52

CHAPTER 4

MOVING CHARGES AND MAGNETISM 1. How are electricity and magnetism related ?

Electricity and magnetism were considered to be independent of one another. Later on, it was realized

that they are related to one another and one give rise to the other ie, electricity produces magnetism and

magnetism produces electricity

2. Who observed the magnetic effect of electric current ? How ?

In 1820, the Danish physicist Oersted accidently observed that a pivoted magnetic needle kept near a

conductor was deflected when a current is passed through the conductor. This made Oersted to suggest

that there is a magnetic field surrounding a current carrying conductor

3. Describe an experiment to demonstrate the magnetic effect of electric current

Keep a straight conductor AB along the N-S direction. Keep a pivoted magnetic needle below it. It

comes to rest along the N-S direction ie, parallel to the conductor. When a current is passed through the

conductor, the needle gets deflected indicating the presence of a magnetic field other than earth’s field. It

is the magnetic field due to the current carrying conductor.

If a current is passed from A to B, the N pole will be deflected into the plane of the paper ie towards west.

The direction in which the N pole moves gives the direction of the magnetic field. Therefore the direction

of the magnetic field below the wire is into the plane of the paper. If the current is reversed the direction

in which the N pole is deflected also gets reversed showing that the magnetic field gets reversed

4. Motion of Charged Particle in a Magnetic Field.

If a particle carrying a positive charge q and moving with velocity v enters a magnetic field B then it

experiences a force F which is given by the expression

)( BvqFrr

×= ⇒⇒⇒⇒ θsinqvBF =

Here =vr

velocity of the particle, =Br

magnetic field

(1) Zero force

Force on charged particle will be zero (i.e. F = 0) if

(i) No field i.e. B = 0 ⇒ F = 0

(ii) Neutral particle i.e. q = 0 ⇒ F = 0

(iii) Rest charge i.e. v = 0 ⇒ F = 0

(iv) Moving charge i.e. θ = 0o or θ = 180

o ⇒ F = 0

(2) Direction of force

The force Fr

is always perpendicular to both the velocity vr

and the field Br

in accordance with Right

Hand Screw Rule, through vr

and Br

themselves may or may not be perpendicular to each other.

vq θ = 0o

θ = 180o

q

B

× × × × × ×

× × × × × ×

× × × × × ×

× × × × × ×

q, m v B

→

v →

B →

Fm

→

θ 90°

v → B

→

Fm

→

53

Direction of force on charged particle in magnetic field can also be find by Flemings Left Hand

Rule (FLHR).

Here, First finger (indicates) → Direction of magnetic field

Middle finger →→→→ Direction of motion of positive charge or direction,

opposite to the motion of negative charge.

Thumb → Direction of force

(3) Circular motion of charge in magnetic field

Consider a charged particle of charge q and mass m enters in a uniform magnetic field B with an initial

velocity v perpendicular to the field.

θ = 90o, hence from F = qvB sinθ particle will experience a maximum magnetic force Fmax = qvB which

act's in a direction perpendicular to the motion of charged particle. (By Flemings left hand rule).

(i) Radius of the path : In this case path of charged particle is circular and magnetic force provides the

necessary centripetal force i.e. r

mvqvB

2

= ⇒ radius of path qB

mvr =

If p = momentum of charged particle and K = kinetic energy of charged particle (gained by charged

particle after accelerating through potential difference V) then mqVmKmvp 22 ===

So q

mV

BqB

2mK

qB

p

qB

mvr

21====

Kpvr ∝∝∝ i.e. with increase in speed or kinetic energy, the radius of the orbit increases.

Note : � Less radius (r) means more curvature (c) i.e. r

c1

∝

(iii) Time period : As in uniform circular motion v = rω, so the angular frequency of circular motion,

called cyclotron or gyro-frequency, will be given by m

qB

r

v==ω and hence the time period,

qB

mT π

ωπ

22

==

i.e., time period (or frequency) is independent of speed of particle and radius of the orbit and depends

only on the field B and the nature, i.e., specific charge

m

q, of the particle.

(4) Motion of charge on helical path

When the charged particle is moving at an angle to the field (other than 0o, 90

o, or 180

o).

Less : r

More : c

More : r

Less : c

r = ∞

c = 0

F

B

v

× × × × × ×

× × × × × × ×

× × × × × × ×

× × × × × × ×

× × × × × × ×

× × × × × × ×

v v

v v

F

+

+ +

+

54

In this situation resolving the velocity of the particle along and perpendicular to the field, we find that the

particle moves with constant velocity v cosθ along the field (as no force acts on a charged particle when

it moves parallel to the field) and at the same time it is also moving with velocity v sinθ perpendicular to

the field due to which it will describe a circle (in a plane perpendicular to the field) of radius.

qB

vsinmr

)( θ=

Time period and frequency do not depend on velocity and so they are given by qB

mT

π2= and

m

qB

πν

2=

So the resultant path will be a helix with its axis parallel to the field Br

as shown in figure in this situation.

The pitch of the helix, (i.e., linear distance travelled in one rotation) will be given by

)cos(2)cos( θπθ vqB

mvTp ==

Note : � 1 rotation ≡ 2π ≡ T and 1 pitch ≡ 1 T

� Number of pitches ≡ Number of rotations ≡ Number of repetition = Number of helical turns

� If pitch value is p, then number of pitches obtained in length l given as

Number of pitchesp

l= and time reqd.

θcosv

lt =

(5) Lorentz force

When the moving charged particle is subjected simultaneously to both electric field Er

and magnetic

field Br

, the moving charged particle will experience electric force EqFe

rr= and magnetic force

)( BvqFm

rrr×= ; so the net force on it will be )]([ BvEqF ×+= . Which is the famous ‘Lorentz-force

equation’. Depending on the directions of Ev,r

and Br

following situations are possible

(i) When Evrr

, and Br all the three are collinear : In this situation as the particle is moving parallel or

anti parallel to the field, the magnetic force on it will be zero and only electric force will act and so

m

Eq

m

Fa

rrr

==

The particle will pass through the field following a straight line path (parallel field) with change in its

speed. So in this situation speed, velocity, momentum kinetic energy all will change without change in

direction of motion as shown

(ii) When E is parallel to B and both these fields are perpendicular to v then : eF is perpendicular

to mF and they cannot cancel each other. The path of charged particle is curved in both these fields.

Er

Br

v q

q

Er

Br

v

θ q, m

v →

B →

θ

v

p

r

B → Y

X

Z

v sinθ

v cosθ

55

(iii) E,v and B are mutually perpendicular : In this situation if Er

and Br

are such that

0=+= me FFFrrr

i.e., 0)/( == mFarr

as shown in figure, the particle will pass through the field with same velocity.

And in this situation, as me FF = i.e., qvBqE = BEv /=

This principle is used in ‘velocity-selector’ to get a charged beam having a specific velocity.

Note : � From the above discussion, conclusion is as follows

� If E = 0, B = 0, so F = 0.

� If E = 0, B ≠ 0, so F may be zero (if o0=θ or o180 ).

� If E ≠ 0, B ≠ 0, so F = 0 (if |||| me FFrr

= and their directions are opposite)

� If E ≠ 0, B = 0, so F ≠ 0 (because constant≠vr

).

5. CYCLOTRON.

Cyclotron is a device used to accelerated positively charged particles (like, α-particles, deutrons etc.) to

acquire enough energy to carry out nuclear disintegration etc. t is based

on the fact that the electric field accelerates a charged particle and the

magnetic field keeps it revolving in circular orbits of constant

frequency. Thus a small potential difference would impart if

enormously large velocities if the particle is made to traverse the

potential difference a number of times.

It consists of two hollow D-shaped metallic chambers D1 and D2 called

dees. The two dees are placed horizontally with a small gap separating

them. The dees are connected to the source of high frequency electric

field. The dees are enclosed in a metal box containing a gas at a low

pressure of the order of 10–3

mm mercury. The whole apparatus is placed between the two poles of a

strong electromagnet NS as shown in fig. The magnetic field acts perpendicular to the plane of the dees.

Note : � The positive ions are produced in the gap between the two dees by the ionisation of the

gas. To produce proton, hydrogen gas is used; while for producing alpha-particles, helium

gas is used.

(1) Cyclotron frequency : Time taken by ion to describe q semicircular path is given by qB

m

v

rt

ππ==

If T = time period of oscillating electric field then qB

mtT

π22 == the cyclotron frequency

m

Bq

T πν

2

1==

(2) Maximum energy of position : Maximum energy gained by the charged particle 222

max2

rm

BqE

=

where r0 = maximum radius of the circular path followed by the positive ion.

Note : � Cyclotron frequency is also known as magnetic resonance frequency.

� Cyclotron can not accelerate electrons because they have very small mass.

Hall effect : The Phenomenon of producing a transverse emf in a current carrying conductor on applying

a magnetic field perpendicular to the direction of the current is called Hall effect.

Target

High frequency

oscillator

Energetic

proton beam W

N

D1

S

D2

v

z

x

Fe

+ q

Fm

+ q

y Er

Br

56

6. Force on a Current Carrying Conductor in Magnetic Field.

In case of current carrying conductor in a magnetic field force experienced by its small length element is

BlidFd ×= ; lid = current element )( BldlFd ×=

Total magnetic force ∫∫ ×== )( BldiFdF

If magnetic field is uniform i.e., Br

= constant

)( BLiBldiFrrrrr

×′=×

= ∫

==∫ Lldrr

vector sum of all the length elements from initial to final point. Which is in accordance with

the law of vector addition is equal to length vector Lr

′ joining initial to final point.

(1) Direction of force : The direction of force is always perpendicular to the plane containing lidr

and Br

and is same as that of cross-product of two vectors )( BArr

× with ldiA =r

.

The direction of force when current element ldi and Br

are perpendicular to each other can also be

determined by applying either of the following rules

Fleming’s left-hand rule Right-hand palm rule

Stretch the fore-finger, central finger and thumb left

hand mutually perpendicular. Then if the fore-finger

points in the direction of field Br

and the central in

the direction of current i, the thumb will point in the

direction of force

Stretch the fingers and thumb of right hand at right

angles to each other. Then if the fingers point in the

direction of field Br

and thumb in the direction of

current i, then normal to the palm will point in the

direction of force

(2) Force on a straight wire : If a current carrying straight conductor (length l) is placed in an uniform

magnetic field (B) such that it makes an angle θ with the direction of field then force experienced by it is

θsinBilF =

If o0=θ , 0=F

If o90=θ , BilF =max

B →

dF →

i dl → P

θ

× × × × × ×

× × × × × × ×

× × × × × ×

× × × × × × ×

× × × × × × ×

× × × × × × ×

dl i

dF

B

B →

dF →

i dl →

P θ

Force

Magnetic

field

Current

i l

B

F

Current

Force

Magnetic

field

57

(3) Force on a curved wire

The force acting on a curved wire joining points a and b as shown in the figure is the same as that on a

straight wire joining these points. It is given by the expression BLiF ×=

7.Force Between Two Parallel Current Carrying Conductors.

When two long straight conductors carrying currents 1i and 2i placed parallel to each other at a distance

‘a’ from each other. A mutual force act between them when is given as

la

iiFFF ×⋅=== 210

21

2

4πµ

where l is the length of that portion of the conductor on which force is to be calculated.

Hence force per unit length a

ii

l

F 210 2

4⋅=

πµ

m

N or

a

ii

l

F 212=

cm

dyne

Direction of force : If conductors carries current in same direction, then force between them will be

attractive. If conductor carries current in opposite direction, then force between them will be repulsive.

Note : � If a = 1m and in free space mNl

F/102 7−×= then Ampii 121 == in each identical wire.

By this concept S.I. unit of Ampere is defined. This is known as Ampere’s law.

Hall effect helps us to know the nature and number of charge carriers in a conductor.

Biot Savart's Law.

It is used to determine the magnetic field at any point due to a current carrying conductors.

This law is although for infinitesimally small conductors yet it can be used for long conductors. In order

to understand the Biot-Savart’s law, we need to understand the term current-element.

Current element

It is the product of current and length of infinitesimal segment of current carrying wire.

The current element is taken as a vector quantity. Its direction is same as the direction of current.

Current element AB = dli

i2 i1

a

• • × ×

• • × ×

• •

× ×

• • × ×

• • × ×

• • × ×

i2 i1

2 1

F2 F1

× ×

× ×

× ×

× ×

× ×

× ×

i2

2

F2

× ×

× ×

× ×

× ×

× ×

× ×

i1

1

F1

× × × × × × ×

× × × × × × ×

× × × × × × ×

× × × × × × ×

× × × × × × ×

× × × × × × ×

L →

B →

a

b × × × × × × ×

× × × × × × ×

× × × × × × ×

× × × × × × ×

× × × × × × ×

× × × × × × ×

B →

L →

i

F

a

b

A B

dl i

58

In the figure shown below, there is a segment of current carrying wire and P is a point where magnetic

field is to be calculated. ldi is a current element and r is the distance of the point ‘P’ with respect to the

current element ldi . According to Biot-Savart Law, magnetic field at point ‘P’ due to the current element

ldi is given by the expression, 2

r

dlsin θikdB = also ∫∫ ==

2

0 sin.

4 r

dlidBB

θπ

µ

In C.G.S. : k = 1 ⇒ 2

sin

r

idldB

θ= Gauss

In S.I. : π

µ4

0=k ⇒ 2

0 sin

4 r

idldB

θπ

µ⋅= Tesla

where 0µ = Absolute permeability of air or vacuum metreAmp

Wb

−×= −7

104π .

8.Obtain expression for the field due to a circular coil carrying current at any point on the axis and

hence find the field at the centre of the coil

Consider a circular coil of centre O, radius r having one turn and carrying a current I. Let P be a point

along the axis at a distance x from the centre O. Required to find the field at P. Consider an element AB

of length dl and centre M. Let a be the distance of the element from the point P. The field at P due to the

element AB is dB = .

Here � is the angle between the element and the line MP. � = 90 sin � = 1 .

dB = . It is perpendicular to the line PM.

It can resolved into dB cos � perpendicular to OP and dB sin � parallel to OP

Consider another element A’ B’ diametrically opposite to AB. The field at P due to the element is

dB =

It has components dB cos � perpendicular to OP and dB sin � parallel to OP.

The perpendicular components cancel each other and parallel components add up. So to get the field at P

due to the entire coil we have to integrate dB sin � between the limits 0 to 2 �r

B = = x sin

x( ) =

B =

dl

i

P

r

59

B = =

= . But a2 = r

2 + x

2. a = . =

. This is for one turn.

If there are ‘N’ turns. .

This field is parallel to the axis and is from left to right for the current in the figure

To find the field at the centre of the coil :

At the centre of the coil x = 0

B0 = = = .

9.AMPERE’S CIRCUITAL LAW.

The line integral of magnetic field vector taken over a closed path is equal to µ 0 times the current

enclosed by the path.

Explanation :

Integral taken over a closed path = µ0 I if there is only one current.

Integral taken over a closed path = µ0 (I1 + I2 + ......)

(If there are a number of currents passing through the space enclosed by them. Integral taken over a

closed path,

10. Apply Amperes circuital theorem to find the field due to an infinitely long conductor carrying

current

Consider a straight conductor carrying a current I. Let P be a point distant r from the conductor.

Required to find the field at P. For this consider a circle of a radius r passing through P and perpendicular

to the conductor, with its centre on the conductor. Let B be the field at P. Consider a short element of

length dl around P. B is parallel to dl = B dl = Bdl

= = B = B 2 .

By Ampere’s circuital law =

60

B 2 = .

B = .

11.Apply Ampere’s circuital theorem to find the field due to an infinitely long cylindrical conductor

carrying current

1) Point outside :

Let P be a point distant r from the axis of cylindrical conductor carrying a current I and of radius R.

Required to find the field at P. Consider a circle passing through P with its centre on the axis of the

conductor and plane of the circle perpendicular to the conductor. Let B be the field at P. B is tangential

to the circle at P. Consider an element of length dl around P.B is parallel to dl.

= B dl = Bdl

= = B = B 2 .

By Ampere’s circuital law =

B 2 = .

B = . B ∝ 1 / r

2) Point inside :

Consider a cylindrical conductor of radius R carrying a current I. Let P’ be a point inside the conductor at

a distance r1 from its axis. Required to find the field at P′. For this construct a circle passing through P′ with its centre on the axis and its plane perpendicular to the circle. Consider a short element of length dl

around P′. B1 is parallel to dl. B

1. dl = B

1dl cos 0 = B

1 dl

= = = 2 .

Let I’ be the current enclosed by the path.

Then by Ampere’s circuital law

= .

. But I1 = = .

) =

12.Apply Amperes circuital theorem to find the field due o a solenoid

A solenoid consists of a hollow cylinder made of a non-magnetic material on which a large number of

turns of insulated wire are closely wound. When a current is passed through it there will be a magnetic

field inside. One end will behave like a north pole and the other as south pole. there will be a magnetic

61

field inside the solenoid. Required to find the magnetic field along the axis of the solenoid.

Consider a short element AB of length ‘d’ along the axis. With AB as one side complete the rectangle

ABCD. Let B1 be the field along the axis.

= =

For any outside point the field is 0. Therefore is equal to 0.

For half length of BC and DA the field is zero and for the other half, the fields are travelled in opposite

directions.2nd and 4th integrals cancel .Therefore only the 1st integral remains.

= = = = .

By Ampere’s circuital law =

where I1 is the current enclosed by the rectangle ABCD. If there are N turns over the length AB then I

1 =

NI, where I is the current through 1 turn.

B1 d = µ0 NI

B1 = µ0 I But = n, the number of turns per unit length.

B1 = µ0 n I . This is for air core. For a solenoid with iron core B = µ0µr nI.

The field is independent of the radius. The field will be more or less uniform near teh centre of the

solenoid along the axis. Towards the ends, the field will decrease and at the ends the field is .

Solenoid is used for magnetizing iron rods. The iron rod is placed along the axis, near the centre and a

strong current is passed through the solenoid. According to the figure in q. no: 135 the right end becomes

south pole & left end becomes north pole

13.What is a toroid. Obtain expression for the field due to a toroid

It consists of a ring made of iron cylinder, on which a large number of turns of insulated wire are closely

wound. When a current is passed through the coil it produces a magnetic field inside. Instead of an iron

core, we can also use an air core. In that case the magnetic field will be reduced to 1/µr times the field

due to iron ring where µr is the relative permeability of iron

Derivation for Air-cored toroid :

1) Point outside (P1) …. path 3

Let P1 be point outside the toroid at a distance r1 from the centre of the toroid. Let B1 be the field at P1.

Consider a circle passing through P1 and concentric with the toroid .

62

By Ampere’s circuital law = where I1 is the current enclosed

ButI1 = (NI - NI) = 0 where N = total number of turns

B1 2πr1 = µo (0) i.e. B1 = 0

2) Point inside (P2)….. path 1

Let P2 be a point inside the toroid at a distance r2 from the centre. Consider a concentric circle passing

through P2. If B2 is the field at P2

By Ampere’s circuital law = where I1 is the current enclosed. But here I

1 = 0 .

B2 2πr2 = µo (0) i.e. B2 = 0

3) Field along the axis of the toroid :…..path 2

Let P be a point along the axis of the toroid. Let B be the field along the axis.

By Ampere’s circuital law = where I1 is the current enclosed

ButI1 = (NI) where N = total number of turns and I is the current through each turn.

B 2πr = µo (NI)

i.e. B = But = n, the number of turns per unit length.

So B = µo nl.

With a core of relative permeability µr the field is B = µoµr nI the field is independent of the radius

14.What are the advantages of toroid over straight solenoid ?

1) In the case of the straight solenoid the field is uniform only along the axis and near the centre. But in

the case of toroid the field is uniform throughout the toroid. 2) In the case of toroid there are no free

poles. The magnetic field is confined only to the core. Toroid is a magnet with no poles.

15.How is the field due to a solenoid related to the radius ?

Field is independent of the radius , B = µo nI

16.Current Loop As a Magnetic Dipole.

A current carrying circular coil behaves as a bar magnet whose magnetic moment is M = NiA; Where

N = Number of turns in the coil, i = Current through the coil and A = Area of the coil

Magnetic moment of a current carrying coil is a vector and it's direction is given by right hand thumb rule

M

→ Magnetic

moment

Current

S N

63

17.Derive expression for the torque acting on a current carrying coil placed in a uniform magnetic

field and discuss the various cases.

Consider a rectangular coil PQRS carrying a current I placed in a uniform magnetic field B with the plane

of the coil making an angle α [ or plane normal to coil make angle � ]with the field. Then the normal to

the plane of the coil will make an angle θ = 90 - α with the field. Let PQ = RS = l and QR = PS = b

Force on PS anQR are parallel to the plane of the coil.[F1 and F2] .They either compress the coil or stretch

it. They do not contribute to the torque.

Force on PQ F3= BI (dl) sin θ = BIlsin 90= BIl out of the plane of the paper

Force on RS = BIl into the plane of the paper. These two forces constitute a couple. The perpendicular

distance between the forces is PN = b cos α So Torque τ = BI l x b cos α = BI (l x b) cos α But l x b = A, the area of the coil.

τ = BI A cos α ,This is for one turn

For N turns, τ = BI A N cos α -------- (1)

But (90 - 0) = θ , α = 90 - θ

cos α = cos (90 - θ) = sin θ

So τ = BI A N sin 0 ------- (2)

But IAN = m, the magnetic dipole moment of the coil.

τ = (m x B) ----------- (3)

The magnitude of the torque is independent of the shape of the coil. It depends only on the dipole

moment which depends on the area of the coil. A rectangular coil and a circular coil of equal area will

experience the same torque in the same magnetic field, when the same current passes through them

Discussion of Equation No.1 :

τ = BI A N cos α

Case I : α = 0 ie plane of the coil parallel to the field , cos α = cos 0 = 1

τ = BI A N, a maximum , when α ie., the normal to the coil is perpendicular to the field. sin

τ = BIAN.

Thus the torque on a coil is maximum when the plane of the coil is parallel to the field or normal to the

coil is perpendicular to the field

Case II : α = 90 ie. plane of the coil perpendicular to the field or normal to the coil is parallel to the field ,

cos α = cos 90 = 0 τ = 0

Thus a current carrying coil will experience no torque when the plane of the coil is perpendicular to the

field or the normal to the coil is parallel to the field

18.What is meant by radial magnetic field ? What is its advantage ?

When the pole pieces are plane, the magnetic field will be as in figure (in the class). When the coil is

parallel to the field, the torque will be ,

it makes an angle α with field and the torque on it becomes =

Therefore the torque is not a constant. It varies as cos

If the pole pieces are concave and if a soft iron cylinder is placed at the centre, then the field will be as

shown in figure .It is known as ‘radial

magnetic field. Therefore the torque is a constant equal to BI A N

58.Describe with theory, the suspended type moving coil galvanometer

The apparatus consists of a rectangular coil

magnet N-S by means of a phosphor bronze fibre F from a torsion head

inside the coil without touching it. The lower end of the coil is connecte

spring[S]. The whole arrangement is enclosed in a case with leveling screws

connected to the terminals T1 & T2 ,

64

α = 0 θ = 90

ie., the normal to the coil is perpendicular to the field. sin θ = sin 90 = 1

Thus the torque on a coil is maximum when the plane of the coil is parallel to the field or normal to the

= 90 ie. plane of the coil perpendicular to the field or normal to the coil is parallel to the field ,

Thus a current carrying coil will experience no torque when the plane of the coil is perpendicular to the

field or the normal to the coil is parallel to the field

What is meant by radial magnetic field ? What is its advantage ?

plane, the magnetic field will be as in figure (in the class). When the coil is

parallel to the field, the torque will be ,τ = BI A N. Under the action of this torque the coil rotates. Then

with field and the torque on it becomes = BI A N cos α.

Therefore the torque is not a constant. It varies as cos α.

If the pole pieces are concave and if a soft iron cylinder is placed at the centre, then the field will be as

It is known as ‘radial magnetic field’. In this field the coil is always parallel to the

magnetic field. Therefore the torque is a constant equal to BI A N

Describe with theory, the suspended type moving coil galvanometer

The apparatus consists of a rectangular coil PQRS suspended between the poles of a powerful horse shoe

S by means of a phosphor bronze fibre F from a torsion head H. A soft iron cylinder

the coil without touching it. The lower end of the coil is connected to a phosphor bronze

The whole arrangement is enclosed in a case with leveling screws. The ends of the coil are

,

Thus the torque on a coil is maximum when the plane of the coil is parallel to the field or normal to the

= 90 ie. plane of the coil perpendicular to the field or normal to the coil is parallel to the field ,

Thus a current carrying coil will experience no torque when the plane of the coil is perpendicular to the

plane, the magnetic field will be as in figure (in the class). When the coil is

= BI A N. Under the action of this torque the coil rotates. Then

If the pole pieces are concave and if a soft iron cylinder is placed at the centre, then the field will be as

magnetic field’. In this field the coil is always parallel to the

poles of a powerful horse shoe

A soft iron cylinder L is kept

phosphor bronze

The ends of the coil are

65

When a coil is placed in a uniform radial magnetic field it will experience a torque τ = B I A N

Under the action of this external torque τe = B I A N, the coil will rotate and the fibre will get twisted.

Internal elastic torque will be developed within the fibre which opposes the applied torque. Therefore

equilibrium will be reached after rotating through a certain angle.

Let θ be the equilibrium deflection. Then the internal torque τi = C θ , where C is the torque required to

produce unit twist of the fiber. For equilibrium τe = τi

B I A N = C θ

So I = .

is a constant for the galvanometer denoted by k. I = k θ

Thus the current is directly proportional to the deflection.

Current Sensitivity :

Current sensitivity is the deflection for unit current.

Methods of increasing current sensitivity :

1) Increase A

2) Increase B

3) Increase the number of turns

4) Decrease the torque per unit twist

But C = , where G is rigidity modulus r = radius , l = length of suspension fibre

For C to be small

(1) G must be small (2) r must be small (3) l must be large.

A thin and long wire will make the galvanometer more sensitive

Voltage sensitivity :

Voltage sensitivity = . .

Divide both sides by R .

But IR = V, .

Voltage sensitivity =

1) Increase B 2) Increase A 3) Decrease, the torque per unit twist

Voltage sensitivity is independent of number of turns because when the number of turns increases, the

resistance increases in such a way that is a constant

59.Magnetic dipole moment of a revolving electron.

Magnetism of materials arises from magnetism of atoms and molecules with which they are composed of.

The magnetism of atoms and molecules arises from the orbital and spin motion of electrons.

Consider an electron revolving in a circular orbit of radius r with a speed v. It has got an angular

momentum. L = mvr ……… (1)

It is equivalent to circulating current , I = .

66

Area of the orbit A = �r2

Magnetic moment m0 = IA

m0 = x �r2 , m0 = …… (2)

From (1) v r = , m0 =

Here m is the electron mass, m0 is the magnetic moment associated with the orbital motion of the

electron. L is the orbital angular momentum

m0 = ……… (3)

For an electron the charge is negative m0 = - x L

The direction of magnetic moment of an electron is opposite to the direction of angular momentum

According to the postulate of Bohr theory L = .

m0 = x

mo = .

m0 has minimum value when n = 1

Minimum value of mo is .

It is a fundamental constant. It is known as Bohr magneton represented by �B

� B = =

Bohr magneton = 9.27 x 10-24

Am2

This is the minimum magnetic moment that can occur .In addition to this, every electron possesses a spin

magnetic moment

ms = x S

where S is the spin angular momentum. ms is the spin magnetic moment ms = x S

Therefore the total magnetic moment is the vector sum of mo + ms .

m = mo + ms = + = ( L+2S) .

This is the total magnetic moment of an electron. An atom will have a number of electrons in it. The total

magnetic moment of an atom is the vector sum of the magnetic moments of all the electrons in the atom.