Question (2g50001)

Potential dierence between points near a thin, very long, uniformly charged

rod

A thin rod of length L has a uniform charge Q. (a) Write an expression for the potential dierence VB −VA,if points A and B are very close to the rod in comparison to the length of the rod. (b) Find the potentialdierence VC − VA if the rod's length is 1 m, the rod's charge is 1 µC, point A is 1 mm from the rod, pointB is 3 mm from the rod, and point C is 3 mm from the rod and 1 mm from point B.

Figure 1: Points near a thin, very long, uniformly charged rod.

Solution

(a) Dene a coordinate system along the perpendicular bisector of the rod with the +x axis through pointsA and B as shown below and the origin at the center of the rod.

Figure 2: Potential dierence along a path from point A to point B.

The magnitude of the electric eld due to the rod, for points near the rod (r << L), along the perpendicularbisector of the rod is

E =1

4πǫ0

2|Q|/L

r

which decreases from point A to point B, as shown above. Since we've dened the +x axis to be perpendicularto the rod with the origin at the rod, then

Ex =1

4πǫ0

2Q/L

x

The potential dierence along a path from point A to point B is

∆V = −

xB∫

xA

~E ·

~dL

Because the electric eld decreases from point A to point B, it is not constant, thus an integral is required. Fora straight-line path from point A to point B, ~dL =< dx, 0, 0 >. The dot product ~E ·

~dL = Exdx+Eydy+Ezdz

simplies to Exdx since ~E and ~dL are both in the +x direction. Thus,

∆V = −

xB∫

xA

Exdx

VB − VA = −

xB∫

xA

1

4πǫ0

2Q/L

xdx

= −1

4πǫ0

2Q

L

xB∫

xA

1

xdx

= −1

4πǫ0

2Q

Lln(x)

∣

∣

∣

∣

∣

xB

xA

= −1

4πǫ0

2Q

L

(

ln(xB) − ln(xA)

)

=1

4πǫ0

2Q

L

(

ln(xA) − ln(xB)

)

=1

4πǫ0

2Q

Lln

(

xA

xB

)

Note that to simplify the expression, we used the fact that the dierence in natural logs can be written asthe log of the ratio of the quantities.

(b) The potential dierence from A to C can be calculated along any path. Since we already know thepotential dierence between A and B, then let's choose a path from A to B and B to C, along the legs ofthe right triangle shown below.

Point C is close enough to the perpendicular bisector that the electric eld at point C is the same as theelectric eld at point B. That is, the electric eld due to a long rod is independent of the vertical distancefrom the perpendicular bisector, as long as the point is near to the perpendicular bisector and near to therod. Thus, from point B to point C, the electric eld is constant and to the right.

The potential dierence (VC − VA) = (VB − VA) + (VC − VB) along the path A to B and B to C. We alreadyknow that

Figure 3: Potential dierence along a path from point A to point C.

VB − VA =1

4πǫ0

2Q

Lln

(

xA

xB

)

From B to C, ~E is perpendicular to ~dL along the path, thus ~E ·

~dL = 0 and ∆V = VC − VB = 0. In otherwords, point C is at the same potential as point B. As a result, the potential dierence from A to C is

VC − VA = (VB − VA) + (VC − VB)

=1

4πǫ0

2Q

Lln

(

xA

xB

)

+ 0

=1

4πǫ0

2Q

Lln

(

xA

xB

)

=1

4πǫ0

2(1 × 10−6 C)

1 mln

(

0.001 m

0.003 m

)

= −1.98 × 104V

Question (../graphics/2d40002)

1-D Superposition of dipole elds along axis

Two dipoles are oriented as shown below. One dipole is made of opposite charges of magnitude q, andthe other dipole is made of opposite charges of magnitude 2q. Both dipoles have a separation s. Derivean equation for the net electric eld at the point shown if the distance d is much larger than the chargeseparation s. (i.e. d >> s)

Figure 1: Two dipoles.

Solution

Sketch a picture of the electric eld at the given point due to to each dipole, and sketch the net electric eld.

Figure 2: Net electric eld due to two dipoles, along the axis of the dipoles.

The electric eld due to a dipole along the axis of the dipole, if the distance r is much greater than thecharge separation s, is

| ~E|on axis ≈1

4πǫ0

2qs

r3

The electric eld a distance 2d from the rst dipole has a magnitude

| ~E|1 ≈1

4πǫ0

2qs

(2d)3

| ~E|1 ≈1

4πǫ0

2qs

8d3

| ~E|1 ≈1

4πǫ0

qs

4d3

Expressed as a vector,

~E1 ≈ <1

4πǫ0

qs

4d3, 0, 0 >

Similarly, the electric eld a distance d from the second dipole has a magnitude

| ~E|2 ≈1

4πǫ0

2(2q)s

d3

| ~E|2 ≈1

4πǫ0

4qs

d3

Expressed as a vector,

~E2 ≈ <1

4πǫ0

4qs

d3, 0, 0 >

The net electric eld is given by the Superposition Principle:

~Enet = ~E1 + ~E2

= <1

4πǫ0

qs

4d3, 0, 0 > + <

1

4πǫ0

4qs

d3, 0, 0 >

= <1

4πǫ0

qs

4d3+

1

4πǫ0

4qs

d3, 0, 0 >

= <1

4πǫ0

qs

d3(1

4+ 4), 0, 0 >

= <1

4πǫ0

17qs

4d3, 0, 0 >

Note that it has a positive x-component, thus agreeing with our initial sketch of the net electric eld at thegiven point.

Question (2d40001)

Dipole model of HCl

The HCl molecule is polar with a dipole moment 3.5 × 10−30 C · m and a negative partial charge δ− at thechlorine atom and a net positive partial charge δ+ at the hydrogen atom. From infrared spectroscopy, thebond length of a HCl molecule is calculated to be 0.13 nm. If you model the HCl molecule as a dipole ofcharges δ+ and δ− separated a distance 0.13 nm, what is the charge |δ| in the model?

Solution

Sketch a picture of the polarized molecule, showing the HCl molecule and a dipole model of the molecule.

Figure 1: Dipole model of an HCl molecule.

The magnitude of the dipole moment of the molecule is dened as

|~p| = qs

where q is the magnitude of each charged particle in the dipole and s is the separation of the chargedparticles. In this case, the magnitude of the charge of each particle in the dipole is δ, thus

|~p| = δs

Solving for the partial charge of each atom δ gives

δ =|~p|

s=

3.5 × 10−30 C · m

0.13 × 10−9 m

δ = 2.7 × 10−20

C

Note that this is less than the magnitude of the charge of an electron (1.6× 10−19 C). That's why it's calleda partial charge.

Question (2ia0001)

Potential dierence, electric eld, drift speed, and current for two wires con-

nected in series.

A 1.5 V battery is connected to two copper wires that are connected together in series, as shown below.Wire 1 has a diameter of 1 mm and a length of 0.5 m. Wire 2 has a diameter of 2 mm and the same lengthas Wire 1. The mobile electron density of copper n is 8.4× 1028 m−3, and the electron mobility in copper u

is 4.5 × 10−3 m/s

N/C.

Figure 1: Circuit with a battery and two copper wires of dierent diameters connected in series.

(a) What is the potential dierence across each wire?

(b) What is the electric eld in each wire?

(c) What is the drift speed of mobile electrons in each wire?

(d) What is the current through each wire?

Solution

First, it's important to think conceptually about the problem before applying mathematics to solve theproblem. Because the wires are connected in series, conservation of charge tells us that the amount ofcharge per second that ows out of wire 1 must ow into wire 2. In other words, I1 = I2.

Since I = neAv, where A is the cross-sectional area of the wire, and since I1 = I2, then A1v1 = A2v2.This means that a larger area wire will have a smaller drift speed; therefore, the drift speed in wire 1 willbe greater than the drift speed in wire 2. The cross-sectional area of a wire is A = πR2. Therefore, sinceR2 = 2R1, then v1 = 4v2.

The drift speed of mobile electrons in the wire is related to the electric eld within the wire, v = uE.Therefore, there must be a larger electric eld in wire 1 since the drift speed is greater in wire 1. Since driftspeed and electric eld are directly proportional and since v1 = 4v2, then E1 = 4E2.

The potential dierence across each wire has a magnitude |∆V | = EL. Since the wires have the same lengthand since the electric eld in wire 1 is 4 times greater than in wire 2, then the potential dierence acrosswire 1 is 4 times the potential dierence across wire 2. That is, |∆V1| = 4|∆V2|.

Conservation of Energy applied to the circuit means that the gain in electric potential energy as chargeows through the battery is lost as the charge ows through the wires. (The electric potential energy isconverted to thermal energy, causing the wires to warm up.) Stated mathematically,

energy gained through the battery = energy lost through the wires

∆Uel,bat = ∆Uel,1 + ∆Uel,2

q∆Vbat = q∆V1 + q∆V2

∆Vbat = ∆V1 + ∆V2

Thus, the statement of conservation of energy can be cast in terms of electric potential dierence across thebattery and each wire. The potential dierence across the battery is its emf ǫ if it's an ideal battery.

ǫbat = ∆V1 + ∆V2

ǫbat = 4∆V2 + ∆V2

ǫbat = 5∆V2

Thus,

5∆V2 = 1.5 V

∆V2 =1.5 V

5∆V2 = 0.3 V

which makes

∆V1 = 4∆V2

∆V1 = 4(0.3 V)

∆V1 = 1.2 V

Note that we can check if the calculations are correct. According to Conservation of Energy, the sumof the potential dierences across the wires must equal the potential dierence across the battery. Indeed,1.2 V + 0.3 V = 1.5 V, as expected.

The electric eld in each wire is

∆V1 = E1L

E1 =∆V1

L

E1 =1.2 V

0.5 m= 2.4 V/m

E2 =0.3 V

0.5 m= 0.6 V/m

Note that E1 = 4E2 exactly as expected!

The drift speed of mobile electrons in each wire is

v1 = uE1

= (4.5 × 10−3m/s

N/C)(2.4 N/C)

= 0.0108 m/s

and

v2 = uE2

= (4.5 × 10−3m/s

N/C)(0.6 N/C)

= 0.0027 m/s

Again, note that v1 = 4v2 exactly as expected!

The current through each wire is the same and is given by I = neAv. Either wire can be used for thecalculation. Let's choose wire 1.

I = neAv

= (8.4 × 1028 m−3)(1.6 × 10−19 C)π

(

0.001 m

2

)2

(0.0108 m/s)

= 114 A

Wow, this is a big current! Short-circuiting a battery with a copper wire like this indeed produces a highcurrent. However, internal resistance of the battery will increase as it warms up which will limit the currentso that it does not get this big. A commercial power supply (which is closer to being an ideal voltage source

typically has a current limiter which will prevent such high currents that might damage the power supply.It also might have a fuse that will break" before the current can get this high.

Question (2ha0001)

Deection of a compass needle by a dipole magnet

A dipole magnet and compass are arranged as shown below, with the dipole aligned East-West with thecompass. The dipole moment of the magnet is 1.5 A · m2 and its center is 20 cm from the center of acompass. The magnetic eld of Earth is shown.

(a) Sketch and calculate the deection of the compass needle from North.

(b) If you replace the magnetic dipole with a thin coil of radius 2 cm and 20 turns at the same location,what must be the current in the coil to give the same deection of the compass needle as the dipolemagnet?

Figure 1: Dipole magnet and compass.

Solution

(a) The magnetic eld at the location of the compass points toward the S pole of the dipole, and Earth'smagnetic eld points North (that is, approximately geographic North) as shown below.

Figure 2: Magnetic elds at the location of the compass.

Thus, the compass will deect to the left, or toward the West side of the compass, as shown below.

The horizontal component of the magnetic eld of Earth at the latitude of High Point, NC, is approximately22,000 nT, or 2.2 × 10−5 T. (You can calculate the horizontal component of the magnetic eld of Earth atyour latitude at <http://www.ngdc.noaa.gov/geomagmodels/IGRFWMM.jsp> . The magnetic eld at thelocation of the compass due to the dipole has a magnitude

Figure 3: Deection of the compass needle.

Bdipole,axis ≈

µ0

4π

2µ

r3

≈ (1 × 10−7 T m2/(C m/s))2(1.5 A · m2)

(0.2 m)3

≈ 3.75 × 10−5 T

The components of the net magnetic eld are the magnetic eld of the dipole and the magnetic eld ofEarth. They can be used to nd the deection angle, shown below.

Figure 4: The net magnetic eld and the deection angle of the compass needle.

From the right triangle, the deection angle is computed by

tan θ =Bdipole

Bearth

=3.75 × 10−5 T

2.2 × 10−5 T

θ = tan−1

(

3.75 × 10−5 T

2.2 × 10−5 T

)

θ = 59.6

θ ≈ 60

(b) If you replace the dipole magnet with a coil, then the current in the coil will have to ow around thecoil in such a way that it creates a magnetic eld along the axis of the coil that is to the left. Using theright hand rule, with your thumb pointing left, your ngers go into the screen at the top of the coil and outof the screen at the bottom of the coil. A side view of the coil is shown below.

The dipole moment of a coil of N turns is

Figure 5: Orientation of a current-carrying coil with the same dipole moment as the previously used magnet.

µcoil = NIA

which must be the same as the previously used magnet of dipole moment µ = 1.5 A m2. Thus,

I =µ

NA

=µ

NπR2

=1.5 A m2

20π(0.022)

= 60 A

Wow, 60 A is a lot of current going through that wire coil! One way to reduce the current is to increase thenumber of turns in the coil. Also, one can make a coil with a larger radius.

Question (2f30001)

Electric eld along the axis of a uniformly charged thin rod

A thin rod of length L has a uniform charge Q. Derive an expression for the electric eld at point P on theaxis of the rod that is at a distance d from the center of the rod. Show that in the limit as d → ∞, theelectric eld due to the rod is the same as if the rod is a charged particle.

Figure 1: Eeld on the axis of a uniformly charged thin rod.

Solution

Dene a coordinate system along the axis of rod, with the origin at the center of the rod as shown below.

Figure 2: Dene the coordinate system.

Break the rod into pieces of charge dq of width dx at location x on the rod. The electric eld at location Pdue to a piece of the rod is d ~E.

Figure 3: Dene a piece of the rod.

The electric eld at point P due to a piece of the rod of charge dq is

d ~E =1

4πǫ0

dq

|~r|2r

where the vector ~r is the position of point P relative to the charge dq. It is calculated in general by

~r = ~rP − ~rpiece

= < d, 0, 0 > − < x, 0, 0 >

= < d − x, 0, 0 >

= (d − x) < 1, 0, 0 >

Note that the distance between dq and point P is just d − x, and the vector ~r points in the +x-directionwhich is the unit vector < 1, 0, 0 >. Because r only has an x-component, the electric eld also only has anx-component and the y and z-components of the electric eld are zero. So, substituting r into the electriceld and solving only for the x-component of the electric eld gives

dEx =1

4πǫ0

dq

(d − x)2

If the charge dq is positive, then dEx is in the +x direction. If dq is negative, then dEx is in the −x direction.

The rod is uniformly charged. As a result, the charge of a piece divided by its length is the same as the totalcharge divided by the total length of the rod.

dq

dx=

Q

L

This allows us to write the charge of a piece of the rod in terms of its length dx.

dq =Q

Ldx

Substituting this into the electric eld gives

dEx =1

4πǫ0

Q

L

dx

(d − x)2

According to the Superposition Principle, the net electric eld at point P is the sum of the electric eldsdue to all pieces of the rod. Thus, we must sum (i.e. integrate) the electric eld due to each piece, over thelength of the rod. Use algebra to simply your answer.

Ex =

+L/2∫

−L/2

1

4πǫ0

Q

L

dx

(d − x)2

=1

4πǫ0

Q

L

+L/2∫

−L/2

1

(d − x)2dx

=1

4πǫ0

Q

L

1

(d − x)

∣

∣

∣

+L/2

−L/2

dx

=1

4πǫ0

Q

L

(

1

d − L2

−1

d − −L2

)

=1

4πǫ0

Q

L

(

(d + L2)

(d + L2)(d − L

2)−

(d − L2)

(d − L2)(d + L

2)

)

=1

4πǫ0

Q

L

(

(d + L2) − (d − L

2)

d2 − (L2)2

)

=1

4πǫ0

Q

L

(

L

d2 − (L2)2

)

=1

4πǫ0

Q

(d2 − (L2)2)

This can be written more generally, even for points to the left of the rod by calculating the magnitude ofthe electric eld.

| ~E| =1

4πǫ0

|Q|

(d2 − (L2)2)

The direction of ~E along the axis is away from the rod if the rod is positively charged, +Q, and toward the

rod if the rod has a negative charge −Q. To check that the answer makes sense, consider the electric eld ata point P that is VERY far away from the rod so that d >> (L/2). In this case, the term (L/2)2 is negligiblecompared to d2. Then,

| ~E| ≈1

4πǫ0

|Q|

d2

as expected for the eld due to a point particle.

Question (2g30001)

Acceleration of an alpha particle by charged plates

An alpha particle is accelerated by two closely spaced, oppositely charged plates, as shown below.

Figure 1: An alpha particle moving between oppositely charged plates.

The alpha particle has a speed of 1.0 × 106 m/s when it enters a slit in the positively charged plate. Aftertraveling for 1 mm, it passes through a slit in the negatively charged plate. If the magnitude of the charge ofeach plate is 10 µC, and if each plate has an area of 10 cm2, what will be the speed of the alpha particle whenit reaches the negatively charged plate? (Note: the plate separation is small compared to the dimensions ofthe plates.)

Solution

Dene a coordinate system with the +x axis in the direction of the velocity of the alpha particle. Dene twopoints i and f along the path from the positively charged plate to the negatively charged plate.

Figure 2: Charged plates with coordinate system, initial and nal points, and the constant electric eld.

Let's begin by converting all units to m, kg, s, C, and combinations thereof. The charge on each plate hasa magnitude 10 × 10−6 C. The area of each plate is (10 cm2)(1 m2/1002 cm2) = 1 × 10−3 m2. The plateseparation is 1 mm, or 0.001 m.

Choose the Fundamental Principle that can be used to solve this problemConservation of Energy. Thesystem of plates and particle are a closed system. Thus, the change in the total energy of the system is zero.

∆E = 0

∆U + ∆K = 0

∆K = −∆U

Because the force on the alpha particle by the electric eld is in the same direction as the velocity of theparticle, then the particle will speed up as it travels toward the negatively charged plate. As a result, itskinetic energy increases. Conservation of Energy tells us that this must be accompanied by a decrease inpotential energy. It is the loss of electric potential energy that results in a gain in kinetic energy. Conceptually,we think of this as

|∆K|gained = |∆U |lost

If we calculate the initial kinetic energy and the change in electric potential energy, then the nal kineticenergy and nal speed can be determined.

(Kf − Ki) = −∆U

Kf = Ki − ∆U

The alpha particle is a helium nucleus with 2 neutrons and 2 protons; it's atomic mass is 4 g/mol which is6.64 × 10−27 kg. Its initial kinetic energy is

Ki =1

2mv2

i

=1

2(6.64 × 10−27 kg)(1 × 106 m/s)2

= 3.32 × 10−15 J

The change in potential energy of the alpha particle (that has a charge of 2qproton) from point i to point fis:

∆U = q∆V

= q(−Ex∆x)

= q

(

−|Q|/A

ǫ0∆x

)

= −(2)(1.6 × 10−19 C)

(

10 × 10−6 C/1 × 10−3 m2

ǫ0

)

(0.001 m)

= −3.616 × 10−13 J

Thus, the nal kinetic energy is

Kf = Ki − ∆U

= 3.32 × 10−15 J −−3.616 × 10−13 J

= 3.32 × 10−15 J + 3.616 × 10−13 J

= 3.649 × 10−13 J

From this, calculate the nal speed, using m = 6.64 × 10−27 kg.

Kf =1

2mv2

f and solving for the nal speed gives:

vf =

√

2Kf

m

=

√

2Kf

m

= 1.05 × 107 m/s

Question (2k80002)

Torque on a coil

A coil of 20 turns of wire has a radius 0.5 cm and is connected to a power supply so that 0.4 A ows throughthe coil. At t = 0, the coil is at rest, current ows into the page at the top of the coil, and the plane of thecoil makes an angle of 55 with respect to a magnetic eld of 0.5 T.

Figure 1: A coil in a uniform magnetic eld.

(a) At the instant shown (t=0), what is the torque on the coil?

(b) In what direction will the loop rotate if it is released from rest?

(c) What is the magnetic potential energy of the loop at t=0?

(d) If the system is frictionless, what will be the kinetic energy of the coil when it reaches its equilibriumorientation?

(e) Describe the motion of the coil, even after a long time, if the system is frictionless.

Solution

(a) The coil acts as a magnetic dipole with dipole moment µ = NIA in a direction perpendicular to theplane of the coil and in the direction given by the right-hand rule, as shown below.

Figure 2: Magnetic dipole moment of the coil.

The torque on a magnetic dipole is ~τ = ~µ × ~B. In this case, µ = NIA = 20(0.4 A)π(0.005 m)2 =

6.28× 10−4 A m2 and ~B = 〈0.5, 0, 0〉 T. There are two ways to calculate the torque. First, one can calculatethe cross product:

~τ = ~µ × ~B

= 〈−µ cos(35), µ cos(35), 0〉 × 〈0.5, 0, 0〉

= 〈−6.28 × 10−4 cos(35), 6.28 × 10−4 sin(35), 0〉 × 〈0.5, 0, 0〉

= (〈−5.14 × 10−4, 3.6 × 10−4, 0〉 A m2) × (〈0.5, 0, 0〉 T)

= 〈0, 0,−(3.6 × 10−4)(0.5)〉 N m

= 〈0, 0,−1.8 × 10−4〉 N m

The second way to determine torque is to calculate the magnitude of the torque and then use the right-handrule to get the direction. First, sketch the magnetic dipole moment and magnetic eld vector tail to tail andnd the angle between them.

Figure 3: Angle between the magnetic dipole moment and magnetic eld vectors.

The magnitude of the torque on the coil is

|~τ | = |~µ|| ~B| sin θ

= (6.28 × 10−4 A m2)(0.5 T) sin(145)

= −1.8 × 10−4 N m

Using your right hand, point your ngers in the direction of µ and rotate toward ~B. Your thumb will pointin the −z direction. Therefore,

~τ = 〈0, 0,−1.8 × 10−4〉 N m

(b) According to the angular momentum principle, the change in the angular momentum of the coil willbe in the same direction as the torque on the coil. Since it is released from rest, then the nal angularmomentum after a small time interval dt will be in the same direction as the torque, in the −z direction.Thus, the coil will rotate clockwise, toward equilibrium.

(c) The magnetic potential energy of a dipole in a magnetic eld is

Umag = −~µ ·

~B

= −|~µ|| ~B| cos θ

= −(6.28 × 10−4 A m2)(0.5 T) cos(145)

= 2.57 × 10−4J

Figure 4: Equilibrium orientation of the coil.

(d) The equilibrium orientation of the coil is shown below.

The magnetic potential energy at equilibrium is

Umag = −~µ ·

~B

= −|~µ|| ~B| cos θ

= −(6.28 × 10−4 A m2)(0.5 T) cos(0)

= −3.14 × 10−4J

Thus, in rotating from its initial position at t = 0 to the equilibrium position, the change in potential energyof the coil is

∆Umag = Uf − Ui

= −3.14 × 10−4J − 2.57 × 10−4J

= −3.14 × 10−4J − 2.57 × 10−4J

= −5.71 × 10−4J

Since it is a closed system, Conservation of Energy states that:

∆E = 0

∆U + ∆K = 0

∆K = −∆U

∆K = −− 5.71 × 10−4J

∆K = 5.71 × 10−4J

In other words, the potential energy lost by the coil in rotating toward equilibrium goes into and increase inkinetic energy. As a result, its angular speed of the coil increases as it rotates toward equilibrium.

(e) The coil will rotate past equilibrium and will then slow down until coming to rest with the orientationshown below.

It will then rotate back toward equilibrium, speeding up as it rotates. In this way, it will oscillate back andforth around equilibrium.

Figure 5: Coil has rotated past equilibrium.

Question (2h30001)

Acceleration of an alpha particle by charged plates

A proton is at the origin and is moving in the +y direction with a speed of 1 × 104 m/s, as shown below.What is the magnetic eld at each of the points shown? Note: the points are symmetric, the distancebetween points C and E is 0.5 µm, and the distance between points B and C is 1.0 µm.

Figure 1: Find the magnetic eld at various points around a moving charged particle.

Solution

Magnetic eld at Point A. To calculate the magnetic eld at any point, apply the Biot Savart Law

~B =µ0

4π

q~v × r

|~r|2

For point A, sketch the vector ~r from the particle to point A and the velocity of the particle ~v.

Figure 2: Position vector for point A and velocity vector for the particle.

Write the vectors ~r and ~v in bracket notation.

~r = < −0.5, 1.0, 0 > µm =< −0.5 × 10−6, 1 × 10−6, 0 > m

~v = < 0, 1 × 104, 0 > m/s

Find the magnitude of ~r and its unit vector.

|~r| =√

0.52 + 12 × 10−6 m

|~r| = 1.118 × 10−6 m

r =~r

|~r|=< −0.4472, 0.8944, 0 >

Calculate ~v × r.

~v × r =

∣

∣

∣

∣

∣

∣

i j kvx vy vz

rx ry rz

∣

∣

∣

∣

∣

∣

~v × r = < (vy rz − vz ry),−(vxrz − vz rx), (vxry − vy rx) >

= < (vy rz − vz ry), (vz rx − vxrz), (vxry − vy rx) >

Substituting components of the vectors gives

~v × r = < 0, 0,−vy rx >

= < 0, 0,−(1 × 104 m/s)(−0.4472) >

Substitute known quantities into the Biot-Savart law, including µ0

4π= 1 × 10−7 T m2/(C m/s).

~B =µ0

4π

(1.6 × 10−19 C) < 0, 0, (1 × 104 m/s)(0.4472) >

(1.118 × 10−6m)2

~Bat point A = < 0, 0, 5.72 × 10−11 > T

Note that the magnetic eld is in the +z direction, in agreement with using the right hand rule. If you usethe nger of your right hand to rotate" ~v into ~r, then you rotate your ngers counterclockwise and yourthumb points in the +z direction.

Figure 3: Rotate the velocity vector counterclockwise; the magnetic eld at point A thus points in the+z-direction.

Magnetic Field at point C. Because point C is the same distance from the particle as point A, and theright-hand rule gives you the same direction for the magnetic eld ~B, then ~Bat point C = ~Bat point A.

Figure 4: Rotate the velocity vector counterclockwise; the magnetic eld at point C thus points in the+z-direction.

Magnetic Field at point B. Note that the angle between the velocity and position vectors is 90. Thus,use the fact that the magnitude of a cross product is

~v × r = |~v||r| sin θ

Figure 5: The angle between the velocity of the particle and the position vector for point B is 90 degrees.

The magnitude of the magnetic eld, from the Biot-Savart law, at point B is

| ~B| =µ0

4π

|q~v × r|

|~r|2

=µ0

4π

|q||~v||r| sin θ

|~r|2

=µ0

4π

|q||~v|(1) sin(90)

|~r|2

=µ0

4π

|q||~v|

|~r|2

= (1 × 10−7)(1.6 × 10−19)(1 × 104)

(0.5 × 10−6)2

= 6.4 × 10−6 T

To get the direction, use the right-hand rule. The velocity vector rotates counterclockwise into the positionvector; therefore, the magnetic eld is in the +z direction.

Figure 6: Rotate the velocity vector counterclockwise; the magnetic eld at point B thus points in the+z-direction.

The magnetic eld at point B is

~Bat point B = < 0, 0, 6.40 × 10−10 > T

Magnetic Field at points D and E. The position vector to point D and the position vector to point Eare parallel (or antiparallel) to the velocity vector. As a result, the angle between the position vector is 0

(or 180).

Figure 7: Cross product is zero for points E and D.

Calculating the magnitude of the magnetic eld from the Biot-Savart law for point D gives

| ~B| =µ0

4π

|q~v × r|

|~r|2

=µ0

4π

|q||~v||r| sin θ

|~r|2

=µ0

4π

|q||~v|(1) sin(0)

|~r|2

| ~Bat point D| = 0

The magnitude of the magnetic eld at point E will also be zero because sin(180) = 0.

Magnetic Field at point F, G, and H. These points are symmetric with points A, B, and C. That is,points F, G, and H are are at the same distances from the particle as points A, B, and C, respectively.However, when using the right-hand rule for each point, the velocity vector must be rotates clockwise to theposition vector, such as for point F shown below.

Figure 8: Rotate the velocity vector clockwise; the magnetic eld points in the -z direction.

As a result, the magnetic eld vector points into the plane, in the −z direction. The magnetic elds at pointsF, G, and H will have the same magnitudes as the magnetic elds at points A, B, and C, respectively, butwill be in the opposite direction. Therefore,

~Bat point F = < 0, 0,−5.72 × 10−11 > T

~Bat point G = < 0, 0,−6.40 × 10−10 > T

~Bat point H = < 0, 0,−5.72 × 10−11 > T

Check that the answers make sense. The magnetic eld should decrease at greater distances. As a result,the magnitude of the magnetic eld should be least at points A, C, F, and H, and this is indeed the case.Also, note that the magnetic eld circulates" around the direction of motion of the particle, pointing outof the plane on the left side and pointing into the plane on the right side, in accordance with the right-handrule.

Question (2m20001)

Induced emf in a coil due to a changing magnetic eld

A coil of 100 turns and radius 2 cm is in a uniform magnetic eld as shown below. The magnetic eldchanges with time according to the graph.

Figure 1: A coil in a uniform magnetic eld.

(a) What is the emf induced around the coil?

(b) If the coil has a resistance of 10 Ω, what is the current in the coil?

(c) In what direction will current ow in the coil?

(d) What is the direction of the electric eld (i.e. the curly electric eld) that is induced by the changingmagnetic eld, at point P (which is located in a +z plane, out in front of the page)?

Solution

(a) Faraday's Law is:

ǫ = −

dΦ

dt

= −

d

dt(NBA cos θ)

Since the normal to the plane of the coil is parallel to the magnetic eld, θ = 0 and cos θ = 1. Thus,

ǫ = −NAdB

dt

From the graph, you can see that the magnetic eld as a function of time is linear. The equation of a straightline is y = mx + b; therefore, since the intercept is 2 T and the slope is −2

0.1= −20 T/s, the magnetic eld is

Bx = −20t + 2. You can check this by substituting t = 0.1 s and calculating Bx which comes out to zero, asexpected from the graph.

Applying Faraday's Law gives:

ǫ = −NAdB

dt

= −(100)π(0.02)2d

dt(−20t + 2)

= (100)π(0.02)2(−20) V

= 2.5V

(b) To calculate the current through the coil, apply Ohm's law. The voltage around the coil is equal to theemf.

∆V = IR

I =∆V

R

=2.5 V

10 Ω= 0.25 A

(c) Because the emf and the current are positive, this means that from the perspective of the normal to thecoil, the current ows counterclockwise around the loop. Or consider Lenz's Law, the induced electric eldcurls around −∆ ~B. The magnetic eld at the surface of the coil is in the +x direction. It is decreasing (as

shown by the graph), thus ∆ ~B is opposite ~B and points to the left, in the -x direction. As a result, −∆ ~B

is in the +x direction. Point your thumb in the direction of −∆ ~B and your ngers curl around the coil,pointing out of the page at the top of the coil and into the page at the bottom of the coil, as shown below.

Figure 2: Direction of induced current in the coil.

(d) Point P is in a +z plane. The electric eld within the coil curls around the magnetic eld in the direction

given by the right-hand rule. Point your thumb in the direction of −∆ ~B and your ngers curl around thecoil, pointing out of the page at the top of the coil and into the page at the bottom of the coil, as shownbelow.

Figure 3: Direction of the electric eld at point P

Note that the curly electric eld exists at point P, whether or not their is a coil present. The changingmagnetic eld induces an electric eld regardless of the presence of the coil. Having a coil means that theelectric eld will exert a force on electrons in the coil, which causes current to ow.