dc voltage and the processes vary slowly in time,we can introduce an ac voltage as Uac D Udc sin.!t/.

Although we cannot prove it in this course, the condition of slow timevariation is equivalent to the condition that the size of the system is muchsmaller than the wavelength of electromagnetic radiation.

68 max w el l ' s e q uat ion s

Dielectrics in time-varying fields

We have considered materials with dielectric and magnetic proper-ties. It turns out that as the frequency increases, the magnetic response ofmost materials decreases. We will therefore consider only how materialsrespond to time-varying electric fields.

E.t/

x.t/

We already have a model for an atom placed in a static electric field (seep. 29). In this model, the field displaces negative electrons relative to thenucleus. Thus an electric dipole forms.

For time-varying fields, we need to refine the model. First, the displacedelectrons will experience a force from the nucleus. This force will opposethe field and try to bring the electrons back. Intuition suggests that thefarther away the electrons are from the nucleus, the stronger will be theattraction. So, this force can be written as �kx, where k is a constant. Theminus sign means that the force opposes the displacement. The force isthe same as the one that acts on a mass attached to a spring.

Cq�q

x.t/p.t/ D qx.t/

There will be another process, familiar to you from the course on semicon-ductors. The dipole will experience forces from the neighbouring atoms,which we can see as an atomic analogue of friction. The friction is higherif the charges move fast and if they are heavy, so we can write this force as�mv=� D �m=� dx=dt . The total force on the negative charge will be then

F D qE � kx �m

�

dx

dt

According to Newton’s equation of motion the force, is equal to the mass,m, times the acceleration a D d2x=dt2, so that

md2x

dt2D qE � kx �

1

�

dx

dt

Introducing two frequencies !20 D k=m and D 1=� and rearranging the

above equation, we can writed2x

dt2C

dx

dtC !2

0x Dq

mE

It is the equation of an oscillator, which you should have already seen inother courses. We are interested in the forced solution under an ac electricfield of the form E.t/ D E0 exp.j!t/. Then, the displacement will varyin time in the same manner, x.t/ D x0 exp.j!t/, and we can rewrite theequation of motion as

.�!2C !2

0 C j! /x0 Dq

mE0

from which the dipole moment p.t/ D qx.t/ is

p D �q2

m

1

!2 � !20 � j!

E

di e l e ct r i c s i n t ime - vary ing f i e ld s 6 9

If the number of the dipoles per unit volume is N , we can introduce arelative permittivity "r in the same way we did it for dc electric fields as

"r D 1 �q2N 2

m"0

1

!2 � !20 � j!

Introducing a frequency �2 D q2N 2=.m"0/, we obtain"r D 1 �

�2

!2 � !20 � j!

As we see, the relative dielectric permittivity is a complex number and canbe written in the form "r D "0

r � j"00r (note the sign of the imaginary part).

!!0

"0r

"00r

We have derived the expression for the permittivity from the equation foran oscillator. One of the properties oscillators is resonance, and not sur-prisingly, the relative permittivity shows a resonant behaviour. In partic-ular, its imaginary part "00

r has a maximum at !0.

To see what the complex permittivity means, you might recall that therefractive index of a material n is defined as n D

p"r D

p"0r � j"00

r . If "00r � "0

r,we can expand the square root into the Taylor series and leave only the firsttwo terms, so that

n D n0� jn00

Dp

"0r

�1 �

"00r

2"0r

�Dp

"0r �

"00r

2p

"0r

If a wave with a wavenumber k propagates inside the material, say, alongthe x-axis, we can write its electric field in the form

E D E0e�jnkxD E0e�n00kxe�jn0kx

The wave amplitude will decay with distance if n00 ¤ 0, and thus "00r ¤ 0.

Therefore, the imaginary part of the permittivity shows how quickly thewave loses its energy inside the material. To reduce the loss, dielectricwaveguides operate far from material resonances.

70 d i e l e ct r i c s i n t ime - vary ing f i e ld s