magnetism

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SF027 1

A group ofA group of

phenomenaphenomena

associated withassociated with

magnetic field.magnetic field.

UNIT 6: MAGNETISMUNIT 6: MAGNETISM

SF027 2

Definition is defined as the region around a magnet wherethe region around a magnet where

a magnetic force can be experienced.a magnetic force can be experienced.

A stationary electric charge is surrounded by an electric field only.

When an electric charge moves, it is surrounded by an electric fieldand a magnetic field. The motion of the electric charge producesmotion of the electric charge producesthe magnetic fieldthe magnetic field.

Magnetic field has two poles, called north (N)north (N) and south (S)south (S). Thismagnetic poles are always found in pairsfound in pairs whereas a single magneticpole has never been found.

Like poles (NLike poles (N--N or SN or S--S) repelS) repel each other.

Opposite poles (NOpposite poles (N--S) attractS) attract each other.

6.1.1 Magnetic field lines

Magnetic field lines are used to represent a magnetic field.

By convention, magnetic field lines leave the north poleleave the north pole and entersentersthe south polethe south pole of a magnet.

Magnetic field lines can be represented by straight lines or curves.The tangent to a curved field line at a point indicates the directionof the magnetic field at that point as shown in figure 6.1.a.

6.1 Magnetic Field

Fig. 6.1aFig. 6.1a

direction of magneticdirection of magnetic

field at point P.field at point P.PP

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Magnetic field can be represented by crossescrosses of by dotted circlesdotted circlesas shown in figures 6.1b and 6.1c.

A uniform field is represented by parallel lines of force. This means thatthe number of lines passing perpendicularly through unit area at allcross-sections in a magnetic field are the same as shown in figure6.1d.

A non-uniform field is represented by non-parallel lines. The number ofmagnetic field lines varies at different unit cross-sections as shown infigure 6.1e.

Fig. 6.1b : magnetic field linesFig. 6.1b : magnetic field lines

enterenterthe page perpendicularlythe page perpendicularly

XX XX XX XX

XX XX XX XX

XX XX XX XX

Fig. 6.1c : magnetic field linesFig. 6.1c : magnetic field lines

leaveleave the page perpendicularlythe page perpendicularly

unit crossunit cross--sectional areasectional area

Fig. 6.1dFig. 6.1d

SF027 4

The number of lines per unit crossnumber of lines per unit cross--sectional area is proportionalsectional area is proportionalto the magnitude of the magnetic fieldto the magnitude of the magnetic field.

Magnetic field lines do not intersectdo not intersect one another.

6.1.2 Magnetic field lines Pattern

The pattern of the magnetic field lines can be determined by using twomethods.

Using compass needlesUsing compass needles (shown in figure 6.1f)

stronger field instronger field in AA11

Fig. 6.1eFig. 6.1e

AA11

AA22

weaker field inweaker field in AA22

Fig. 6.1f : plotting a magneticFig. 6.1f : plotting a magnetic

field line of a bar magnet.field line of a bar magnet.

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SF027 5

Using sprinkling iron filings on paperUsing sprinkling iron filings on paper (shown in figure 6.1g).

Figure below shows the various pattern of magnetic field lines aroundthe magnets.

Fig. 6.1g : Thin iron filing indicate the magneticFig. 6.1g : Thin iron filing indicate the magnetic

field lines around a bar magnet.field lines around a bar magnet.

a. Bar magneta. Bar magnet

SF027 6

b. Horseshoe or U magnetb. Horseshoe or U magnet

c. Two bar magnets (unlike pole)c. Two bar magnets (unlike pole) -- attractiveattractive

d. Two bar magnets (like poles)d. Two bar magnets (like poles) -- repulsiverepulsive

Neutral point (Neutral point (point wherepoint where

the resultant magneticthe resultant magnetic

force is zeroforce is zero).).

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6.1.3 Magnetization of a Soft Iron

There are two methods to magnetized the soft iron.

Using the permanent magnet.Using the permanent magnet.

One permanent magnet

A permanent magnet is bring near to the soft iron andtouching the surface of the soft iron by following the path in

the figure 6.1h.

This method is called induced magnetizationinduced magnetization.

The arrowsarrows in the soft iron represent the magnetization

direction with the arrowhead being the north pole andarrow tail being the south pole. It is also known as

domainsdomains ( the tiny magnetized region because of spinmagnetic moment of the electron).

Fig. 6.1hFig. 6.1h

NN SS

SF027 8

In an unmagnetized piece of soft iron, these domains arearranged randomly but it is aligned in one direction whenthe soft iron becomes magnetized.

The soft iron becomes a temporary magnet with its southpole facing the north pole of the permanent magnet andvise versa as shown in figure 6.1h.

Two permanent magnets

Bring and touch the first magnet to one end of the soft ironand another end with the second magnet as shown infigure 6.1i and 6.1j.

Fig. 6.1iFig. 6.1i NN SS

Fig. 6.1jFig. 6.1j NN NN SSSS

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Using Electrical circuit.Using Electrical circuit.

A soft iron is placed inside a solenoid (a long coil of wireconsisting of many loops of wire) that is connected to thepower supply as shown in figure 6.1k.

When the switch S is closed, the currentIflows in the solenoidand produces magnetic field.

The directions of the fields associated with the solenoid can befound by viewing the current flows in the solenoid fromviewing the current flows in the solenoid from

both endboth end as shown if figure 6.1k or applying the right handright handgrip rulegrip rule below.

NN SS

II II

Fig. 6.1kFig. 6.1k

SSNN

Switch,Switch, SS

ThumbThumb north polenorth pole

Other fingersOther fingers direction of current in solenoid.direction of current in solenoid.

Important

CurrentCurrent --

anticlockwiseanticlockwiseCurrentCurrent -- clockwiseclockwise

SF027 10

a.a.

NNSS

Other examples:

Note :

If you dropdrop a permanent magnet on the floor or strikestrike it with ahammer, you may jar the domains into randomnessdomains into randomness. The magnetcan thus lose some or alllose some or all of its magnetism.

HeatingHeating a magnet too can cause a loss of magnetism.

II II

b.b.

II II

SS

NN

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The permanent magnet also can be demagnetized by placing itplacing itinside a solenoid that connected to an alternating sourceinside a solenoid that connected to an alternating source.

Example 1 : (exercise)

Sketch the magnetic field lines pattern around the bar magnets asshown in figures below.

a.

6.2.1 Magnetic flux density,B Definition is defined as the magnetic flux per unit area across anthe magnetic flux per unit area across an

area at right angles to the magnetic fieldarea at right angles to the magnetic field.

Mathematically,

It also known as magnetic inductionmagnetic induction (magnetic field intensitymagnetic field intensity)

6.2 Magnetic Flux Density and Magnetic Flux

=A

B Bwhere

fluxmagnetic:Bfieldmagneticthetoanglesrightatarea:A

b.

SF027 12

It is a vector quantity and its direction follows the direction of themagnetic field.

Its unit is weber per metre squared (Wb m-2) or tesla (T).

Unit conversion :

6.2.2 Magnetic flux, B Magnetic flux of a uniform magnetic fielduniform magnetic field,

Definition is defined as the scalar product between the magneticthe scalar product between the magnetic

flux density, B with the vector of the surface area,flux density, B with the vector of the surface area, A.A.

Mathematically,

It is a scalar quantity and its unit is weber (Wb).

Consider a uniform magnetic field B passing through a surface area Aas shown in figures 6.2a and 6.2b.

From the fig. 6.2a, =0, thus

BAABB cos== rr

)(Ggauss10Wb m1T1 42- ==

A.andBofdirectionebetween thangle:where fluxmagnetic:B

o0BAB cos=BAB=

Br

r

area, A

Fig. 6.2aFig. 6.2a

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The magnetic flux is proportional to the number of field linesmagnetic flux is proportional to the number of field lines

passing through the area.passing through the area.

Let us consider an element of area dA on an arbitrarily shaped surfaceas shown in figure 6.2c.

From the fig. 6.2b, the angle

betweenB andA is , thus

Note that the direction of vector

A is always perpendicularperpendicular

(normal) to the surface area, A.

BAB cos=

area, A

Fig. 6.2bFig. 6.2b

r

Br

Fig. 6.2cFig. 6.2c

Br

Adr

If the magnetic field at this element is

, the magnetic flux through the element

is

Where is a vector that is

perpendicular to the surface and has a

magnitude equal to the area dA.

Therefore, the total magnetic flux

through the surface is given by

AdBrr

Adr

Br

== BdAAdBB cosrr

SF027 14

Br

o30 Example 2 :

Figure above shows a flat surface with an area of 3.0 cm2 is placed in

the uniform magnetic field. The plane surface makes an angle 30 withthe direction of the magnetic field. If the magnetic flux through the

surface is 0.90 mWb, calculate the magnitude of the magnetic field.

Solution:A=3.0x10-4 m2, B

=0.90x10-3 Wb

T06B .=

Br

r

o30 o30

From the figure, the angle betweenB

andA is

By applying the equation of magnetic

flux for uniform B hence

ooo 603090 ==

BAB cos=

AB B

cos

=

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The law can be stated as :

The magnitude of magnetic flux densityThe magnitude of magnetic flux density dBdB at a point P which is aat a point P which is a

distancedistance rrfrom a very short lengthfrom a very short length dldlof a conductor carrying aof a conductor carrying a

currentcurrent IIis given byis given by

wherewhere is the angle between the short length and the line joiningis the angle between the short length and the line joining

it to point P.it to point P.

This law can be summarized by using the diagram shown in figure 6.3a.

6.3 Biot-Savart Law

2r

IdldB

sin

Fig. 6.3aFig. 6.3a

PPI

I

ldr

r

into the paperinto the paperBdr

r The direction of dB is given by theright hand grip rule (figure 6.3b).

(6.3a)(6.3a)

ThumbThumb direction of currentdirection of current

Other fingersOther fingers direction of magnetic field.direction of magnetic field.

Important Fig. 6.3bFig. 6.3b

SF027 16

If the medium around the conductor is vacuum or airvacuum or airthen the equation6.3a can be written as

In vector form, using the unit vector , we have

To find the total magnetic fieldB at any point in space due to the currentin a complete circuit, eq. 6.3b needs to be integrated over all segmentthat carry current, symbolically

2

0

r

Idl

4

dB

sin

=

r

(6.3b)(6.3b) Magnitude formMagnitude form

2

0

r

rIdl

4

dB

= (6.3c)(6.3c) Vector formVector form

spacefreeoftypermeabili:0

where elementcurrent:Idl

17 AmT10x4 =

ldr

= 20

r

Idl

4

B

sin(6.3d)(6.3d) Magnitude formMagnitude form

or

=

2

0

r

rIdl

4

B

(6.3e)(6.3e) Vector formVector form

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SF027 17

Amperes law is an alternative to the Biot-Savart law.

For conductors in a vacuum or air the law states

This law can be summarized by using the diagram shown in figure 6.4a.

6.4 Amperes Law

enc0IldB = rr

(6.4a)(6.4a)or enc0IBdl = coswhere

pathclosedaarounddensityfluxtheofintegralline:cos Bdl

ldBrr

andofrectiondiebetween thangle:

It is the sum of the termsBdl cos for every

very short length dlof the closed path.densityfluxmagnetictheofmagnitude:B

pathby theenclosedcurrent:encI

ldr

Br

Part of closed pathPart of closed path

Fig. 6.4aFig. 6.4a

clockwiseclockwiseoror

anticlockwiseanticlockwise

SF027 18

Note :

Both Biot-Savart and Amperes law can be used to determine themagnetic field of a long straight conductor, a circular coil and a longthin solenoid.

When a current flows in a conductor wire or coil, the magnetic field willbe produced.

The direction of magnetic fielddirection of magnetic field around the wire or coil can bedetermined by using the right hand grip ruleright hand grip rule as shown in figure 6.3b.

6.5.1 Magnetic field of a long straight conductor (wire) carrying current The magnetic field lines pattern around a straight conductor carrying

current is shown in figures 6.5a and 6.5b.

6.5 Magnetic field produced by the electrical

current

or

Fig. 6.5aFig. 6.5a

I

Br

B

r

I

Br

Br

I

Current out of the pageCurrent out of the page

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SF027 19

The equation of magnetic flux density at any point from a long straightwire carrying current can be determined by using Biot-Savart law.

Consider a straight conductor with length 2a carrying a currentIasshown in figure 6.5c.

or

I

Fig. 6.5bFig. 6.5b

Br

Br

ICurrent into the pageCurrent into the page

Br

B

r

I

XX

XX

00

aa

--aa

PPy

I

ldr

22 yxr +=

2a2a

Fig. 6.5cFig. 6.5c

2

0

r

Idy

4

dB

sin

=

XX

Bd

r

To find the field dB at point P caused bythe element of the conductor of length

dl=dy shown in figure 6.5c, firstly we useBiot-Savart law.

SF027 20

From the figure 6.5c,

Therefore dB at point P produced by element dl is given by

To find the magnitude of the total magnetic flux densityB at point Psupplied by the whole conductor, integrate eq. 6.5a from -a to a then

22 yxr +=22 yx

x

+== )sin(sin and

22

22

0

yx

yx

xdy

4

IdB

+

+=

( ) 23

22

0

yx

xdy

4

IdB

/

+

=

(6.5a)(6.5a)

( ) +=

a

a 2322

0

yx

xdy

4

IdB

/

22

0

axx

a2

4

IB

+=

(6.5b)(6.5b)

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SF027 21

I IXX

SS

NN

When the length of the conductor is infinitely long then a is muchlarger thanx so that

Therefore theB at point P produced by whole conductor is

6.5.2 Magnetic field of a Circular Shaped Coil

The magnetic field lines pattern around a circular shaped coil carryingcurrent is shown in figures 6.5d.

.aax 22 +

x2

IB 0

=where

conductorthefrompointanyofdistance:x

or

R

I

I I Fig. 6.5dFig. 6.5d

SS

NN

SF027 22

The equation of magnetic flux density at any point from a circularshaped coil carrying current can be determined by using Biot-Savart law.

Consider a circular shaped conductor with radius R that carries a

currentIas shown in figure 6.5e.

To find the field dB at a point P on the axis of the circular (loop) atdistancex from the centre O, we need to apply Biot-Savart law.

z

y

PP

ydB Bdr

ldr

I

I

I

OO

R

x

22 Rxr +=

xdB

Fig. 6.5eFig. 6.5e

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SF027 23

From the figure 6.5e, dland rare perpendicular and the direction ofthe field dB caused by the element dllies in theyz-plane.

Since

dlis given by

The components of the vector dB arex-component :

22 Rxr +=

( )220

Rx

dl

4

IdB

+=

, the magnitude dB of the field due to element

2

0

r

Idl

4

dB

sin

= o90=and

( ) 2122 RxR

/cos

+== cosdBdBx and

( ) 23220

x

Rxdl

4IRdB /

+=

( )( ) 2122220

x

Rx

R

Rx

dl

4

IdB

/++

=

SF027 24

y-component :

By symmetry, when the componentcomponent dBdByy are summed over allelements around the loop, the resultant component is zerozero. Itbecause the current in any element on the one side of the loopcurrent in any element on the one side of the loop

sets up a dBy that cancelscancels the dBy set up by the current throughthe current throughthe element diametrically opposite itthe element diametrically opposite it. (figure 6.5f)

Therefore the resultant field at point P must along the x axisresultant field at point P must along the x axis and

is given by

( ) 2122 Rxx

/sin

+== sindBdBy and

( ) 23220

y

Rx

dl

4

IxdB

/+

=

I

IXX

y

PP

Bdr

Bdr

OO

Fig. 6.5fFig. 6.5f

( ) +=

2322

0x

Rx

dl

4

IRdB

/

( ) += dl

Rx4

IRB

2322

0x /

R2dl =andCircumference ofCircumference of

the circular coilthe circular coil

( ) 23222

0x

Rx2

IRB

/+

=

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SF027 25

In general ,

The magnitude of magnetic fieldmagnetic fieldBB at point O (centre of thecentre of the

circular coil or loopcircular coil or loop) ,x=0 is given by

or

On the axis ofOn the axis of

a circular coila circular coil( ) 23222

0

Rx2

IRB

/+

=

On the axis of NOn the axis of N

circular coilscircular coils( ) 23222

0

Rx2

NIR

B /+=

( ) 2322

0

R02

IRB

/+

=R2

NIB 0= At the centre of NAt the centre of N

circular coilscircular coils

wherecoilcirculartheofradius:R

(loops)coilsofnumber:N

spacefreeoftypermeabili:0 17 AmT10x4 = current:I

coiltheofcentreapoint withabetweendistance:x

SF027 26

Fig. 6.5gFig. 6.5g

I Ior

6.5.3 Magnetic field of a Solenoid

The magnetic field lines pattern around a solenoid carrying current isshown in figures 6.5g.

SSNN

I

IXX XX XX XX

I

I

I

I

I

I

SSNN

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SF027 27

Fig. 6.5hFig. 6.5h

The equation of magnetic flux density around a solenoid carrying

currentI can be determined by using Amperes law. Consider a very long solenoid with closely packed coils, the field is

nearly uniform and parallel to the solenoid axes within the entirecross section, as shown in figure 6.5h.

To find the magnetic field inside the solenoid and at the centre, wechoose and draw the rectangle closed path abcd as shown in figure6.5h (clockwise) for applying Amperes law.

By considering this path consists of four segment : ab, bc, cd andda, then Amperes law becomes

)()( smallvery0outsideB =r

Br

d

ab

c

l

ldr

ldr

ldr

ldr

SF027 28

The magnetic flux density at the end of the solenoidat the end of the solenoid is given by

nIB 0=

l

NIB 0=

NIldldldld 0a

d

d

c

c

b

b

a=+++

rrrrrrrr

BBBB

oror

where

n

l

N=

encirclespathour(loops)coilsofnumber:N

NI90dl0dl90dl 0a

d

d

c

c

b=++ ooo coscoscos BBB

NIdl 0d

c=B ldl

d

c=and

and

where lengthunitpercoilsofnumber:n

= enc0Ildrr

B NIIenc=and

At the centre/ midAt the centre/ mid--point/ inside of Npoint/ inside of N

turn solenoidturn solenoid

( )nI2

1B 0=

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Example 3 :

Two long straight wires are placed parallel to each other and carrying

the same currentI. Sketch the magnetic field lines pattern around bothwires

a. when the currents are in the same direction.

b. when the currents are in opposite direction.

Solution:

a.

I I

I

I Ioror

SF027 30

b.

oror

I

I

I XX

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SF027 31

Example 4 :

A long wire (X) carrying a current of 30 A is placed parallel to and 3.0

cm away from a similar wire (Y) carrying a current of 6.0 A.

a. Find the magnitude and direction of the magnetic flux density midway

between the wires :

i. when the current are in the same direction.

ii. when they are in opposite direction.

b. When the currents are in the same direction there is a point

somewhere between X and Y at which the magnetic flux density is

zero. How far from X is this point ?

(Given 0 = 4 x 10-7 H m-1)

Solution:IX=30 A, IY=6.0 A, d=3.0x10-2 m

a. i.

XBr

YBr

X

IYI

Xr Yr

XBr

YBr

XI YI

d

X

r Y

r oror

m10x512

drr 2YX .

===

SF027 32

By using the equation of magnetic field at any point near thestraight wire, then at point A

Magnitude of BX :

Magnitude of BY :

Therefore the total magnetic flux density at point A is

a.ii.

T10x23B 4A . =

YXA BBBrrr

+=

X

X0X

r2

IB

=

Direction : into the pageDirection : into the page

T10x04B 4X . =

YXA BBB =

(upwards)(upwards)

Y

Y0Y

r2

IB

=

Direction : out of pageDirection : out of page

T10x08B 5Y . =

(downwards)(downwards)

Direction : into the pageDirection : into the page(upwards)(upwards)

Into the pageInto the page + (positive)+ (positive)Out of pageOut of page -- (negative)(negative)

Sign convention of B

XI YI

d

Xr Yr

XBr

YBr

XBr

YBr

oror

m10x512

drr 2YX .

===

XI YIXr Yr XX

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SF027 33

By using the equation of magnetic field at any point near the

straight wire, then at point A

Magnitude of BX :

Magnitude of BY :

Therefore the total magnetic flux density at point A is

b.

T10x84B 4A . =

YXA BBBrrr

+=

X

X0X

r2

IB

=


T10x04B 4X . =

YXA BBB +=

(upwards)(upwards)

Y

Y0Y

r2

IB

= T10x08B 5Y . =



XI YI

d

X

rY

r

XBr

YBr

YBr

oror

rrX=

XBr

X

IYI

Xr Yr

rdry =

SF027 34

By using the equation of magnetic field at any point near the

straight wire, then at point A

Magnitude of BX :

Magnitude of BY :

Since the total magnetic flux density at point A is zero, hence

m10x52r 2. =

YXA BBBrrr

+=

r2

IB X0X= Direction : into the pageDirection : into the page

YX BB0 =

(upwards)(upwards)

r)(d2

IB Y0Y

= Direction : out of pageDirection : out of page(downwards)(downwards)

r)(d2I

r2I Y0X0

=

YX

X

II

dIr

+=

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SF027 35

Example 5 :

a. A 2000 turns solenoid of length 40 cm and resistance 16 is

connected to a 20 V supply. Find the magnetic flux density at the mid

point of the axis of the solenoid.

b. A solenoid 1.30 m long and 2.60 cm in diameter carries a current of

18.0 A. The magnetic field inside the solenoid is 23.0 mT. Find the

length of the wire forming the solenoid. (Halliday,Resnick&Walker,p.708,no.42)

(Given 0=4 x 10-7 H m-1)

Solution:

a. Given N=2000 turns, l=40x10-2 m, R=16 , V=20 VBy applying the equation of magnetic flux density at the centre of the

solenoid, thus

T10x97B 3. =

l

NIB 0=

R

VI=

lR

NVB 0=

and

SF027 36

b. Given l=1.30 m, d=2.60x10-2 m, I=18.0 A, B=23.0x10-3 TBy applying the equation of magnetic flux density inside the solenoid,

thus

Since the shaped of each coil forms the solenoid is circle, then the

circumference of one coil is

Therefore the length of the wire forming the solenoid,L is

Example 6 :

A closely wound circular coil a diameter of 4.00 cm has 600 turns andcarries a current of 0.500 A. Determine the magnitude of the magneticfield

a. at the centre of the coil.

b. at a point on the axis of the coil 8.00 cm from its centre.

(Given 0=4 x 10-7 H m-1) (Young&Freedman,p.1098,no.28.28)

m108L =

l

NIB 0=

turns1322N =I

BlN

0

=

dncecircumfere =

m10x178ncecircumfere

2

= .

)( ncecircumfereNL=

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SF027 37

Solution:N=600 turns, d=4.00x10-2 m, I=0.500 Aa. By using the equation of magnetic flux density at the centre of the

coil, thus

b. Givenx=8.00x10-2 m.By applying the equation of magnetic flux density at a point of

distancex on the axis of the circular coil from its centre, thus

T10x439B 3. =

R2

NIB 0=

d

NI

B

0

=

2

dR=and

( ) 23222

0

Rx2

NIRB

/+

=2

dR=and

232

2

2

0

4dx8

NIdB

/

+

=

T10x351B 4 . =

SF027 38

Example 7 :

The segment of wire in figures a and b carry a current ofI=5.00 A,

where the radius of the circular arc isR=3.00 cm.

For each figure, find the magnetic flux density at origin (point O)

(Given 0=4 x 10-7 H m-1)

Solution:I=5.00 A, R=3.00x10-2 mFigure a : Sections (1) and (3) are straight line and the angle

between dland the line joining dlto point O is 0or 180.

O

R

I

Fig. aFig. a

OR

I

Fig. b : exerciseFig. b : exercise

O

R

I 1

3

2

ldr

ldr

Hence B1 and B3 at point O is

= 20

1r

0Idl

4

B

osin

0BB 31 ==

= 20

3r

180Idl

4

B

osinand

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SF027 39

Section (2) is a quarter of circular coil hence the B2 at point O is

Therefore the total magnetic flux density at point O is

Figure b :Ans. :5.24 x 10-5 T into the page.

Example 8 :

Four long, parallel power wires each carry 100 A current. A cross

sectional diagram for this wires is a square, 20.0 cm on each side. For

each case in figures a, b and c,

R2

NI

4

1B 02

=

R8

IB 02=

1N=and

T10x622B 52 . =

321O BBBB ++=T10x622B

5

O . =



Fig. aFig. a

XX XX

XXXX

Fig. b :exerciseFig. b :exercise

XX

XX

Fig. c : exerciseFig. c : exercise

XX XX

Use right hand gripUse right hand grip

rulerule

SF027 40

i. sketch the magnetic field lines pattern on the diagram.

ii. calculate the magnetic flux density at the centre of the square.

(Given 0=4 x 10-7 H m-1)

Solution:I1=I2=I3=I4=I=100 A, l=20.0x10-2 m

Figure (a) :

i.

Since r1=r2=r3=r4 andI1=I2=I3=I4 then the magnitude ofB1,B2, B3andB4 at point C are the same and given by

XX XX

XXXX

1r

1I 2I

4I 3I

2r

4r 3rC

XX XX

XXXX

ii.l

l1B

r

2Br

4Br

3Br

2

llrrrrr

22

4321

+=====

m1410r .=

r2

IBBBBB 04321 =====

T10x411B 4. =

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SF027 41

From the diagram, B1 cancel the B3 and B2 cancel the B4, thus

Figure (b) :Ans. : 0

Figure (c) :Ans. : 4.0 x 10-4 T direction : to the left (180)

A stationarystationary electric charge in a magnetic field will not experience anynot experience any

forceforce. But if the charge is movingcharge is moving with a velocity, v in a magnetic field,

B then it will experience a force. This force known as magnetic forcemagnetic force.

The magnitudemagnitude of the magnetic force can be calculated by using the

equation below :

In vector formvector form,

4321C BBBBBrrrrr

+++=0BC=

6.6 Magnetic Force on a Moving Charge

BqvF sin=

( )BvqFrrr

=

where forcemagnetic:F

densityfluxmagnetic:Bchargeaofvelocity:v

chargetheofmagnitude:q

Bvrr

andbetweenangle:

(6.6a)(6.6a)

(6.6b)(6.6b)

SF027 42

Fig. 6.6bFig. 6.6b

Br

vr

Fr

Fig. 6.6aFig. 6.6a

Br

vr

Fr

The direction of the magnetic force can be determined by using theFlemings hand rule.

FlemingFlemings right hands right hand rule : - for negativenegative charge

FlemingFlemings left hands left hand rule : - for positivepositive charge

Example 9 :

Determine the direction of the magnetic force, exerted on a charge in

each problems below.

a. b.

shown in figures

6.6a and 6.6b

ThumbThumb direction ofdirection of ForceForce

First fingerFirst finger direction ofdirection of FieldField

Second fingerSecond finger direction ofdirection of VelocityVelocity..

Important

Fr

+

Br

vr

Br

vr

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SF027 43

c. d. e.

Solution:a. By using Flemings left hand rule,

thus

c. By using Flemings right hand rule,

thus

Br

vr

XX XX XX XX

XX XX XX XX

XX XX XX XX

vr

+

I

vr

+

Br

vr

Fr

(into the(into the

page)page)

b. By using Flemings right hand rule,

thus

Br

vr

Fr

(to the left)(to the left)

Br

vr

XX XX XX XX

XX XX XX XX

XX XX XX XX

Fr


d. Using right hand grip rule to

determine the direction of magnetic

field forms by the currentIon thecharge position. Then apply the

Flemings right hand rule, thus

I

vr

Br

XX XX XX XX

XX XX XXFr


SF027 44

e. Using right hand grip rule to determine the direction of magnetic field

forms by the currentIon the charge position. Then apply the Flemingsleft hand rule, thus


Determine the sign of a charge in each problems below.

a. b.

Ans. : positive charge, positive charge


Determine the direction of the magnetic force exerted on a positivecharge in each figures below when a switch S is closed.

vr

+

I

Fr

(upwards)(upwards)

Br

XXXX

XX

XX

XX

XX

XX

XX

Br

vr

Fr B

r

vrF

r

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SF027 45

a. b.

Ans. :a. into the page, b. out of page

Example 12 :

Calculate the magnitude of the force on a proton travelling 3.0x107 m s-1 in the uniform magnetic flux density of 1.5 Wb m-2, if :

a. the velocity of the proton is perpendicular to the magnetic field.

b. the velocity of the proton makes an angle 50 with the magnetic field.(Given the charge of the proton is +1.60 x 10-19 C)

Solution: v=3x107m s-1, B=1.5 T, q=1.60x10-19 Ca. Given = 90 then by applying the equation of magnetic force on a

moving charge, thus

Switch, SSwitch, S

+ vr

Switch, SSwitch, S

+vr

sinBqvF=N10x27F 12. =

SF027 46

b. Given = 50 then by applying the equation of magnetic force on amoving charge, thus


An electron experiences the greatest force as it travels 2.9 x 106 m s-1 ina magnetic field when it is moving north. The force is upward and ofmagnitude 7.2 x 10-13 N. Find the magnitude and direction of themagnetic field. (Giancolli, pg.705, no.22)

(Given the charge of the electron is -1.60 x 10-19 C)

Ans : 1.6 T to the east.


An electron is moving in a magnetic field. At a particular instant, thespeed of the electron is 3.0 x 106 m s-1. The magnitude of the magneticfield on the electron is 5.0 x 10-13 N and the angle between the velocityof the electron and the magnetic force is 30. Calculate the magnitudeof the magnetic flux density on the electron in the field.

(Given the charge of the electron is -1.60 x 10-19 C)

Ans : 1.2 T

sinBqvF=N10x55F 12 . =

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SF027 47

Fi . 6.7aFi . 6.7a

Br

I

Fr

When a current-carrying conductor is placed in a magnetic field B, thusa magnetic force will acts on that conductor.

The magnitudemagnitude of the magnetic force exerts on the current-carryingconductor is given by

In vector formvector form,

The direction of the magnetic force can be determined by using the

FlemingFlemings left hand rules left hand rule as shown in figure 6.7a.

6.7 Magnetic Force on a current-carrying

conductor

BILF sin=

( )BLIFrrr

=

where forcemagnetic:Fdensityfluxmagnetictheofmagnitude:B

current:Iconductortheoflength:L

BIr

andofdirectionbetweenangle:

ThumbThumb direction ofdirection of ForceForce

First fingerFirst finger direction ofdirection of FieldField

Second fingerSecond finger direction ofdirection of CurrentCurrent..

Important

(6.7a)(6.7a)

(6.7b)(6.7b)

SF027 48

Note :

It is clear from eq. (6.7a),

the magnetic force on the conductor has its maximum valuewhen the conductor (and therefore the current) and themagnetic field are perpendicular (at right angles) to each other

then =90(shown in figure 6.7b). the magnetic force on the conductor is zero when the

conductor (and therefore the current) is parallel to the magnetic

field then =0(shown in figure 6.7c).

OneOne teslatesla is defined as the magnetic flux density of a fieldas the magnetic flux density of a fieldin which a force of 1in which a force of 1 newtonnewton acts on a 1acts on a 1 metremetre length of alength of a

conductor which carrying a current of 1 ampere and isconductor which carrying a current of 1 ampere and isperpendicular to the field.perpendicular to the field.

o90BILF sinmax=BILF =max

Br

o90=I

Fig. 6.7bFig. 6.7b

o0BILF sin= 0F=

Br

o0=

Fig. 6.7cFig. 6.7c

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SF027 49

Example 15 :

Determine the direction of the magnetic force, exerted on a conductor

carrying current,Iin each problems below.

a. b.

Solution:

For both problems, use Flemings left hand rule :

a.

Br

I

XX XX XX XX

XX XX XX XX

XX XX XX XX Br

I

XX XX XX XX

XX XX XX XX

XX XX XX XX

Br

I

XX XX XX XX

XX XX XX XX

XX XX XX XX B

r I

XX XX XX XX

XX XX XX XX

XX XX XX XX

b.

Fr

Fr


(to the right)(to the right)

SF027 50

Example 16 :

A wire of 20 cm long is placed perpendicular to the magnetic field of

0.40 Wb m-2.

a. Calculate the magnitude of the force on the wire when a current 12 A

is flowing.

b. For the same current in (a), determine the magnitude of the force on

the wire when its length is extended to 30 cm.

c. If the force on the 20 cm wire above is 60 x 10-2 N and the current

flows is 12 A, find the magnitude of magnetic field was supplied.

Solution:L=20x10-2 m, B=0.40 T, =90

a. GivenI = 12 A.By applying the equation of magnetic force on a current-carrying

conductor, thus

b. GivenI = 12 A andL = 30x10-2 m

By applying the equation of magnetic force on a current-carrying

conductor, thus

BILF sin=N960F .=

BILF sin=N41F .=

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SF027 51d

2I

2I

1I

1I

X Y

Fig. 6.8aFig. 6.8a

c. GivenI = 12 A,L = 20x10-2 m , =90andF = 60x10-2NBy applying the equation of magnetic force on a current-carrying

conductor, thus

Consider two identical straight conductors X and Y carrying currentsI1andI2 with lengthL are placed parallel to each other as shown in figure6.8a.

BILF sin=

T250B .=IL

FB

sin=

6.8 Forces between two current-carryingconductors

1Br

2Br

The conductors are in vacuum and

their separation is d. The magnitude of the magnetic flux

density,B1 at point P on conductor Ydue to the current in conductor X isgiven by

Conductor Y carries a currentI2 and in

the magnetic fieldB1 then conductor Ywill experiences a magnetic force,F .

P

d2

I

B

10

1 =Direction : into theDirection : into the

page/paperpage/paper

12Fr

21F

r

Q

SF027 52

The magnitude ofF12 is given by

The magnitude ofF21 is given by

Conclusion :

sinLIBF 2112= o90=and

o90LId2

IF 2

1012 sin

=

Direction : to the left (towards X)Direction : to the left (towards X)LId2

IF 2

1012

=

Use FlemingUse Flemings lefts left

hand rulehand rule

sinLIBF 1221= o90=and

o90LId2

IF 12012 sin

=

Direction : to the right (towards Y)Direction : to the right (towards Y)LId2

IF 1

2012

=

d2

LIIFFF 2102112

===

rr

The properties of this force : Attractive forceThe properties of this force : Attractive force

(6.8a)(6.8a)

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SF027 53

If the direction of current in conductor Y is change to upside down as

shown in figure 6.8b.

Note :

The currents are in the same directionsame direction 2 conductors attractattract each

other.

The currents are in opposite directionopposite direction 2 conductors repelrepel eachother.

The magnitude ofF12 andF21 canbe determined by using eq. 6.8a andits direction by applying Flemings lefthand rule.

Conclusion :

The properties of this force :The properties of this force :

Repulsive forceRepulsive force

2I

2I

1I

1I d

X Y

Fig. 6.8bFig. 6.8b

21Fr

12F

r1Br

P

2Br

Q

SF027 54

Example 17 :

Two very long parallel wires are placed 2.0 cm apart in air. Both wires

carry a current of 8.0 A and 10 A respectively. Find

a. the magnitude of the magnetic force in newton, on each metre length

of wire.

b. the magnetic flux density at point P, midway between the wires if the

currents (exercise)

i. in the same direction.

ii. In opposite direction.

(Given 0=4 x 10-7 H m-1)

Solution:I1=8.0 A, I2=10 A, d=2.0x10-2 ma. GivenL = 1.0 m

By applying the equation of force for two parallel current-carrying

conductors, thus

b. Ans. : 0.4 x 10-4 T out of page (downwards), 3.6 x10-4 T into the page

(upwards)

Hint : Example 4a.

d2

LIIF 210=

N10x08F 4. =

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SF027 55


Two long, straight, parallel wires in a vacuum are 0.25 m apart.

i. The wires each carry a current of 2.40 A in the same direction.

Calculate the force between the wires per metre of their length. Draw

a sketch showing clearly the direction of the force on each wire.

ii. The current in one of the wires is reduced to 0.64 A. Calculate the

current needed in the second wire to maintain the same force

between the wires per metre of their length as in (i).

(Given 0=4 x 10-7 H m-1)

Ans. : 4.6 N, 9.0 A

6.9.1 Definition of Ampere

From the eq. (6.8a), if two long, straight, parallel conductors , 1.0 m

apart in vacuum carry equal 1.0 A currents hence the force per unit

length that each conductor exerts on the other is

6.9 Definition of Ampere and Ampere Balance

d2II

LF 210

= )0.1(2)0.1)(0.1)(10

x4(

LF

7

=

17 mN10x02L

F = .

SF027 56

m

Fig. 6.9aFig. 6.9a

DD

QQ

AA

EE

FF

BB

CC

GG

HH

PP

I

d

l

The ampereThe ampere is defined as the constant current that, when it isthe constant current that, when it isflowing in each of two infinitely long, straight, parallel conduflowing in each of two infinitely long, straight, parallel conductorsctors

which have negligible of cross sectional areas and are 1.0which have negligible of cross sectional areas and are 1.0 metremetre

apart in vacuum, would produce a force per unit lengthapart in vacuum, would produce a force per unit length

between the conductors of 2.0 x 10between the conductors of 2.0 x 10--77 N mN m--11..

6.9.2 Ampere (current) Balance

An instrument used to measure a current absolutely, on the basis of the

definition of the ampere (due to the forces between two long, straight,

parallel conductors).

Figure 6.9a shows a schematic diagram for a current balance where the

current can be determined by measuring the force between two

conductors carrying the same current.

Fr

gmW rr

=

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SF027 57

The plane ABCD and light rod PQ are adjusted so that the plane is

initially horizontal.

Conductors BC and EF are placed parallel to each other, equal in length

land separated by a distance d.

When a current I flows, conductors BC and EF repels, hence the plane

ABCD is unbalanced.

The mass m needed to restore the plane ABCD in torque equilibriumwhere Torque due to the mass = Torque due to the force on BCTorque due to the mass = Torque due to the force on BC

PQAB ll =and

ABPQ FlWl =

d2

lIImg 210

= III 21 ==and

d2

lImg

2

0

=

then

lmgd2I

0=

where )(onacceleratilgravitiona: 2-m s81.9g

SF027 58

6.10.1 A charged particle moves perpendicular to the magnetic field.

Consider a charged particle moving in a uniform magnetic field with its

velocity perpendicular to the magnetic field.

As the particle enters the region, it will experiences a magnetic force

which the force is perpendicular to the velocity of the particle. Hence the

direction of its velocity changes but the magnetic force remains

perpendicular to the velocity.

This magnetic force,FB makes the path of the particle is a circular asshown in figures 6.10a, 6.10b, 6.10c and 6.10d.

6.10 Motion of a Charged Particle in a

Uniform Magnetic Field

+ vr

vr

+BFr

+

vr

BFr

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

Fig. 6.10aFig. 6.10a

+ vr

vr

+BFr

+

vr

BFr

Fig. 6.10bFig. 6.10b

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SF027 59

Since the path is circle therefore the magnetic forceFB contributes the

centripetal forceFc (net force) in this motion. Thus

vr

vr

BFr

vr

BFr

vr

vr

BFr

vr

BFr

Fig. 6.10dFig. 6.10dFig. 6.10cFig. 6.10c

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

cB FF =

r

mvBqv

2

=sin o90=and

Bq

mvr=

where

particlechargedtheofmass:mvelocitytheofmagnitude:v

pathcirculartheofradius:rparticlechargedtheofmagnitude:q

SF027 60

The period of the circular motion, Tmakes by the particle is given by

Since

6.10.2 A charged particle moves not perpendicular to the magnetic field.

If the direction of the initial velocity is not perpendicular to the uniformmagnetic field, the velocity component parallel to the field is constantbecause there is no force parallel to the field.

Therefore the particle moves in a helix pathhelix path as shown in figure 6.10e.

rv=

v

r2T

=

T

2=and

Bq

m2T

=

or

f

1

T=

Bq

mvr=and

thus the frequency of the circular motion makes by theparticle is

m2

Bqf

=

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SF027 61

The radius of the helix, ris given by Example 19 :

An electron at point A in figure below has a speed v0 of 1.41 x 106 m s-1.

Given e = 1.60 x10-19 C

me= 9.11 x 10-31 kg

vr

Br

z

y

xFig. 6.10eFig. 6.10e

//vvv rrr

+=

Bq

mvr =

vr

//vr

where

Bvrr

thelar toperpendicucomponentvelocity:Bvrr

thetoparallelcomponentvelocity://

0v

Bcm010 .

Find

a. the magnitude and direction of the

magnetic field that will cause the electron

to follow the semicircular path from A to B.b. the time required for the electron to move

from A to B. (Young&Freedman,p.1055,no.27.15)

SF027 62

Solution: v0=1.41x106m s-1, d=10.0x10-2 ma. Since the path makes by the electron is semicircular thus the

magnitudemagnitude of the magnetic field is given by

The directiondirection ofB : electron (use Flemings right hand rule)

b. Since the path is semicircular then the time required for the electron

moves from A to B is half of the period and given by

s. 710x111t =

;=rv0T2

1t= where

Be

vmr 0e=

2

dr=and

ed

vm2B 0e= T10x611B 4 . =

0v

BFr Into the page.Into the page.

T

2= and

2

dr=

0v

dT

=

0v2

dt

=

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SF027 63

6.11.1 Lorentz Force

Definition is defined as the total force acting on a chargethe total force acting on a charge qq movingmoving

with a velocitywith a velocity vv in the presence of both an electric fieldin the presence of both an electric field

EEand a magnetic fieldand a magnetic field BB. Its formula is given by

It also known as electromagnetic forceelectromagnetic force.

6.11.2 Determination of q/m

The value of q/m is constant for any charged particle.

Consider a positive charged particle with mass m, charge q and speedv enters a region of space where the electric and magnetic fields areperpendicular to the particles velocity and to each other as shown infigure 6.11a.

6.11 Lorentz Force and Determination of q/m

BE FFFrrr

+= EqFErr

=where ( )BvqFBrrr

=and

)( BvEqFrrrr

+= where forceLorentz:Fr

SF027 64

Er

The charged particle will experiences the electric forceFEis downwards

with magnitude qEand the magnetic forceFB is upwards with

magnitudeBqv (fig. 6.11a). If the particle travels in a straight line with constant velocity hence the

electric and magnetic forces are equal in magnitude. Therefore

Only particles with speeds equal to E/B can pass through without beingdeflected by the fields. Eq. (6.11a) also works for electron or othernegative charged particles.

+ ++ + + ++++ ++ + + ++++ +


XX XX XX XX XX XX

XX XX XX XX XX XX

XX XX XX XX XX XX

XX XX XX XX XX XX

Br

vr

+ vr

+ vr

+

BFr

EFr

BE FFFrrr

+= 0F=r

and

qEBqv=

B

Ev= (6.11a)(6.11a)

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SF027 65

If no electric field in fig. 6.11a, then the particle will makes a circular pathwhere

In the J.J. Thomsons experiment , the speed v of electron with mass m

is determined by an accelerating potential difference Vbetween twoplates where

Therefore by substituting eq. (6.11c) into eq. (6.11a) then the value ofe/m is given by

CB FF =

rB

v

m

q= and

r

mvBqv

2

=

B

Ev=

(6.11b)(6.11b)2rB

E

m

q=therefore

eVmv2

1 2 =

m

eV2

v=

Kinetic energy of theKinetic energy of the

electronelectron== Electric potentialElectric potential

energyenergy

(6.11c)(6.11c)

SF027 66

From the experiment the value of e/m is 1.758820174 x 1011 C kg-1.

Example 20 :

An electron with kinetic energy of 5.0 keV passes perpendicular through

a uniform magnetic field of 0.40 x 10-3 T. It is found to follow a circular

path. Calculate

a. the radius of the circular path.

b. the time required for the electron to complete one revolution.

(Given 1 eV=1.60x10-19 J, e/m =1.76x1011 C kg-1, me =9.11x10-31 kg)

Solution:B=0.40x10-3 T, K=5.0x103(1.60x10-19)=8.0x10-16Ja. The speed of the electron is

2

B

E

m

eV2

=

(6.11d)(6.11d)2

2

VB2

E

m

e=

2mv2

1K=

m

K2v= 17 sm10x24v = .

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SF027 67

Since the path made by the electron is circular, thus

b. The time required for the electron to complete one revolution is

Example 21: (exercise)

A proton moving in a circular path perpendicular to a constant magnetic

field takes 1.00 s to complete one revolution. Determine the magnitudeof the magnetic field. (Serway & Jewett,pg.921,no.32)

(Given mp =1.67x10-27 kg, charge of the proton, q=1.60 x 10-19 C)

Ans. : 6.56 x 10-2 T

CB FF =

r

mvBev

2

=sin o90=and

Bme

vr

=

m600r .=

=rv

v

r2T=

T

2=and

s10x09T 8

. =

SF027 68

Q

P

b

a


Consider a rectangular coil (loop) of wire with side lengths a and b that

it can turn about axis PQ. The coil is in a magnetic field of flux densityB

and the plane of the coil makes an angle with the direction of the

magnetic field. A currentIis flowing round the coil as shown in figure6.12a.

6.12 Torque on a Coil in a Magnetic Field

BrB

r

Br

Br

BrF

r

Fr

1Fr

I I

I

I1F

r

r

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SF027 69

sin2b

sin2

b

From the figure 6.12a, the forceF1 on the right side of the coil is to the

right andB is perpendicular to the current thus the magnitude of the

forceF1 is given by

Another forceF1 with the same magnitude but opposite direction actson the opposite side of the coil. (fig. 6.12a)

The forcesFact on the side length b where the lines of action of bothforces lies along the PQ. (fig. 6.12a)

Br

Br

Br

1Fr

1Fr

r

Q

2

brotationrotation

rotationrotation

o90BILF1 sin= aL=andIaBF1= (6.12a)(6.12a)

Fig. 6.12b : Side viewFig. 6.12b : Side view

SF027 70

The total forcetotal force on the coil is zerozero but the net torquenet torque is not zeronot zero

because the forcesF1 are not lie along the same line thus the rotationof the coil about an axis PQ is clockwise (fig. 6.12b).

The magnitude of the net torque about the axis PQ ( fig. 6.12b) is given

by

= sinsin2

bF

2

bF 11

IaBF1=

= sin2

bF2 1 and

( )

= sin2

bIaB2

= sinIabB coil)ofarea(Aab=and

= sinIAB90 = oand

or

IAB cos=

(6.12b)(6.12b)

where coilon thetorque:densityfluxmagnetic:B

coilin theflowscurrent:I

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SF027 71

For a coil of N turns, the torque is given by

Note :

The torque is zero when =90or =0 but the magnetic flux ismaximum as shown in figure 6.12c.

The torque is maximum when =0or =90 but the magneticflux is zero as shown in figure 6.12d.

= sinNIAB

or

NIAB cos=

where BArr

andareactorbetween veangle:Br

andcoiltheofplaneebetween thangle:

o0=

o90=Br

r

Fig. 6.12cFig. 6.12c

o0NIAB sin= o90NIAB cos=or0=

but o0BAB cos=

BAB= maximummaximum

(6.12c)(6.12c)

(coils)turnsofnumber:N

SF027 72

If

In vector formvector form is

where is called magnetic momentmagnetic moment or electromagnetic momentelectromagnetic moment. Magnetic moment is a vector quantity.

Its direction can be determined by using right hand gripright hand grip rule.

E.g.

ANIr

r

= then eq. (6.12c) can be written as

o90=

o0=

Br

r

o90NIAB sin= o0NIAB cos=orNIAB= maximummaximum

but o90BAB cos=

0B=Fig. 6.12dFig. 6.12d

= sinB Magnitude formMagnitude form

Brrr =

r

I

IThumbThumb direction of magnetic momentdirection of magnetic moment

Other fingersOther fingers direction of current in thedirection of current in the

coilcoil

Important

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SF027 73

In a radial fieldradial field, the plane of the coilplane of the coil is always parallelalways parallel to the

magnetic fieldmagnetic field for any orientation of the coil about the vertical axis

as shown in figure 6.12e.

Hence the torquetorque on the coil in a radial fieldradial field is always constantconstant

and maximummaximum given by

Radial field is used in moving coil galvanometer.

o0= o90=or

o90NIAB sin= o0NIAB cos=or

NIAB= maximummaximum

SSNNcoilcoilfixed softfixed soft

iron cylinderiron cylinder

Fig. 6.12e : Plan view of moving coil meterFig. 6.12e : Plan view of moving coil meter

radial fieldradial field

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6.13.1 Moving-Coil Galvanometer

A galvanometer consists of a coil of wire suspended in the magnetic

field of a permanent magnet. The coil is rectangular and consists of

many turns of fine wire as shown in figure 6.13a.

6.13 Moving-Coil Galvanometer and DirectCurrent (DC) Motor


When the currentIflows throughthe coil, the magnetic field exerts

a torque on the coil as given by

This torque is opposed by aspring which exerts a torque, sgiven by

The coil and pointer will rotate

only to the point where the spring

torque balances the torque due to

magnetic field, thus

NIAB=

ks=where

constanttorsional:kcoiltheofanglerotation:

radianin

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6.13.2 Direct-Current (DC) Motor

A motor is an instrument that converts electrical energy to mechanical

energy.

A motor works on the same principle as a galvanometer, except that

there is no spring so the coil can rotate continuously in one direction.

When a current flows in the coil, a torque is produced, which causes the

coil PQRS to rotate as shown in figure 6.13b.

s=kNIAB=

NAB

kI=

CC11, C, C22 :: CommutatorsCommutators

BB11, B, B22 : Brushes: Brushes

PQRSPQRS : Rectangular coil: Rectangular coil

SSNNPP

QQRR

SS

CC22CC11BB11 BB22

I I

I IFig. 6.13bFig. 6.13b

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The commutators also rotate with the coil PQRS but the brushes are

stationary with the circuit.

When the coil rotates half revolution (180), each commutator changesits connection to the other brush where C1 touching B2 and C2 touching

B1. This arrangement will cause the direction of the current through the

coil to be reversed after every half revolution and ensures that the

direction of the torque is always the same. Therefore the coil can turn

continuously.

The torque on the coil PQRS of the motor is given by

Example 22:

A 20 turns rectangular coil with sides 6.0 cm x 4.0 cm is placed verticallyin a uniform horizontal magnetic field of magnitude 1.0 T. If the current

flows in the coil is 5.0 A, determine the torque acting on the coil when

the plane of the coil is

a. perpendicular to the field,

b. parallel to the field,

c. at 60 to the field.

= sinNIAB NIAB cos=or

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Solution:N=20 turns, A=24x10-4 m2,B=1.0 T, I=5.0 A

a.

b.

c.

From the figure, =90and =0,thus the torque on the coil is

= sinNIABNIAB cos= or

0=

BrA

r

o90=From the figure, =0and =90,thus the torque on the coil is

= sinNIABNIAB cos= ormN240 .=

Br

Ar

o90=

Br

Ar

o

30=

o60=From the figure, =60and =30,thus the torque on the coil is

= sinNIABNIAB cos= ormN120 .=

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y+

x+

z+

h

Q

d

P

Definition is defined as the production of a potential differencethe production of a potential difference

within a conductor or semiconductor through which awithin a conductor or semiconductor through which a

current flowing when there is a strong transversecurrent flowing when there is a strong transverse

magnetic field.magnetic field.

6.14.1 Explanation of Hall Effect

Consider a flat conductor (such as copper) carrying a currentIin the

direction of +x-axis and is placed in a uniform magnetic fieldB (-z axis)perpendicular to the plane of the conductor as shown in figure 6.14a.

6.14 Hall Effect

+ + + + + + + ++ + + +

Br

Br

dvr

IBFr

EFr


dd: width of the conductor: width of the conductorhh : thickness of the conductor: thickness of the conductor

Er

HV

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In metal the charge carrier is electron. The electrons drift with a drift

velocity vd in the opposite direction of the currentI(shown in figure6.14a).

The magnetic forceFB acts on the electron in the direction upwards(Flemings right hand rule) and cause the electron deflected to the upper

surface (P).

As time passes, more and more electrons will accumulate on the upper

surface (P) and left behind positive charges at the lower surface (Q).

This results in an electric fieldEacting in the direction upwards and the

electrons will experience electric forceFEin the direction downwards.

The electric forceFEwill gradually increase as more electronsaccumulate at the upper surface.

An equilibrium will be reached when the magnitude of the electric force

FEbecomes equal to the magnitude of the magnetic forceFB and thenthe electrons will drift along the x axis without any more deflection.

At this instant, the upper surface (P) will be negative potential and the

lower surface (Q) will be positive potential. Hence a potential difference

will exist and known as Hall potential difference (voltage)Hall potential difference (voltage) VVHH.

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If the flat conductor is replaced with the flat semiconductor and thecharge carrier is positive charge, explain the phenomenon of hall effect

in this flat based on figure 6.14b. (exercise)

6.14.2 Equation of Hall Voltage ,VH According to the figures 6.14a and 6.14b, the Hall potential difference

(voltage) is

When the equilibrium is reached between the electric and magnetic

force then

Fig. 6.14bFig. 6.14b

y+

x+

z+

I+

Br

Br

P

HV

h

Q

d

dd: width of the conductor: width of the conductorhh : thickness of the conductor: thickness of the conductor

EB FF =

EdVH=

qEFE= o90BqvF dB sin=where and

qEBqvd=d

VE H=and

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From the definition of the drift velocity vd,

Therefore eq. 6.14a can be written as

h

BI

nq

1VH

=

dBvV dH=

nAe

Ivd= qe=where and

=d

VqBqv Hd

dhA =

( )qdhnI

vd=

dndhq

IBVH

=

(6.14a)(6.14a)

(6.14b)(6.14b)

where tcoefficienHall:nq1

torsemiconducconductor/flattheofthickness:heunit volumpercarrierschargeofnumber:n

density)carrier(charge

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6.14.3 Uses of the Hall Effect

The Hall effect can be used to

determine the sign of the charge carriers in a conductor /

semiconductor.

determine the density of the charge carriers in a conductor/

semiconductor.

measure the magnetic flux density B that is known as Hall probe.

Example 23:

A current of 3.00 A flows through a piece of metal of width 0.800 cm and

thickness 625 m. The metal is placed in a magnetic field of flux density5.00 T perpendicular to the plane of the metal. If the Hall voltage across

the width of the metal is 24.0 V, determinea. the drift velocity of the electron in the metal.

b. the density of the conduction electron in the metal.

(Given charge of the electron, e= 1.60 x 10-19 C)

Solution:I=3.00 A, d=0.800x10-2 m, h=625x10-6m,

B=5.00 T, VH=24.0x10-6V.

a. When the equilibrium is reached between the electric and magnetic

force thenEB FF =

eEBevd= andd

VE H=

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b. By applying the equation of the Hall voltage, thus

Example 24: (exercise : unit 6.12)

A rectangular coil of 10 cm x 4.0 cm in a galvanometer has 50 turns and

a magnetic flux density of 5.0 x 10-2 T. The resistance of the coil is 40 and a potential difference of 12 V is applied across the galvanometer,

calculate the maximum torque on the coil.

Ans. : 3.0 x 10-3 N m.

Bd

Vv Hd=

14

d sm10x006v = .

ehV

BIn

H

=

h

BI

ne

1VH =

328 m10x306n = electron.

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An electron moving at a steady speed of 0.50 x 106 m s-1 passes

between two flat, parallel metal plates 2.0 cm apart with a p.d. of 100 V

between them. The electron is kept travelling in a straight line

perpendicular to the electric field between the plates by applying a

magnetic field perpendicular to the electrons path and to the electric

field. Calculate :

a. the intensity of the electric field.

b. the magnetic flux density needed.

Ans. : 0.50 x 104 V m-1, 0.010 T


A moving coil meter has a 50 turns coil measuring 1.0 cm by 2.0 cm. It is

held in a radial magnetic field of flux density 0.15 T and its suspension

has a torsional constant of 3.0 x 10-6 N m rad-1. Find the current is

required to give a deflection of 0.5 rad.

Ans. :1.0 mA

magnetism

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