chapter - 4 moving charges and magnetism · tesla (t). i.e. “when the force acting on a unit...

4.1 Introduction

Both Electricity and Magnetism have been known for more than 2000 years. The

branch of physics which envelops a comprehensive study of electricity and

magnetism is called electrodynamics. In the modern technology of

communication, electrodynamics is of prime importance.

In the present chapter, we will study (1) how magnetic field exerts forces on

moving charged particles and current-carrying wires, (2) how particles can be

accelerated to very high energies in a cyclotron, (3) how currents and voltages

are detected by a galvanometer.

We can take the following convection: A current or a field (electric or magnetic)

emerging out of the plane of the paper is indicated by dot (ʘ). A current or a

field going into the plane of the paper is indicated by a cross ().

In the year 1820, the Danish physicist Hans Christian Oersted noticed that “a

current in a straight wire caused a noticeable deflection in the nearby magnetic

compass needle.”

He found that the alignment of the needle is tangential to an imaginary circle

which has the straight wire as its centre and has its plane perpendicular to the

wire. This is shown in fig. (a).

When the current is large and the needle sufficiently close to the wire so that the

earth’s magnetic field may be ignored. When the direction of the current is

reversed, so the orientation of the needle is also reverse. This is shown in fig. (b).

The deflection increases on increasing the current or bringing the needle closer

to the wire. Iron filing sprinkled around the wire arrange themselves in concentric

circles with the wire as the centre. This is shown in fig. (c).

Oersted concluded that “moving charges or currents produced a magnetic field

in the surrounding space.”

Short Questions

1. Write Oersted’s observation.

A. Oersted concluded that, “moving charges or currents produced a magnetic field

in the surrounding space”.

2. Write the rule used to find the direction of magnetic field acting at a point

near a current carrying straight conductor.

Chapter - 4

Moving Charges and Magnetism

MQ : Explain the concept of

magnetic field.

Important Note : (1) Electrostatic field lines start

from positive charge and

end at negative charge while

magnetic field lines always

form a closed loops.

(2) In this chapter we discuss only

for steady currents which do

not vary with time.

MQ : State and explain

Oersted’s observation.

Hans Christian Oersted

(1777 - 1851)

Danish physicist and

chemist, professor at

Copenhagen. He observed

that a compass needle

suffers a deflection when

placed near a wire carrying

an electric current. This

discovery gave the first

empirical evidence of a

connection between electric

and magnetic Phenomena.

Modern Physics Made Easy (Part : 1) : Class XII 4.2

A. (1) A current or a field emerging out of the plane of the paper is depicted by a

dot i.e. (ʘ) (2) A current or a field going into the plane of the paper is depicted

by a cross i.e. ().

Curiosity Question

1. What is the basic difference between magnetic field and electric field ?

A. Whether a charged particle is at rest or in motion, an electric field always exerts

a force on it and changes its speed and kinetic energy. While a magnetic field

exerts a force on a changed particle only when it is in motion. Also there is no

change in the speed or the kinetic energy of the charged particle.

4.2 Magnetic Force

4.2.1 Sources and Fields

We know that static charges produce an electric field, while the current or

moving charges produce a magnetic field, denoted by ,B r which is a vector

field. It has several basic properties identical to the electric field. It is defined at

each point in space and can in addition depend on time.

The super position principle for magnetic field : The magnetic field of several

sources is the vector addition of magnetic field of each individual source.

4.2.2 Magnetic Field, Lorentz force

Consider a electric charge q moving with velocity v in a magnetic field B ,

making an angle θ with B . It is found from experiments that the moving charge

q experiences a force F , and the force is proportional to

(i) magnitude of magnetic field i.e. F B

(ii) moving electric charge i.e. F q

(iii) component of the velocity in the perpendicular to direction of magnetic field

i.e. sinF v

Combining all factors we get,

sinF qBv

sinF kqBv …… (1)

The unit of magnetic field is so defined that the constant k becomes unity and

dimensionless. So equation (1) can written as,

sinF qBv …… (2)

Since the direction of F is perpendicular to V and B , so we can write equation

(2) in terms of vector product as,

F q v B …… (3)

Important Point :

A moving charge produces a

magnetic field which exerts

a force on another moving

charge.

If in a field, the force

experienced by a moving

charge depends on the

strength of the field and not

on the velocity of the charge,

then the field must be an

electric field and if force is

depend only on velocity, not

on the strength of the field

must be magnetic field.

LQ : Obtain F = q v × B

and discuss its cases. Also

define 1 tesla.

Fleming’s Left hand Rule :

Chapter-4 : Moving Charges and Magnetism 4.3

Figure (a) shows the direction of F , can be find with the help of right-hand

screw rule.

Special Cases :

(1) If velocity and magnetic field are parallel or antiparallel i.e. θ 0 or 180

then 0F . Thus, if electric charge moving parallel or antiparallel to a magnetic

field does not experiences any force in the magnetic field.

(2) If velocity of charge particle is zero i.e. 0v . Thus, the magnetic force is

zero if electric charge is not moving (i.e. remain stationary).

(3) If velocity and magnetic field are perpendicular to each other i.e. θ 90

then sin90 .F qvB qvB Thus, the electric charge experiences the

maximum force when it moves perpendicular to the magnetic field.

If θ 90 , 1 , 1F N q C and 11v ms then magnetic field is said to be 1

tesla (T).

i.e. “when the force acting on a unit charge (1 C), moving perpendicular to

magnetic field with a speed 1 m/s, is 1 newton then magnetic field is said to be 1

tesla.”

This unit is called tesla (T) named after Nikola Tesla. Tesla is a large unit, while

smaller unit (non - SI) is called gauss. 1 gauss = 10-4 tesla.

Dimensional formula for magnetic field,

1 1 21 2 0 1

1 1 1 1

F M LTB M L T A

qv A T LT

4.2.2.A Lorentz force

Suppose a point charge q moving with a velocity ,v located at r at a given time

t in presence of both electric field E r and the magnetic field B r .

Electric charge q in an Electric field E experiences the electric force eF qE

and magnetic force experienced by the charge q moving with velocity v in the

magnetic field B is .mF q v B

The force on an electric charge q due to both field can be written as,

meF F F

qE q v B

F q E v B

…… (1)

This force was first given by H.A. Lorentz based on the extensive experiments

of Ampere and others. It is called the Lorentz force.

Lorentz force depends on electric charge q, velocity v and the magnetic field .BForce on a negative charge is opposite to that on a positive charge.

Short Questions

1. What is magnetic Lorentz force?

A. Magnetic force experienced by the charge q moving with velocity v in the

magnetic field B is called magnetic Lorentz force i.e. mF q v B .

Right hand palm Rule:

MQ : Obtain Lorentz force

equation.

Hendrik Antoon Lorentz

(1853 - 1928)

Dutch theoretical physicist,

professor at Leiden. He

investigated the relationship

between electricity,

magnetism, and mechanics.

In order to explained the

observed effect of magnetic

fields on emitters of light i.e.

Zeeman effect, he

postulated the existence of

electric charge in the atom,

for which he was awarded


2. What is the force experienced by a stationary charge in a magnetic field?

A. For stationary charge 0v

Hence, sin 0F qvB

3. Write the expression for the Lorentz force on a charged particle.

A. F q E v B

4. What is the work done by magnetic field on a moving charge?

A. Zero, because F v

5. An electron beam is moving vertically downwards. If it passes through a

magnetic field which is from South to North in a horizontal plane, then in

which direction beam would be deflected?

A. Towards west

6. Two particles A and B having masses m and 2m, charges q and 2q

respectively. If the ratio of the force acting on charge in magnetic field is

1:2, find the ratio of their velocities respectively. Take o

θ = 90 .

A. 1 1

2 2

sin 90

sin 90

A

B

F q v B

F q v B 1 2

2 1

1 2

2

A

B

v F q q

v F q q

= 1 1 2: 1:1v v

7. An electron performs circular motion of radius r, perpendicular to a

uniform magnetic field B. The K.E gained by this electron in the half

revolution is …… (Fill the blank)

A. . constantK E

0W K E

8. At a place, an electric field and a magnetic field are in the downward

direction. There an electron moves in the downward direction. Hence the

motion of electron will be …… (Fill the blank)

A. sin 0mF qvB and F qE i.e. opposite in the direction of E hence it

will lose velocity.

Curiosity Question

1. When north pole of a magnet is brought near a stationary negatively

charged conductor, will the pole experience any force?

A. No, because a stationary charge does not produce any magnetic field.

Higher Order Thinking Skills (HOTS)

1. Write the equation of magnetic force acting on a particle moving through a

magnetic field. Using it obtain Newton’s equation of motion and find K.E of

the particle with respect to time.

4.2.3 Magnetic force on a current-carrying conductor

Z

Y

X o q

l area = A

MQ : Obtain an expression for

the force experienced by a

current carrying straight

conductor placed in a magnetic

field.

the Nobel Prize in 1902. He

derived a set of

transformation equations

(known as Lorentz

transformation) by some

tangled mathematical

arguments, but he was not

aware that these equations

hinge on a new concept of

space and time.


Consider a rod of a uniform cross-sectional area A and length l, current I carrying

along +X direction. The magnetic field B along +Y direction, mobile charges

(here electrons) moves with drift velocity d

.

Total volume of rod = Al., now the number density (i.e number of charges per

volume) of these mobile charges is = n. So the total number of mobile charge

carriers in it is N = n volume

N = nAl.

For a steady current I, each mobile carries has an average drift velocity d

in this

conducting rod. In the presence of an external magnetic field B , the force on

these charge carriers is,

dF Nq B

dnAlq B n

dAl q B …… (1)

Now, d

nq j = current density

equation (1) become,

F Al j B …... (2)

Since j A electric current = I

So, equation (2) become,

F I l B …… (3)

Here, l = vector of magnitude l, the length of the rod and with a direction

identical to the current I. Note that current I is not a vector, in equation (3) we

have transferred the vector sign from j to I l (the current element).

Equation (3) holds for a straight rod. Here, B is the external magnetic field. It is

not the field produced by the current-carrying rod.

If the wire has an arbitrary shape, we can calculate the Lorentz force on wire by

considering it as a collection of linear strips jdl then the force acting on a current

element jI dl due to the magnetic field B is jdF Idl B and total force on

wire is summing,

jF Idl B …… (4)

This summation can be converted to an integral in most cases.

The direction of force can be determined using right hand screw rule.

Short Questions

1. The force experienced by charge depends upon its velocity and become zero.

When it is at rest, the field is …… while the force experienced by a charge

depends only upon the magnitude of the field and does not depend upon the

velocity. The field is …… (Fill the blank)

A. magnetic, electric

2. What is the force that a conductor dl , carrying a current I experiences

when placed in a magnetic field B . What is the direction of the force?

Important Note :

If current carrying conductor

placed parallel to the

direction of the magnetic

field does not experience

any force.

If a current carrying

conductor placed

perpendicular to the

direction of a magnetic field

experiences a maximum

force.

Direction of force:


A. dF I dl B Using right hand screw rule, direction of force perpendicular

to plane of dl B .

3. Which rule that gives the direction of force on a current-carrying conductor

placed perpendicular to the magnetic field. A. Fleming’s left hand rule.


1. As shown in the figure, very long

conducting wire carrying current I1 is

arranged in y direction. Another

conducting wire of length l carring

current I2 is placed on X-axis at a

distance from this wire. Find the torque

acting on this wire with respect to O.

4.3 Motion in a Magnetic Field

Consider the motion of a charge in a

magnetic field. In the case of motion of

a charge in a magnetic field, the

magnetic force is perpendicular to the

velocity of the particle. So no work is

done and no change in the magnitude of

the velocity is produced. (though the

direction of momentum may be

changed).

When q charged particle, moving with

velocity v enters in a uniform magnetic

field B , it experiences a force

F q v B

sinF qvB …… (1)

Special Cases :

(1) 0 i.e. / /v B

sin 0 0F qvB

i.e. the parallel magnetic field does not exert any force on the moving charged

particle. The charged particle will continue to move along the line of force.

(2) 90 i.e. v B

sin90F qvB qvB

As the magnetic force acts on a particle perpendicular to its velocity, it does not

do any work on the particle. The perpendicular force F q v B , acts as a

centripetal force and produces a circular motion perpendicular to the magnetic

field. The particle will describe a circle if v and B are perpendicular to each

other, which is shown in fig. (4.5).

MQ : Write the equation of

force on electric charge moving

in magnetic field. And discuss

(i) If v B (ii) If v B

Important Note :

Electric force on a charge

eF qE

Magnetic force on a charge

sinmF qvB

If charge enters

perpendicularly to magnetic

field then, Radius :

mv pr

qB qB

Periodic time 2 m

TqB

Path : Circular

If charge enters at angle to

the magnetic field then,

Radius : sinmv

rqB

Periodic time : 2 m

TqB

Path : Helical


If r is the radius of the circular path of a particle, then a force of

2

,mv

racts

perpendicular to the path towards the centre of the circle, and is called the

centripetal force. Thus, the magnetic force provides the centripetal force.

So,

2mvqvB

r

mv

rqB

=p

qB …… (2) (Where p = mv momentum)

The radius of the circular orbit is inversely proportional to the specific charge

(i.e q/m) and to the magnetic field. The larger the momentum, the larger is the

radius and bigger the circle described.

If w is the angular frequency then, v r

So from (2), mr

rqB

qB

m

2

qBv

m ….... (3)

which is independent of the velocity or energy.

The time taken for one revolution is,

1 2 2 m

Tv qB

…… (4)

(3) When the initial velocity makes an arbitrary angle with the field direction.

If velocity has a component along B , this component remains unchanged as the

motion along the magnetic field will not be affected by the magnetic field.

The motion in a plane perpendicular to B is as before a circular one, there by

producing a helical motion fig. (4.6).

If there is a component of the velocity parallel to the magnetic field (i.e 11v ), it

will make the particle move along the field and the path of the particle would be

a helical one. (fig. 4.6).

The distance moved along the magnetic field in one rotation is called pitch p.

MQ : Discuss the motion of a

charged particle in a uniform

magnetic field with initial

velocity at an arbitrary angle

with the field direction.


We know time taken for one revolution is

2 mT

qB

so, the pitch (distance along

B ) p = (velocity parallel to the magnetic field) (time taken for one

revolution)

11v T 112 mv

qB

The radius of the circular component of motion is called the “radius of the helix.”

Short Questions

1. What will be the path of a charged particle moving perpendicular to a

uniform magnetic field? A. Circular

2. What will be the path of a charged particle moving in a uniform magnetic

field at any arbitrary angle? A. Helix

3. What will be the path of a charged particle moving along the direction of a

uniform magnetic field? A. The charged particle will move along a straight line.

4. When a charged particle moving with a velocity is subjected to a magnetic

field B the force acting on it is non-zero would the particle gain any energy? A. The magnetic force acts perpendicular to the direction of motion of the charged

particle. No work is done by the magnetic force on it. The particle does not gain

any energy.

5. If electron enter magnetic field perpendicular to the uniform magnetic field.

Write the expression for the radius of the path of it follows.

A.

2

sin 90mv

qvBr

mv

rqB

6. If electron moving + x direction, force on it is in +y direction. What is the

direction of its magnetic field.

A. From ,F q v B B is in +z direction.

7. A charged particle moving in a uniform magnetic field penetrates a layer of

lead and there by loses ½ of its K.E. How does the radius of curvature of its

path change?

A. Radius of curvature 2 2mv m K Km

rqB qB m qB

i.e. r K

So, its K.E is halved, radius of curvature is reduced to 1

2times its initial value.

8. A particle of mass m has an electric charge q. This particle is accelerated

through a p.d. V and then entered normally in a uniform magnetic field B.

It performs circular motion of radius R. Find the ratio of q/m.

A.

2mvqvB

R

qBRv

m


Now, 21

2mv qV

2 2 2

2

1

2

q B Rm qV

m

2 2

2q V

m B R

Curiosity Questions

1. A charged particle moves through a region of uniform magnetic field. Is the

momentum of the particle affected?

A. Magnetic force deflects the charged particle continuously from its path, so its

momentum charges due to the change in its direction of motion.

2. An electron does not suffer any deflection while passing through a region.

Are you definite there is no magnetic field in that region?

A. No, it may possible that the magnetic field is present but the electron is moving

parallel or anti-parallel to the magnetic field and magnetic force is zero.

4.4 Motion in Combined Electric and Magnetic Fields

4.4.1 Velocity selector

Suppose a charge q moving with velocity v in

presence of electric field E and magnetic field Bexperiences a Lorentz force,

i.e. E BF F F q E v B

Now consider the simple case in which electric

and magnetic fields are perpendicular to each

other and also perpendicular to the velocity of the

particle.

As shown in fig. ˆˆ,E Ej B Bk and ˆv vi also ˆEF qE qEj and

ˆ ˆˆ ˆBF q v B q vi Bk qvB i k ˆ,qvBj thus electric and

magnetic forces are in opposite directions as shown in fig. Now Lorentz force

ˆ ˆE BF F F q Ej qvBj ˆq E qvB j

Now, suppose we adjust the value of E and B such that magnitude of the two

forces are equal. Then total force on the charge is zero and the charge will in the

fields undeflected.

i.e. when E BF F

qE qvB E

vB

…… (1)

This condition can be used to select charged particles of a particular velocity out

of a beam containing charges moving with different speeds (irrespective of their

charge and mass). The crossed E and B fields (As the fields are perpendicular

to each other they are called crossed fields), therefore serve as a velocity selector.

Only particles with speed E/B pass undeflected through the region of crossed

fields. This method was employed by J.J. Thomson in 1897 to measure the charge

to mass ration (e/m) of an electron.

MQ : Show that the velocity of

charged particle is E

v = ,B

for

electric and magnetic fields

applied mutually

perpendicular to each other.


The principle is also employed in mass spectrometer-a device that separates

charged particles (usually ions) according to their charge to mass ratio.

Short Question

1. An electron passes through a region of crossed electric and magnetic fields

of intensities E and B respectively. What is the value of electron speed will

the beam remain unchanged?

A. e mF F

qE qvB

E

vB

Curiosity Question

1. What will be the path of a charged particle moving in a region of crossed

uniform electrostatic and magnetic field with initial velocity zero?

A. Cycloid with its forward motion normal to both E and B .

4.4.2 Cyclotron

The cyclotron is a machine to accelerate charged particles or ions to high

energies. It was invented by E.O. Lawrence and M.S. Livingston in 1934 to

investigate nuclear structure.

Principle : The cyclotron uses both crossed electric and magnetic fields (both

the fields are perpendicular to each other means they are crossed fields) in

combination to increase the energy of charged particles, the frequency of

revolution of the charged particle in a magnetic field is independent of its energy.

The operation of the

cyclotron is based on

the fact that the time

for one revolution of an

ion is independent of

its speed or radius of its

orbit.

Construction : The

particles move most of

the time inside two

semi-circular disc-like

metal containers D1

and D2 , which are

called dees as they look

like the letter D. Fig.

shows a schematic

view of the cyclotron.

Inside the metal boxes

the particle is shielded

and is not acted on by

the electric field. The magnetic field, acts on the particle and makes it go round

in a circular path inside a dee.

Every time the particle moves from one dee to another it is acted upon by the

electric field. The sign of the electric field is changed alternately in tune with the

circular motion of the particle. This ensures that the particle is always accelerated

by the electric field.

Each time the acceleration increases the energy of the particle. As energy

increases, the radius of the circular path increases. So the path is a spiral one.

LQ : What is Cyclotron?

Discuss principle, construction,

working of a cyclotron. What is

maximum K.E acquired by the

accelerated charged particles?

Cyclotron :


The whole arrangement is evacuated to minimize collisions between the ions and

the air molecules. A high frequency alternating voltage is applied to the dees.

Working : As shown in figure, positive ions or positively charged particles (eg.

protons) are released at the centre P. They move in a semicircular path in one of

the dees and arrive in the gap between the dees in a time interval T/2; here T is

the period of revolution.

2 m

TqB

or frequency

2c

qB

m

…… (1)

This frequency is called the cyclotron frequency ( c ) and is independent of

velocity of the particle and the radius of the orbit.

The frequency a of the applied voltage is adjusted so that the polarity of the dees

is reversed in same time that it takes the ions to complete one half of the

revolution. The requirement a c is called the resonance condition.

The phase of the supply voltage is adjusted so that when the positive ions arrive

at the edge of 1 2,D D is at a lower potential and the ions are accelerated across

the gap. Inside the dees the particles travel in a region free of the electric field.

The increase in their kinetic energy is qV each time they cross from one dee to

another (here V = Voltage across the dees at that time)

We know the radius of circular path is, mv

rqB

, it is clear that the radius of their

path goes on increasing each time their kinetic energy increases. The ions are

repeatedly accelerated across the dees until they have the required energy to have

a radius approximately that of the dees.

They are then deflected by a magnetic field and leave the system via an exit slit

and hits the target.

Maximum K.E of Ions : The ions will attain maximum velocity near the

periphery of the dees.

The maximum velocity is, qBR

vm

where R is the radius of the trajectory at exit and equals the radius of the dee.

The maximum kinetic energy of the ions will be 21

2K mv

21

2

qBRm

m

2 2 2

2

q B R

m

Uses :

1. It is used to bombard nuclei with energetic particles, so accelerated by it and

study the resulting nuclear reactions.

2. It is also used to implant ions into solids and modify their properties or even

synthesise new materials.

3. It is used in hospitals to produce radioactive substances which can be used in

diagnosis and treatment.

Limitations :

1. To accelerate very light particles like electrons, A.C. of very high frequency

(GHZ) is required

2. Size of Dee is not small.

3. It is not very easy to maintain uniform magnetic field over such a large region.

4. Neutrons, which is electrically neutral, cannot be accelerated in a cyclotron.

Important Point :

In a cyclotron, it is the

electric field which

accelerates the charged

particles. The magnetic field

does not change the speed.

It only makes the charged

particle to cross the same

electric field again and again

by making it move along a

circular path.

As the magnetic force on a

charged particle acts


velocity, it does not do any

work on the particle. As a

result the K.E of the particle

does not change due to the

magnetic force.

Maximum K.E of the

accelerated charged particle

is, 2 2 2

2max

q B Rk

m

MQ : Write uses and

limitations of Cyclotron.


Short Questions

1. Why cyclotron is placed in evaculated chamber ?

A. Cyclotron is kept in an evaculated chamber in order to avoid the possible

collision of charged particle with the air molecules.

2. State the principle of a cyclotron.

A. The operation of the cyclotron is based on the fact that the time for one revolution

of an ion is independent of its speed or radius of its orbit, while the frequency of

revolution of the charged particle in a magnetic field is independent of its energy.

3. Two identical charged particles moving with same speed enter a region of

uniform magnetic field. If the one of these enters normal to the field

direction and the other enters along a direction at o

45 with the field. Find

the ratio of their angular frequencies.

A. Angular frequency is independent of angle so 1 2: 1:1

4. An α- particle and a proton enter in uniform magnetic field normally. If

both have equal linear momenta. Find the ration of their radii.

A. Radius mv p

rqB qB

1

2 2

p

p d

qr e

r q e

Curiosity Questions

1. Can we accelerate neutrons by a cyclotron? Give reason.

A. No, neutrons are electrically neutral. They cannot be accelerated by electric

fields or deflected by magnetic field.

2. Why is a cyclotron not suitable for accelerating electron? A. Electrons are very light particles. Even a small increase in the energy of the

electron increases its speed to a very large value. Due to high speed, the electrons

get quickly out of step with the oscillating electric field.


1. Two particles of masses M1 and M2 and having the equal electric charge are

accelerated through equal potential difference and then move inside a

uniform magnetic field, normal to it. If the radii of their circular paths are

R1 and R2 respectively find the ratio of their masses.

2. A proton, a deuteron ion and an α-particle of equal K.E perform circular

motion normal to a uniform magnetic field B. Find the relation between

their radii.

4.5 Magnetic field due to a current Element Biot-Savart law

All magnetic fields are due to the currents

(or moving charges) and due to intrinsic

magnetic moments of particles. The

relation between current and the magnetic

field it produces, is given by Biot-Savart’s

law.

Fig. Shows a finite conductor XY carrying

current I. Consider an infinitesimal

element dl of the conductor. The magnetic

field dB due to this element is to be

determined at a point P which is at a

distance r from it, and be the angle

MQ : State and explain Bolt-

Savarat law for magnetic field.


between dl and the displacement vector r .

Statement : The magnitude of the magnetic field d B is proportional to the

current I, the element length dl , and inversely proportional to the square of the

distance r.

Its direction is perpendicular to the plane containing dl and r .

According to statement,

2

sinIdldB

r

2

sinIdldB k

r

0

3

sin

4

Idl r

r

…… (1)

where 0

4k

constant of proportionality and

0 permeability of vacuum

74 10 .

T m

A

The above expression holds for vacuum medium

The vector form of above equation can be written as,

0

34

Idl rdB

r

……. (2)

The Biot-Savart law for the magnetic field has certain similarities as well as

differences with the coulomb’s law for the electrostatic field.

Some of these are :

(1) Both are long range, since both depend inversely on the square of distance

from the source to the point of interest. The principle of super position applies

to both fields.

(2) The magnetic field is linear in the source Idl just as the electrostatic field is

linear in its source, the electric charge.

(3) The electrostatic field is produced by a scalar source i.e. electric charge, while

the magnetic field is produced by a vector source i.e. .Idl

(4) The electrostatic field is along the displacement vector joining the source and

the field point. The magnetic field is perpendicular to the plane containing

the displacement vector r and the current element Idl .

(5) There is an angle dependence in the Biot-Savart’s law which is not present in

the electrostatic case. The magnetic field at any point in the direction of dl

(i.e. axial line of element) is zero. Along this line 0 sin 0 and

hence 0dB

Relation Between 0 0, and C :

We know

29

2

0

19 10

4

Nm

C and

7

0 4 10T m

A

Now, 0

0 0 044

Jean-Baptise Biot

(1774 - 1862)

He was French physicist,

astronomer and

mathematician who

established the reality of

meteorites, made an early

balloon flight and studied

the polarization of light. The

mineral BIOTITE was named

in his honor.

Felix Savart

(1791 - 1841)

He was a physicist,

mathematician who is

primarily known for Biot-

Savart law of

electromagnetism, which he

discovered together with his

colleague Biot. His main

intrest was in acoustics and

the study of vibrating bodies.

A particular intrest in the

violin led him to create an

experimental trapenzodial

model.

MQ : Write some points of

similarities and differences

between Biot-Savart law and

coulomb’s law.

MQ : Obtain relation between

0 0 and C.


7

9

1 4 10

9 10 4

2

8

1

3 10

2

1

c

where c = speed of light in vacuum

83 10 /m s

0 02

1

c

2

0 0

1c

0 0

1c

Short Questions

1. Why Biot-Savart’s law is inverse square law?

A. 0

2

sin

4

IdldB

r

2

1,dB

r hence we say that Biot-Savart’s law is inverse square law.

2. Write Si unit and value of 0μ .

A. 7

0 4 10T m

A

3. What is the unit of

0 0

1

μ ε?

A.

0 0

1

= c = velocity of light so unit is m/s.

4.6 Magnetic field on the axis of a circular current loop

Suppose radius of a circular

loop of wire is R and carrying

current I, as shown in figure.

The loop is placed in the Y-Z

plane with its centre at the

origin 0, the X-axis is the axis

of the loop. We wish to

calculate the magnetic field at

the point P on the X-axis. Let

x be the distance of P from the

centre O of the loop.

Consider a current element

dl of the loop. This is shown

in fig. According to Biot-

Savart’s law, the magnetic field due to dl is given by,

0

34

Idl rdB

r

0

3

sin

4

IdlrdB

r

Any element of the loop will be perpendicular to the displacement vector from

the element to the axial point. Here the element dl is in the Y-Z plane, whereas

the displacement vector r from dl to the axial point P is in the X-Y plane.

Hence, , 90dl r So,

Important Point :

Electric field

2

0

1

4

dqdE

r and

magnetic field

0

24

I dldE

r

Both fields are inversely

depends on square of the

distance form the source to

the point.

Both are long range fields

The magnetic field is

produced by current element

Id l while electric field by

electric charge (dq).

MQ : Using Biot-Savart law

obtain magnetic field on the

axis of a circular current loop.

Write its special cases and give

the rule for finding direction of

magnetic field.

Important Note : 1. Magnetic field at O is

0

4

IB

R


0

24

IdldB

r

0

2 24

Idl

x R

…… (1)

2 2 2( from fig r )x R

The direction of dB is perpendicular to the plane formed by dl and r . It has

two components.

1. xdB x - component = dB cos

2. ydB y - component (perpendicular component to x axis) = d B sin .

When the components perpendicular to the X- axis are summed over, they cancel

out and we obtain a null result. i.e. the dB component due to dl is cancelled

by the contribution due to the diametrically opposite dl element, shown in

figure. And their axial components will be in the same direction i.e. along OP

and get added up.

Hence, the net contribution along X-axis can be obtained by integrating

cosxdB dB over the loop.

So, cosxdB dB

0

2 2cos

4

Idl

x R

0

12 22 2 2

4

Idl R

x Rx R

( from fig.

1/2

2 2cos

R

x R

)

0

3/22 24

x

Idl RdB

x R

…… (2)

For total magnetic field at the point P line integration should be taken over the

circumference of the loop.

So,

0

3/22 24

x x

IRB dB dl

x R

0

3/22 2

24

IRR

x R

2dl R

2

0

3/22 22

x

IRB

x R

2

0

3/22 2

ˆ ˆ

2x

IRB B i i

x R

…… (3)

Special Cases :

(1) If a loop consisting N closely wound turns the magnetic field along x-axis is,

2

0

3/22 2

ˆ ˆ

2x

NIRB B i i

x R

(2) At the centre of the loop i.e. x = 0,

0ˆ ˆ2

x

IB B i i

R

2. Resultant magnetic field at O.

1 2B B B

2 2

1 2B B B

2 2

0 1 0 2

1 22 2

I I

R R

3. Variation of magnetic field

along the axis of a circular

loop.

X X = 0

B

X

Y

I1

I2

O


(3) For a point far away from the centre of the loop, as compared to its radius i.e.

x >>R, neglecting R2 in comparison to 2 ,x we get.

2

0

3ˆ ˆ

2x

NIRB B i i

x

Direction of the magnetic field :

The magnetic field lines due to a circular wire form

closed loops and are shown in fig. The direction of the

magnetic field is given by right-hand thumb rule

stated below:

“Curl the palm of your right hand around the circular

wire with the fingers pointing in the direction of the

current. The right-hand thumb given the direction of

the magnetic field”.

Short Questions

1. Find the magnetic field at a distance equal to radius of circular ring.

A. Here, x R

2

0

3/22 22

IRB

x R

2 2

0 0

3/2 3/22 2 22 2 2

IR IR

R R R

2

0 0

5/2 3 5/2

1

2 2

IR I

R R

i.e.

1B

R

2. An electron is moving around nucleus of an atom in an orbit of radius 8 Å

with frequency 1015 Hz. Find the magnetic field at the centre of an atom.

A. 0 0

2 2

I QB

R R T

0 1. ( )

2e

R T

7 19 15

10

4 10 1.6 10 10

2 8 10

24 10 T

3. A circular coil having an average radius of 6 cm has 1000 turns. A current

of 5A pass through it. Find the magnetic field at a point on its axis at 8 cm

from the centre.

A. Here, x = 8 cm = 8 10-2 m, R = 6 cm = 6 10-2 m

7 42

0

3/2 3/24 42 2

4 10 1000 5 36 10

2 36 10 64 102

IN RB

x R

436 10 T

4. Where is the magnetic field due to current through circular loop uniform? A. The centre of circular loop.

5. How does a current carrying coil behave like a bar magnet?

A. Because it possesses a magnetic dipole moment.

6. Two concentric rings are kept in the same plane. Number of turns in both

the rings is 20. Their radii are 40 cm and 80 cm and they carry currents of

0.4 A and 0.6 A respectively, in mutually opposite directions. The magnitude

of the magnetic field produced at their centre is …… T.


(a) 04μ (b) 02μ (c) 0

10μ

4 (d) 0

5μ

4

A. 0 1 0 2

1 2

1 22 2

NI NIB B B

R R

0 01 2

1 2

20 0.4 0.6

2 2 0.4 0.8

N I I

R R

0

10

4

7. A long wire carries a steady current. When it is bent in a circular form, the

magnetic field at its centre is B. Now if this wire is bent in a circular loop of

n turns, what is the magnetic field at its centre?

A. For one loop, 1N R R

0

2

IB

R

…… (1)

For n- circular loop from the same wire

' , 'R

N n Rn

So, 0 0''

2 ' 2 /

M N I nIB

R R n

2 20

2

In n B

R

8. As shown in the figure a circular conducting wire

carries current I. It lies in XY- plane with centre

at O. What the tendency of this circular loop?

Why? A. Expand, here on every small element loop, radial

magnetic field and centrifugal force is exerted.

Hence tension is produced in loop so loop is to

expand. (This can also be solved with the direction

of right hand thumb rule for circular loop).


1. A charge Q is uniformly spread over a disc of radius R made from non-

conducting material. This disc is rotated about its geometrical axis with

frequency . Find the magnetic field produced at the centre of the disc.

2. A spiral coil has N turns. Its inner and outer radii are a and b respectively.

Find magnetic field at its centre when electric current I is passed through it.

4.7 Ampere’s Circuital law

Gauss’s law is an alternating form of

Coulomb’s law in electrostatics. Similarly,

Ampere’s circuital law is an alternative form

of Bio-Savart’s law in electrodynamics.

Ampere’s circuital law considers an open

surface with a boundary (see figure). The

surface has current passing through it.

Consider the boundary to be made up of a

number of small line elements. Consider one

element of length dl. The value of the tangential component of the magnetic field,

Bt at this element and multiply it by the length of element dl. All such products

are added together. We consider the limit as the lengths of elements get smaller

and their number gets larger. The sum then tends to an integral.

MQ : State and explain

ampere’s circuital law.


Statement : “The line integral of magnetic induction over a closed surface in a

magnetic field is equal to the 0 times the total current passing through the

surface”.

i.e. 0B dl I …… (1)

Where I is the total current through the surface. The integral is taken over the

closed loop coinciding with the boundary C of the surface.

The relation above in equation (1) involves a sign-convention, given by the right-

hand rule.

Let the fingers of the right- hand be curled in the sense the boundary is traversed

in the loop integral .B dl Then the direction of the thumb gives the sense in

which the current I is regarded as positive, and the currents in the opposite

direction are considered negative.

For several applications, a much simplified version of equation (1) proves

sufficient we shall assume that, in such cases, it is possible to choose the loop

called an amperial loop such that at each point of the loop either.

(1) B is tangential to the loop and is a non-zero constant B or

(2) B is normal to the loop or

(3) B vanishes

Note that Ampere’s circuital law holds for steady currents which do not fluctuate

with time.

Short Questions

1. State Ampere’s circuital law and give its mathematical form. A. “The line integral of magnetic induction over a closed surface in a magnetic field

is equal to the 0 times the total current passing through the surface”.

i.e. 0B dl I

2. Write sign convention for electric current in Ampere’s circuital law.

A. The sign convention for electric current in Ampere’s circuital law given by the

right-hand rule. Let the fingers of the right hand be curled in the sense the

boundary is traversed in the loop integral .B dl Then the direction of the

thumb gives the sense in which the current I is regarded as positive.

4.7.1 Uses of Ampere’s circuital Law

Magnetic field due to a very long straight conductor carrying current :

Consider a very long (in principle infinitely long)

straight conductor carrying electric current I as

shown in figure.

Suppose L be the length (path) of the loop for

which B is tangential and eI be the current

enclosed by the loop. Hence using Ampere’s

circuital law

0B dl I

0 eBL I …… (1)

Important Note :

Ampere’s circuital law can

be derived from the Biot-

Savart’s law.

Ampere’s circuital law holds

for steady currents which do

not change with time.

Using Ampere circuital law

to find magnetic field for

solenoid and toroid becomes

much simpler.

The closed curve is called

Amperean loop which is a

geometrically entity and not

real wire loop.

MQ : Using Ampere’s circuital

law, obtain magnetic field due

to a very long straight

conductor carrying current.


When there is a system with a symmetry such as for a straight infinite current-

carrying wire in fig; the Ampere’s circuital law enables an easy evaluation of the

magnetic field.

The boundary of the lo-*op chosen is a circle and magnetic field is tangential to

the circumference of the circle i.e. B and dl are in the same direction at every

element so cos cos0 1 and for Amperial loop, is a circle concentric with

the cross-section. So for this loop 2L r ( 2 )dl L r

Hence equation (1) become,

02B r I

0

2

IB

r

…… (2)

Above equation is the magnetic field at a distance r outside the wire is tangential.

Magnetic field due to infinite long straight conductor carrying electric current is,

0

2

IB

r

…… (3)

The above result for the infinite wire is interesting from several points of view.

(1) It implies that the field at every point on a circle of radius r (with the wire along

the axis), is same in magnitude, i.e. magnetic field possesses a cylindrical

symmetry. The field that normally can depend on three co-ordinates depends

only on r, whenever three is symmetry, the solutions simplify.

(2) The field direction at any point on this circle is tangential to it. Thus, the lines of

anstant magnitude of magnetic field from concentric circles. These lines called

magnetic field lines form closed loops. The expression for the magnetic field of

a straight wire provides a theoretical justification to Oersted’s experiments.

(3) Even through the wire is infinite, the field due to it at a non-zero distance is not

infinite. It tends to blow up only when we come very close to the wire. The field

is directly proportional to the current and inversely proportional to the distance

from the current source.

(4) A simple right hand rule is use to determine the direction of the magnetic field

due to a long wire. “Grasp the wire in right hand with extended thumb pointing

in the direction of the current. The fingers will curl around in the direction of

the magnetic field”.

Short Questions

1. The magnetic induction at a point P which is at a distance 4 cm from a long

current carrying wire is 10-8 T. Find the magnetic induction at a distance 12

cm from the same current.

A. 0

2

IB

y

1 2

1 2

1 1B B

y y

2

2

1B

y

2 1

1 2

B y

B y

1

2 1

2

yB B

y

884 10

3.33 1012

T

Important Note : 1. Magnetic field for conductor of

finite length at a distance y is

01 2sin sin

4

IB

y

2. From above fig. magnetic field

at P point is

0

2 22 4

ILB

y y L

MQ : Write the important point

based on magnetic field due to

very long current carrying

wire. Write the rule to find

magnetic field for it.

A I O B X

Y P

L1 L2

y

θ1 θ2

L


2. Write the rule used to find the direction to magnetic field acting at a point

near a current carrying straight conductor.

A. Right hand rule: Grasp the wire in your right hand with your extended thumb

pointing in the direction of the current. Your figures will can around in the

direction of the magnetic field.

3. Write the expression for magnetic field due to infinitely long straight

current currying wire.

A. 0

2

IB

r

4. Draw the graph of B r for very long wire.

A.

4.8 The Solenoid and the Toroid

The solenoid and the toroid are two pieces of equipment which generate magnetic

fields. The television uses the solenoid to generate magnetic fields needed. The

synchrotron uses a combination of both to generate the high magnetic fields.

4.8.1 The solenoid

In practice long and short solenoids are used. When length of a solenoid is very

large as compared to its radius, the solenoid is called long solenoid.

“A long wire wound in the form of a helix where the neighbouring turns are

closely spaced. So each turn can be regarded as a circular loop, is called a

solenoid”.

The net magnetic field is the vector sum of the fields due to all the turns.

Enamelled wires are used for winding so that turns are insulated from each other.

Figure 4.17 displays the magnetic field lines for a finite solenoid we show a

section of this solenoid in an enlarged manner in fig. 4.17 (a). Figure 4.17 (b)

shows the entire finite solenoid with its magnetic field.

In fig. 4.17 (a), it is clear from the circular loops that the field between two

neighbouring turns vanishes. In fig. 4.17 (b), the field at the interior mid-point P

is uniform, strong and along the axis of the solenoid. While the field at the

Inside outside

MQ : Give a qualitative

discussion of magnetic field

produced by a straight

solenoid.


exterior mid-point Q is weak and moreover is along the axis of the solenoid with

no perpendicular components.

Fig. 4.18 represents, the solenoid is made longer it appears like a long cylindrical

metal sheet. The field outside the solenoid is zero. The field inside becomes

everywhere parallel to the axis and uniform.

Consider a rectangular Amperial loop abcd. Along cd the field is zero since it is

outside of solenoid. Along transverse sections i.e. bc and ad, the field component

is zero. (because B dl hence 0B dl ) Thus, these two sections make no

contribution.

Suppose the field along ab is B and the relevant length of the Amperial loop is

L = h

Now suppose that the number of turns per unit length of a solenoid is n.

Therefore, the number of turns passing through the Amperial loop is N = nh.

Current passing through each turn is I, so total current passing through the loop

is eI nIh

From Ampere’s circuital law,

0 eBL I

0B h nIh 0B nI

The direction of the field is given by the right-hand rule. The solenoid is

commonly used to obtain a uniform magnetic field.

Short Questions

1. What is solenoid?

A. A long wire wound in the form of a helix where the neighbouring turns are

closely spaced so each turns can be regarded as a circular loop is called a

solenoid.

2. How much is the flux density B at the centre of a long solenoid?

A. 0B nI

3. There are 100 turns per cm length in a very long solenoid. It carries a

current of 5A. Find the magnetic field the centre of it.

A. 0

0

NIB nI

h

72

2

4 10 100 56.28 10

10T

Important Note :

The magnetic field inside a

toroidal solenoid is

independent of its radius and

depends only on the current

and number of turns per unit

length.

Variation of magnetic field

along the axis of solenoid:

MQ : Calculate magnetic field

inside a long straight solenoid

using Ampere’s circuital law.

Important Note :

Magnetic field at the centre

of solenoid of finite length is

01 2sin sin

2

nIB

Magnetic field due to

straight solenoid

(a) at a point well inside the

solenoid 0B nI

(b) At either end of the

solenoid 0

1

2B nI

B

B12

O centre

end of

solenoid

end of

solenoid


4. What is long solenoid?

A. When length of a solenoid is very large as compared to its radius, the solenoid is

called long solenoid.

Curiosity Question

1. Why does a solenoid contract when a current is passed through it?

A. The current in adjacent turns of the solenoid flows in the same direction. So

different turns attract one another and the solenoid contracts.

4.8.2 The toroid

The toroid is a hollow circular ring on

which a large number of turns of a wire

are closely wound. (the shape of a toroid

is the same as that of an inflated tube, also

called doughnut shape). A solenoid bent

into the form of a closed ring is called a

toroidal.

The magnetic field in the open space

inside (i.e. point P) and exterior to the

toroid (i.e. point Q) is zero. The magnetic

field inside the toroid is constant in

magnitude for the ideal toroid of closely

would turns.

Figure 4.19 (b) is the sectional view of the

toroid. The direction of the magnetic field

inside is clock wise as per the right-hand

thumb rule for circular loops.

Three Amperian loops 1,2 and 3 are

shown by dashed lines. By symmetry the

magnetic field should be tangential to

each of them and constant in magnitude

for each of the loops.

The circular areas bounded by loops 2 and

3 both cut the toroid. So that each turn of

current carrying wire is cut once by the

loops 2 and twice by the loop 3.

For points in the open space inside to the toroid (i.e. along loop 1) :

Suppose1B is the magnitude of the magnetic field along loop 1, having radius 1.r

Length of the loop 1, 1 12L r However, the loop encloses no current so

0eI thus, applying Ampere’s circuital law,

1 1 0 eB L I

1 1 02 0B r 1 0B …… (1)

Thus, the magnetic field at any point P in the open space inside the toroid is zero.

For the points in the open space exterior to the toroid (i.e. along loop 3) :

Suppose the magnetic field along loop 3 is 3B . The radius of loop 3 is 3r .

Length of the loop 3 3 32L r

However, the current coming out of the plane of the paper is cancelled exactly

by the current going into it. So 0eI

Thus, applying Ampere’s circuital law,

LQ : Using Ampere’s circuital

law find the magnetic field both

inside and outside of a toroidal

solenoid.

Important notes :

In ideal toroid, the coils are

circular and magnetic field is

zero external to the toroid. In

a real toroid, the turns from

a helix and there is small

magnetic field external to

the field.

Toroids are expected to play

a key role in the TOKAMAK

device, which acts as a

magnetic container for the

fusion of plasma in

thermonuclear power

reactor.


3 3 0 2B L I

3 3 02 0B r 3 0B …… (2)

For the points inside the toroid (i.e. along loop 2) :

Suppose 2B be the magnitude of the magnetic field along the Amperian loop 2

of radius 2r .

Length of loop 2 2 22L r

If N is the total number of turns in the toroid and I the current in the toroid, so

the total current enclosed by the loop 2, eI NI

Applying Ampere’s circuital law, 2 2 0 eB L I

2 2 02B r NI 0

2

22

NIB

r

…… (3)

Suppose r be the average radius of the toroid and n be the number of turns per

unit length. Then,

N (average perimeter of the toroid) (number of turns per unit length)

2 rn

So, ,2

Nn

r thus equation (3) becomes

2 0B nI …… (4)

Now total magnetic field for toroid is 1 2 3 20 0B B B B B

2B

0B nI …… (5)

In an ideal toroid the coils are circular. In reality the turns of the toroidal coil

from a helix and there is always a small magnetic field external to the toroid.

Short Question

1. What is toroid? A. A solenoid bent into the form of a closed ring is called ring is called a toroid.

4.9 Force between two parallel currents, the ampere

Consider two long parallel

conductors “a” and “b”

separated by a distance “d”

and carrying (parallel)

currents aI and bI

respectively.

The conductor ‘a’ produces,

the same magnetic field aB at

all points along the conductor

‘b’. The right-hand rules

gives the direction of

magnetic field is downwards

(when the conductors are

placed horizontally) It is

given by,

Andre-marie

Ampere (1775-1836)

MQ : Obtain the force between

two parallel current carrying

wires Define 1 ampere.


0

2

aa

IB

d

……. (1)

The conductor ‘b’ carrying a current bI will experience a sideways force due to

the field aB . The direction of this force is towards the conductor ‘a’. The force

baF , which is on a segment L of ‘b’ due ‘a’. The magnitude of this force is,

aba bF I L B

sin90ba b BaF I L 0

2

a bI IL

d

…… (2)

Similarly to compute the force on ‘a’ due to ‘b’. i.e. ,abF on a segment of length

L of ‘a’ due to the current in ‘b’. It is equal in magnitude to baF and directed

towards ‘b’.

i.e. ba abF F

Note that currents flowing in the same direction attract each other while for

oppositely directed currents repel each other. Thus, parallel current attract,

and antiparallel currents repel.

Now, baf represent the magnitude of the force baF per unit length, then from

equation (2),

0

2

ba a bba

F I If

L d

…… (3)

Defination of Ampere :

In equation (3), when 1a bI I A and d = 1m we get,

7 10 2 10

2baf Nm

i.e. “the ampere is the value of that steady current which, when maintained in

each of the two very long, straight, parallel conductors of negligible cross-

section, and placed one metre a part in vacuum, would produce on each of these

conductors a force equal to 72 10 newtons per meter of length.

Short Questions

1. Define 1 ampere.

A. “The ampere is the value of that steady current which, when maintained in each

of the two very long, straight, parallel conductors of negligible cross-section, and

placed one metre a part in vacuum, would produce on each of these conductors

a force equal to72 10 newtons per meter of length.

2. What is the direction of force between two parallel wires carrying currents

in opposite directions?

A. Repel each other

3. The force between two parallel wire is F. If the current in each conductor is

doubled. What is the value of the force between them?

A. 0 1 2

2

I I lF

d

Now,

1 202 2

'2

I IF L

d

0 1 24

2

I I l

d

4F

4. Is the force between two parallel current-carrying wires affected by the

nature of the dielectric medium between them?

Important notes :

It was first observed by

Ampere in 1820 that two

parallel straight conductor

carrying currents in the same

direction attract each other

and currents in the opposite

direction repel each other.


A. The force is independent of the nature of the dielectric medium, so it does not

charge.

5. As shown in the figure two very long straight wires are kept parallel to each

other and 2 A current is passed through them in the same direction. In this

condition the force between them is F. Now if the current in both of them is

made 1 A and direction are reversed in both. Find force between them.

A. 0 01 2

2 2

LI I LF

d d

(4)

0 0

' '1 2'

2 2

I I L LF

d d

(1)

' ,4

FF attractive

6. Equal currents are passing through two very long and straight parallel

wires in mutually opposite direction. They will …… (Fill the blank)

A. attract each other

Curiosity Question

1. Two parallel wires carrying currents in the same direction attract each

other, while the two beams of electrons travelling in the same direction repel

each other-Give reason.

A. In case parallel wires only attractive magnetic interaction acts. While in case of

electron beams, the electrostatic repulsion is stronger than the attractive magnetic

interaction.

4.10 Torque on current loop, magnetic dipole

4.10.1 Torque on a rectangular current loop in a uniform magnetic

field

As shown in fig. 4.21 (a)

consider a rectangular

coil ABCD suspended

in a uniform magnetic

field B , with its axis


field.

Let I = Current flowing

through the coil ABCD

a,b = sides of the coil

ABCD

A = a.b = area of the coil

= angle between the

direction of B and

normal to the plane of

the coil.

Case I : Consider, when the uniform rectangular loop is placed such that the

uniform magnetic field B is in the plane of the loop. This is shown in fig. 4.21

(a) Here, / 2

LQ : Derive an expression for

the torque acting on a current

carrying loop suspended in a

uniform magnetic field with

two cases.

Important Point :

Torque on a planar current

loop depends on current,

strength of magnetic field

and area of the loop. It is

independent of the shape of

the loop.

For a planar current loop of

a given perimeter in a

magnetic field, the torque is

maximum when the loop is

circular in shape.


The field exerts no force on the two arm AD and BC of the loop. It is

perpendicular to the arm AB of the loop and exerts a force 1F on it which is

direction into the plane of the loop. Its magnitude is, 1F IbB

Similarly, it exerts a force 2F on the arm CD and 2F is directed out of the plane

of the paper. Its magnitude is 2 1F IbB F

Hence, the net force on the loop is zero, because 1F and 2F

are equal, opposite and collinear. So they constitute a

torque on the loop.

Fig. 4.21 (b) shows a view of the loop from the AD end. It

shows that the torque on the loop tends to rotate it

anticlockwise. This torque is (in magnitude).

1 2

2 2

a aF F

2 2

a aIbB IbB I ab B IAB …… (1)

Case II : Consider the case when the plane of the loop is not along the magnetic

field, but makes an angle with it. The angle between the field and the normal to

the coil to be angle . (see figure 4.22)

The forces on the arms BC and DA are equal, opposite and act along the axis of

the coil, which connects the centres of mass of BC and DA. Being collinear along

the axis they cancel each other, resulting in no net force or torque.

The force on arms AB and CD are 1F and 2F .

They are equal and opposite i.e.

1 2F F IbB , 90F Ib B

1F and 2F are not collinear, hence they produce (torque) couple. The torque is,

less than the case when plane of loop was along the magnetic field (i.e. = 90°).

Hence the perpendicular distance the forces of the couple has decreased.

Fig. 4.22 (b), shows the arrangement from the AD end and

it illustrates these two forces constituting couple.

The magnitude of the torque on the loop is,

1 2sin sin

2 2

a aF F

sin sin2 2

a aIbB IbB


sinI ab B

sinIAB A a b …… (1)

As → 0 the perpendicular distance between the forces of the couple also

approaches zero. This makes the forces collinear and the net force and torque

zero.

In vector notation above equation (1) can be written as,

IA B m B …… (2)

Here m IA magnetic moment of the current loop. and the direction of area

vector A is given by the right-hand thumb rule and is direction in to the plane of

the paper. The direction of the torque is such that it rotates the loop clockwise

about the axis of suspension.

Torque acting on a rectangular current loop in a uniforms magnetic field is,

m B …… (1)

Where m IA magnetic moment of the current loop.

The dimensions of the magnetic moment is 0 2 0 1M L T A and its unit is

2.Am

Direction of m can be determined using right hand screw rule. Keep a right

screw perpendicular to the plane of the coil and rotate it in the direction of

current, the direction in which screw advances give the direction of m .

From equation (1), we see that the torque vanishes when m is either parallel

or antiparallel to the magnetic field B . This indicates a state of equilibrium as

there is no torque on the coil.

When m and B are parallel the equilibrium is a stable one. Any small rotation

of the coil produces a torque which bring it back to its original position. When

they are antiparallel, the equilibrium is unstable as any rotation produces a torque

which increase with the amount of rotation.

If the loops has N closely wound turns, the magnetic moment becomes,

m NIA and the expression for torque is, NI A B m B

This is analogous to the electrostatic case i.e. ep E ,

where ep = electric dipole of dipole moment and E = an electric field.

Short Questions

1. Write an expression for the torque acting on a current carrying coil located

in a uniform magnetic field.

A. sinNAIB

2. Write an expression for the magnitude of torque acting on a current

carrying coil placed in a uniform radial magnetic field.

A. NAIB

3. Under which condition will current carrying loop not rotate in the magnetic

field? A. If the current loop is placed in a magnetic field, with its plane perpendicular to

the field then it will not rotate.

4. Write the unit and dimensional formula for magnetic dipole moment.

A. Unit : 2 0 2 0 1. .Am D F M L T A

MQ : Write the equation of

torque acting on a rectangular

current loop in uniform

magnetic field. And discuss

magnetic moment, also

compare it with electric dipole.

Important Note :

m B and

,ep E from this one

can say that, a current loop

is a magnetic dipole.

In non uniform magnetic

field, the net magnetic force

on a current is non-zero but

torque acting on it may be

zero or non-zero.


4.10.2 Circular current loop as a magnetic dipole

The magnetic field on the axis of a circular loop, of a radius R, carrying a steady

current I at a distance x on the axis of loop from centre is,

2

0

3/22 22

IRB

x R

The direction is along the axis and given by the right-hand thumb rule. Here, x is

the distance along the axis from the centre of the loop.

For x>>R, neglecting 2R in the denominator, we get,

2

0

32

IRB

x

2

0

32

I R

x

0

32

IA

x

2 area ofthe loopA R

0

32

m

x

magnetic momentm IA

0

3

2

4

mB

x

……(1)

Above expression is similar to an expression obtained earlier for the electric field

of a dipole. If we substitute 0 01/ , ,em p and B E we get,

3

0

2,

4

epE

x which is precisely

The field for an electric dipole at a point on its axis.

The electric field on the perpendicular bisector of the dipole is given by,

3

04

epE

x

If we replace 0

0

1,p m

and E B we obtain the result for B for a

point in the plane of the loop at a distance x from the centre. For x>>R,

0

34

mB

x

…… (2)

The results given by equations (1) and (2) become exact for a point magnetic

dipole. The results obtained above can be shown to apply to any planner loop; a

planner current loop is equivalent to a magnetic dipole of dipole moment

,m IA which is the analogue of electric dipole moment ep .

4.10.3 The magnetic dipole moment of a revolving electron

In the Bohr model, the electron revolves around a positively charged nucleus.

Consider the charge of electron is – e 191.6 10e C performs uniform

circular motion around a stationary heavy nucleus of the charge +Ze.

The current I associated with this orbiting electron is e

IT

…… (1)

Where T is the time period of revolution Suppose r be the orbital radius of the

electron, and v the orbital speed then,

MQ : Obtain equation of

magnetic field of circular

current loop in terms of

magnetic dipole.

Important notes :

If the current round any face

of the coil is in anticlockwise

it behaves like a north pole.

If currents flows in clockwise,

it behaves like a south pole.

MQ : Write note on magnetic

dipole moment of a revolving

electron.


distance 2

time

rv

T

2

,r

Tv

substitute this in eqn. (1),

2

evI

r …… (2)

The magnetic moment associated with this circulating current is .l

So, l I A

2

2

evr

r

1

2evr …… (3)

Multiplying and dividing the R.H.S of above expression by the mass of electron

i.e. me, we have

2

l e r

e

em v

m

2 e

el

m …… (4)

Here, l= magnitude of the angular momentum of the electron about the central

nucleus is also called orbital angular moment).

Vectorially, 2

l

e

el

m …… (5)

The negative sign indicates that the angular momentum of the electron is opposite

in direction to the magnetic moment. The direction of magnetic moment is in to

the plane of the paper in figure.

Instead of electron with charge (-e), if we had taken a particle with charge (+q)

the angular momentum and magnetic moment would be in the same direction.

From equation (5),

,2

l

e

e

l m

this ratio is called gyro-magnetic ratio, and its value is 8.8 1010

C/kg for an electron.

Any charge in uniform circular motion would have an associated magnetic

moment given by an expression to equation (5). This dipole moment is called the

orbital magnetic moment. Hence, the subscript ‘l’ in l .

We know the orbital magnetic moment for an electron is,

2l

e

el

m …… (1)

The fact that even at an atomic level there is a magnetic moment, confirms

Ampere’s hypothesis of atomic magnetic moment.

According to Bohr’s hypothesis, the angular momentum assumes a discrete set

of values, it is

2

nhl

…… (2)

Where n = natural number = 1, 2, …… and h is plank’s constant

= 6.626 10-34 J s. This condition of discreteness is called the Bohr quantisation

condition.

Now, to calculate the elementary dipole moment, take the value n = 1, hence

substitute (2) in (1)

min 2 2

l

e

e nh

m

Important Note :

The orbital magnetic

moment of an electron is

2l

l

el

m

and due to its

spinning motion is spin

magnetic moment or

intrinsic magnetic moment

is, 2

s

e

es

m

The total magnetic moment

of the electron is the vector

sum of these two momenta.

Orbital magnetic moment of

electron in nth orbit:

1

2l evr

2 e

el

m

4 e

ehn

m

MQ : Write equation of orbital

magnetic moment, hence obtain

eqn. for spin magnetic moment

and discuss.


4 e

neh

m

19 34

31

1 1.6 10 6.63 10

4 3.14 9.11 10

27 29.27 10 Am …… (3)

Where the subscript ‘min’ stands for minimum. This value is called the Bohr

magneton.

Besides the orbital moment, the electron has an intrinsic magnetic moment,

which has the same numerical value as given in equation (3). It is called the spin

magnetic moment.

The electron is an elementary particle and it does not have an axis to spin.

Nevertheless, it does possess this intrinsic magnetic moment.

Short Questions

1. Write a fundamental difference between an electric dipole and magnetic

dipole.

A. A fundamental difference :- an electric dipole is built up of two elementary units-

the charges; while a magnetic dipole is the most elementary element.

2. No magnetic monopoles are possible – Explain reason.

A. A current loop (i) produces a magnetic field and behaves like a magnetic dipole

at large distances, and (ii) is subject to torque like a magnetic needle. This led

Ampere to suggest that all magnetism is due to circulating currents and no

magnetic monopoles have been seen so far.

3. A conducting wire of L m length is used to form a circular loop. If it carries

a current of I A its magnetic moment will be …… Am2 (Fill the blank)

A. For circle, 2L r 2

Lr

and

area of circle

2 22

24 4

L LA r

Now

2

4

L II A

4. At each of the ends of a rod of length 2r, a particle of mass m and charge q

is attached. If this rod is rotated about its centre with angular speed . Find

the ratio of its magnetic dipole moment to the angular moment of this

particle.

A. Here, 2 2

2

q qvI

T r

2 rT

v

while area 2A r

So magnetic moment I A

22

2

qvr

r

qvr

Now total angular momentum of the system,

L = mvr + mvr =2mvr 2 2

qvr q

L mvr m

5. What is gyromagnetic ratio? Write its value and unit.

A. 2 e

e

mis called gyro-magnetic ratio value

108.8 10 /C kg

q q

m m

2r


4.11 The moving coil galvanometer

Figure 4.24 exhibits a very

useful instrument for detect

and measure small electric

currents; the moving coil

galvanometer (MCG).

The galvanometer consist of

a coil, with many turns, free

to rotate about a fixed axis,

in a uniform radial magnetic

field. There is a cylindrical

soft iron core which not

only makes the field radial

but also increases the

strength of the magnetic

field.

When a current flows

through the coil, a torque

acts on it. sinNAIB

Due to radial field, the angle

between A and B will

always be 90

NAIB …… (1)

which is called deflecting torque. (The torque due to which the coil is deflected).

A spring pS provides a counter torque i.e. k that balances the magnetic torque

NAIB; resulting in a steady angular deflection .

In, equilibrium, k NAIB …… (2)

Where k = torsional constant of the spring; i.e. the restoring torque per unit twist.

(unit = J/rad).

The deflection is indicated on the scale by a pointer attached to be spring.

from equation (2),

NAI

Ik

…… (3)

Since the quantity in brackets is a constant for a given galvanometer so, the scale

of a galvanometer can be appropriately calibrated to measure I by knowing .

Short Questions

1. What is use of soft iron cylindrical core in galvanometer?

A. A small soft iron cylindrical core is placed at the axis of the coil so that uniform

radial magnetic field is produced.

2. Write the unit of torsional constant of spring and define it.

A. Unit : J

rad “the restoring torque per unit twist is known as torsional constant.”

3. State the principle of a moving coil galvanometer.

A. The principle of a moving coil galvanometer is, a current carrying coil placed in

a magnetic field experiences a torque.

4. What is the nature of the magnetic field in a moving coil galvanometer?

A. Radial magnetic field.

MQ : Explain the principle

construction and working of a

moving coil galvanometer.

Important Note :

If the radial field were not

present in a moving coil

galvanometer, i.e. if soft iron

cylinder were removed then

the scale would then be non-

linear and difficult to

calibrate or to read

accurate.

As the coil is wound over a

metallic frame, the eddy

currents produced in the

frame bring the coil to rest

quickly.


5. Why soft iron cylindrical core used in moving coil galvanometer?

A. The cylindrical soft iron core which not only makes the field radius but also

increases the strength of the magnetic field.

6. Why should the spring/suspension wire in a moving coil galvanometer have

low torsional constant? A. Low torsional constant of spring ensures high sensitivity of the moving coil

galvanometer.

7. The coils in certain galvanometers, have a fixed core made of a non-

magnetic metallic material. Why does the oscillating coil come to rest so

quickly in such a core?

A. The eddy current is produced in the metallic material, which oppose the motion

of the coil in the magnetic field so it bring to rest.

Curiosity Questions

1. Why is the coil wrapped on a conducting frame in a galvanometer? A. Eddy currents set up in the conducting frame help in bringing the coil to rest at

once i.e. help in making the galvanometer dead beat.

2. What is the importance of radial magnetic field in a moving coil

galvanometer?

A. Radial magnetic field makes the arm of the couple fixed and hence the torque on

the coil is always same in all positions of the coil in the magnetic field. This

provides a linear current scale.

4.11.1 Measurement of electric current and potential difference

using galvanometer

To measure the parameters related to a circuit component like the electric current

passing through it and the potential difference across its two ends. The

instruments to measure these quantities are called an ammeter and a voltmeter

respectively. The basic instrument to measure electric current or the voltage the

galvanometer.

4.11.1 (A) Ammeter

A galvanometer has to be joined in series with

the component through which the electric current

is to be measured.

The galvanometer cannot as such be used as an

ammeter to measure the values of the current in

a given circuit. This is for two reasons:

(1) Galvanometer is a very sensitive device, it

gives a full-scale deflection for a current of the

order of A. (2) For measuring currents the

galvanometer has to be connected in series, and

as it has a large resistance, this will change the

value of the current in the circuit.

To overcome these difficulties, one attaches a small resistance rs called Shunt resistance, in parallel with the galvanometer coil; so that most of the current

passes through the shunt.

From the figure, the resistance of the arrangement is G se s

G s

R rR r

R r

here G sR r , so GR is neglected. If sr has small value, in relation to the

resistance of the rest of the circuit eR , the effect of introducing the measuring

MQ : Explain how can we

convert a galvanometer in to an

ammeter.


instrument is also small and negligible. This arrangement is shown in fig. 4.25.

The scale of this ammeter is calibrated and then graduated to read off the current

value with ease.

Formula for shunt : Suppose eR = resistance of galvanometer,

GI = the current capacity of galvanometer

I = maximum current passing in ammeter

(or the required current of the ammeter)

sr = shunt resistance

Here galvanometer and shunt are connected in parallel so,

P.D. across galvanometer = P.D. across shunt

G s sI G I r

Gs

s

G

G

I Gr

I

I G

I I

So by connecting shunt across the

given galvanometer, we get an

ammeter of desired range.

Short Questions

1. What is the resistance of an ideal Ammeter?

A. Zero

2. Define Shunt.

A. If one wants to convert a galvanometer in to an ammeter, then a resistance of

proper small value is joined in the parallel to the coil of galvanometer. This

resistance is called shunt.

3. If resistance of shunt is 0.05 Ω is connected across a galvanometer, what can

be said about the resistance of the resulting ammeter?

A. Less than 0.05 Ω

4. If one wants to increase the range of an ammeter from 1 mA to 4 mA. What

should be done to the shunt resistance?

A. The shunt should be reduced to that more

5. What is the effective resistance of ammeter?

A. G se

G s

R rR

R r

Curiosity Questions

1. Why should an ammeter have a low resistance?

A. An ammeter is connected in series in a circuit so the whole of the current, which

it is required to measure, passes through it. In order that it insertion in the circuit

does not effect the current in the circuit, the ammeter must have least resistance.

2. What happens when an ammeter is placed in parallel with a circuit?

A. When it is placed in parallel with circuit, the resistance decreases and current in

the circuit increases to a large extent. Moreover, it measure the current flowing

through it only and not the current in the circuit.

Important Note :

The resistance of an ideal

ammeter is zero. Higher the

range of ammeter to be

prepared from a

galvanometer lower is the

value of the shunt resistance

required for purpose.

The range of ammeter can

be increased but it cannot be

decreased.

MQ : Obtain formula for shunt.

RG IG

I

rs

Is

Is = I - IG

Ammeter


4.11.1 (B) Voltmeter

The galvanometer can also be used as a

voltmeter to measure the voltage across a

given section of the circuit. For this it must

be connected in parallel with that section of

the circuit.

Further, it must draw a very small current,

otherwise the voltage measurement will

disturb the original set up by an amount

which is very large.

To ensure this, a large resistance R is

connected in series with the galvanometer.

This arrangement is shown in fig. 4.26.

Now, the resistance of the voltmeter is, GR R R : large

The scale of the voltmeter is calibrated to read off the voltage value with ease.

Formula for the series resistance :

Suppose, GR = resistance of galvanometer

GI = the current capacity of galvanometer

V = maximum voltage between P and Q (or the required range of the voltmeter)

R = the required high series resistance

According to ohm’s law,

G GV I R R

G

G

VR R

I

G

G

VR R

I

So by connecting a high resistance R in series with the galvanometer, we get a

voltmeter of desired range.

Current sensitivity : We define the current sensitivity of the galvanometer as

the deflection per unit current

i.e. NAB

I k

…… (1)

A convenient way for the manufacturer to increase the sensitivity is to increase

the number of turns.

Voltage Sensitivity : We define the voltage sensitivity of the galvanometer as

the deflection per unit voltage.

By dividing both sides of equation (1) by the resistance of voltmeter R,

1NAB

IR k R

1NAB

V k R

…… (2)

If N = 2N then current sensitivity of galvanometer is double i.e.

'

2I I

However, the resistance of the galvanometer is also likely to double, since it

proportional to the length of the wire.

MQ : Explain how can we

convert a galvanometer in to a

voltmeter.

Important Point :

An ideal voltmeter should

have infinite resistance.

Higher the range of

voltmeter to be prepared

from galvanometer, higher

value of series high

resistance required for the

purpose.

MQ : Obtain formula for the

series resistance.

MQ : Explain current

sensitivity and voltage

sensitivity.

The rage of voltmeter can be

both increased or decreases.

Voltmeter

V

P Q IG


i.e. ' 2N N and ' 2R R the voltage sensitivity

'

V V

remains

unchanged.

So, in general, the modification needed for conversion of a galvanometer to an

ammeter will be different from what is needed for converting it into a voltmeter.

Short Questions

1. What is the resistance of an ideal voltmeter?

A. Infinite

2. Write the order of increasing resistance for voltmeter, ammeter and

galvanometer.

A. Ammeter < Galvanometer < voltmeter

3. Why should an ammeter have a high current capacity?

A. Because of high current capacity an ammeter is not damaged by excessive

current.

4. Why should a voltmeter have a low current carrying capacity?

A. Because of low current capacity, the voltmeter will draw only small part of the

total current.

5. Define voltage sensitivity and give its unit.

A. “The voltmeter sensitivity of the galvanometer is the deflection per unit voltage.”

Unit : radian/volt.

6. Define current sensitivity and give its unit.

A. “The current sensitivity of the galvanometer is the deflection per unit current.”

Unit : radian/ampere

7. Write two factors by which the current sensitivity of a moving coil

galvanometer can be increased.

A. (I) Increasing the number of turns in the galvanometer

(II) Decreasing the torsion constant of the spring.

8. Write two factors by which voltage sensitivity of a moving coil galvanometer

can be increased.

A. (I) Increasing the number of turns of the galvanometer.

(II) Decreasing the torsion constant of the spring.

9. If current sensitivity of moving coil galvanometer is 10 division/ mA and

voltage sensitivity is 20 division/ V. Find the resistance of the galvanometer.

A. Here 10 division/mAI

310 10 /div A 20 division/Vv

We know NAB

I k

and

NAB

V k R

3/ 10 10500

/ 20G

VR

V

Curiosity Question

1. Why is a voltmeter always connected in parallel with a circuit element

across which voltage is to be measured? A. A voltmeter is a high resistance galvanometer. When it is connected in parallel

across any element of a circuit, it draws a very small current from the main

circuit. Most of the current passes through that element.


1. There are 21 marks (zero to 20) on the dial of a galvanometer, that is there

are 20 divisions on passing 10 A current through it, it shows a deflection

of 1 division. Its resistance is 20 Ω. (a) How can it can be converted into an

ammeter which can measure 1A current? (b) How can the original

galvanometer be converted into a voltmeter which can measure a p.d of 1V?

Also find the effective resistance of the above meters.

Modern Physics Made Easy (Part - 1) : Class XII 4.36

Textbook Illustrations

4.1 A straight wire of mass 200 g and length 1.5 m

carries a current of 2 A. It is suspended in mid-

air by a uniform horizontal magnetic field B

(Fig. 4.3). What is the magnitude of the

magnetic field ?

Soln. : From Eq. Il B , we find that there is an upward

force F , of magnitude IlB. For mid-air suspension,

this must be balanced by the force due to gravity:

mg IlB

mg

BIl

0.2 9.8

0.652 1.5

T

Note that it would have been sufficient to specify m/l, the mass per unit length of the wire. The earth’s

magnetic field is approximately 4 × 10–5 T and we

have ignored it.

4.2 If the magnetic field is parallel to the positive

y-axis and the charged particle is moving along

the positive x-axis (Fig. 4.4), which way would

the Lorentz force be for (a) an electron

(negative charge), (b) a proton (positive

charge).

Soln. : The velocity v of particle is along the x-axis, while

B , the magnetic field is along the y-axis, so v B is

along the z-axis (screw rule or right-hand thumb rule).

So, (a) for electron it will be along –z axis. (b) for a

positive charge (proton) the force is along +z axis.

4.3 What is the radius of the path of an electron

(mass 9 × 10-31 kg and charge 1.6 × 10–19 C)

moving at a speed of 3 ×107 m/s in a magnetic

field of 6 × 10–4 T perpendicular to it? What is

its frequency? Calculate its energy in keV.

(1 eV = 1.6 × 10–19 J).

Soln. : Using /r mv qB

31 7 1

19 4

9 10 3 10

1.6 10 6 10

kg ms

C T

226 10 26m cm

6 1/ 2 2 10v r s

62 10 2 .Hz MHz

2 31

14 2 2

1/ 2 1/ 2 9 10

9 10 /

E mv kg

m s

1740.5 10 J

164 10 2.5J keV

4.4 A cyclotron’s oscillator frequency is 10 MHz.

What should be the operating magnetic field

for accelerating protons? If the radius of its

‘dees’ is 60 cm, what is the kinetic energy (in

MeV) of the proton beam produced by the

accelerator.

( -19 -27= 1.60×10 C, = 1.67×10 kg,

pe m

-131 MeV = 1.6×10 J )

Soln. : The oscillator frequency should be same as proton’s

cyclotron frequency.

Using Eqs. /r m qB and 2 /qB m

we have

27 17

19

6.3 1.67 10 102 /

1.6 10B m q

0.66T

Final velocity of protons is

7

7

2 0.6 6.3 10

3.78 10 / .

v r m

m s

2

27 14

13

1/ 2

1.67 10 14.3 10

2 1.6 10

E mv

7 MeV


4.5 An element ˆΔ Δl = xi is placed at the origin and

carries a large current I = 10 A (Fig. 4.10).

What is the magnetic field on the y-axis at a

distance of 0.5 m. Δ =1cmx .

Soln. : 0

2

sin

4

IdldB

r

210 , 10 , 0.5dl x m I A r m ,y

7

0 / 4 10T m

A

90 ;sin 1

7 2

8

2

10 10 104 10

25 10dB T

The direction of the field is in the +z-direction. This

is so since,

ˆˆ ˆ ˆ ˆdl r xi yj y x i j y xk

We remind you of the following cyclic property of

cross-products,

ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ; ;i j k j k i k i j

Note that the field is small in magnitude.

4.6 A straight wire carrying a current of 12 A is

bent into a semi-circular arc of radius 2.0 cm

as shown in Fig. 4.13(a). Consider the

magnetic field B at the centre of the arc. (a)

What is the magnetic field due to the straight

segments? (b) In what way the contribution to

B from the semicircle differs from that of a

circular loop and in what way does it

resemble? (c) Would your answer be different

if the wire were bent into a semi-circular arc

of the same radius but in the opposite way as

shown in Fig. 4.13(b) ?

Soln. : (a) dl and r for each element of the straight segments

are parallel.

Therefore, 0dl r . Straight segments do not

contribute to B .

(b) For all segments of the semicircular arc, dl rare all parallel to each other (into the plane of the

paper). All such contributions add up in magnitude.

Hence direction of B for a semicircular arc is given

by the right-hand rule and magnitude is half that of a

circular loop. Thus B is 41.9 10 T normal to the

plane of the paper going into it.

(c) Same magnitude of B but opposite in direction to

that in (b).

4.7 Consider a tightly wound 100 turn coil of

radius 10 cm, carrying a current of 1 A. What

is the magnitude of the magnetic field at the

centre of the coil?

Soln. : Since the coil is tightly wound, we may take each

circular element to have the same radius

R = 10 cm = 0.1 m. The number of turns N = 100. The

magnitude of the magnetic field is,

7 2

0

1

4 10 10 1

2 2 10

NIB

R

4 42 10 6.28 10 T

4.8 Figure 4.15 shows a long straight wire of a

circular cross-section (radius a) carrying

steady current I. The current I is uniformly

distributed across this cross-section. Calculate

the magnetic field in the region r < a and r > a.

Soln. : (a) Consider the case .r a The Amperian loop,

labelled 2, is a circle concentric with the cross-

section. For this loop,

2L r eI Current enclosed by the loop = I

The result is the familiar expression for a long straight

wire 02B r I 0

2

IB

r

…… (a)

1

B r ar


(b) Consider the case .r a The Amperian loop is a

circle labelled 1. For this loop, taking the radius of the

circle to be r,

2L r

Now the current enclosed eI is not I, but is less than

this value.

Since the current distribution is uniform, the current

enclosed is,

2 2

2 2e

r IrI I

a a

Using Ampere’s law, 2

0 22

IrB r

a

0

22

IB r

a

…… (b)

B r r a

Figure (4.16) shows a plot of the magnitude of B

with distance r from the centre of the wire. The

direction of the field is tangential to the respective

circular loop (1 or 2) and given by the right-hand rule.

This example possesses the required symmetry so that

Ampere’s law can be applied readily.

4.9 A solenoid of length 0.5 m has a radius of 1 cm

and is made up of 500 turns. It carries a

current of 5 A. What is the magnitude of the

magnetic field inside the solenoid ?

Soln. : The number of turns per unit length is,

500

1000 turns/m0.5

n

The length 0.5l m and radius 0.01 .r m

Thus, / 50l a i.e., .l a

Hence, we can use the long solenoid formula,

0B nI 7 34 10 10 5 36.28 10 T

4.10 The horizontal component of the earth’s

magnetic field at a certain place is 3.0 ×10–5 T

and the direction of the field is from the

geographic south to the geographic north. A

very long straight conductor is carrying a

steady current of 1A. What is the force per

unit length on it when it is placed on a

horizontal table and the direction of the

current is (a) east to west; (b) south to north?

Soln. : F Il B sinF IlB

The force per unit length is / sinf F l IB

(a) When the current is flowing from east to west,

90

Hence, f IB5 5 11 3 10 3 10 Nm

This is larger than the value 7 12 10 Nm quoted in

the definition of the ampere. Hence it is important to

eliminate the effect of the earth’s magnetic field and

other stray fields while standardising the ampere.

The direction of the force is downwards. This

direction may be obtained by the directional property

of cross product of vectors.

(b) When the current is flowing from south to north,

0 0f

Hence there is no force on the conductor.

4.11 A 100 turn closely wound circular coil of

radius 10 cm carries a current of 3.2 A.

(a) What is the field at the centre of the coil?

(b) What is the magnetic moment of this coil?

The coil is placed in a vertical plane and is free

to rotate about a horizontal axis which

coincides with its diameter. A uniform

magnetic field of 2T in the horizontal direction

exists such that initially the axis of the coil is in

the direction of the field. The coil rotates

through an angle of 90° under the influence of

the magnetic field.

(c) What are the magnitudes of the torques on

the coil in the initial and final position?

(d) What is the angular speed acquired by the

coil when it has rotated by 90° ? The moment

of inertia of the coil is 0.1 kg m2.

Soln. : (a) From Eq. 0

2

NIB

R

Here, 100; 3.2 ,N I A and 0.1 .R m Hence,

7 2 5

1 1

4 10 10 3.2 4 10 10

2 10 2 10B

using 3.2 10

32 10 T

The direction is given by the right-hand thumb rule.

(b) The magnetic moment is given by Eq.

2m NIA NI r

2 2100 3.2 3.14 10 10A m


The direction is once again given by the right-hand

thumb rule.

(c) m B sinmB

Initially, 0 . Thus, initial torque 1 0. Finally,

/ 2 (or 90º).

Thus, final torque 10 2 20f mB Nm

(d) From Newton’s second law,

sind

mBdt

where is the moment of inertia of the coil. From

chain rule,

d d d d

dt d dt d

Using this,

sind mB d

Integrating from 0 to / 2,

/2

0 0

sin

f

d mB d

2

/2

0cos

2

fmB mB

1/2 1/2

1

1

2 2 2020 .

10f

mBs

4.12 (a) A current-carrying circular loop lies on a

smooth horizontal plane. Can a uniform

magnetic field be set up in such a manner that

the loop turns around itself (i.e., turns about

the vertical axis).

(b) A current-carrying circular loop is located

in a uniform external magnetic field. If the

loop is free to turn, what is its orientation of

stable equilibrium? Show that in this

orientation, the flux of the total field (external

field + field produced by the loop) is

maximum.

(c) A loop of irregular shape carrying current

is located in an external magnetic field. If the

wire is flexible, why does it change to a

circular shape?

Soln. : (a) No, because that would require to be in the

vertical direction. But I A B , and since A of

the horizontal loop is in the vertical direction, would

be in the plane of the loop for any B .

(b) Orientation of stable equilibrium is one where the

area vector A of the loop is in the direction of

external magnetic field. In this orientation, the

magnetic field produced by the loop is in the same

direction as external field, both normal to the plane of

the loop, thus giving rise to maximum flux of the total

field.

(c) It assumes circular shape with its plane normal to

the field to maximize flux, since for a given

perimeter, a circle encloses greater area than any

other shape.

4.13 In the circuit (Fig. 4.27) the current is to be

measured. What is the value of the current if

the ammeter shown (a) is a galvanometer with

a resistance RG = 60.00 Ω; (b) is a

galvanometer described in (a) but converted to

an ammeter by a shunt resistance rs = 0.02 Ω;

(c) is an ideal ammeter with zero resistance?

Soln. : (a) Total resistance in the circuit is,

3 63 .GR

Hence, 3/ 63 0.048 .I A

(b) Resistance of the galvanometer converted to an

ammeter is,

60 0.02

0.0260 0.02

G s

G s

R r

R r

Total resistance in the circuit is,

0.02 3 3.02

Hence, 3/ 3.02 0.99I A

(c) For the ideal ammeter with zero resistance

3/ 3 1.00I A

Solutions to Textbook Exercise

4.1 A circular coil of wire consisting of 100 turns,

each of radius 8.0 cm carries a current of 0.40

A. What is the magnitude of the magnetic field

B at the centre of the coil?

Soln. : Here , N = 100, R = 8 cm = 8 10-2 m

I = 0.40 A, B = (?)

7

0

2

4 10 100 0.40

2 2 8 10

NIB

R

4

4

10

3.14 10

T

T


4.2 A long straight wire carries a current of 35 A.

What is the magnitude of the field B at a point

20 cm from the wire?

Soln. : Here, I = 35 A, r = 20 cm = 20 10-2 m, B = (?)

7

50

2

4 10 353.5 10

2 2 20 10

IB T

r

4.3 A long straight wire in the horizontal plane

carries a current of 50 A in north to south

direction. Give the magnitude and direction of

B at a point 2.5 m east of the wire.

Soln. : Here, I = 50 A (N to S), r = 2.5 m (E), B = (?)

0

7

6

2

4 10 50

2 2.5

4 10

IB

r

T

From fig,. current is in N to

S direction and at 2.5 m east of the wire, the direction

of the magnetic field using right hand thumb rule is in

upward direction.

4.4 A horizontal overhead power line carries a

current of 90 A in east to west direction. What

is the magnitude and direction of the magnetic

field due to the current 1.5 m below the line?

Soln. : Here, I = 90 A (E to W), r = 1.5 m

B = (?)

0

7

5

2

4 10 90

2 1.5

1.2 10

IB

r

T

From fig., current is in E to W direction and magnetic

field at 1.5 m below the line i.e. in S direction, so the

direction of the magnetic field using right hand thumb

rule in south direction.

4.5 What is the magnitude of magnetic force per

unit length on a wire carrying a current of 8 A

and making an angle of 30º with the direction

of a uniform magnetic field of 0.15 T?

Soln. : Here, I = 8 A, = 30, B = 0.15 T, Find F/l = (?)

F = IlB sin

Force per unit length i.e.

sinθF

IBl 8 0.15 sin30 = 0.6 Nm-1

4.6 A 3.0 cm wire carrying a current of 10 A is

placed inside a solenoid perpendicular to its

axis. The magnetic field inside the solenoid is

given to be 0.27 T. What is the magnetic force

on the wire?

Soln. : Here, l = 3 cm = 3 10-2 m, I =10 A,

B = 0.27 T, = 90, F = (?)

F = BIl sin = 0.27 10 3 10-2 sin 90

= 8.1 10-2 N.

4.7 Two long and parallel straight wires A and B

carrying currents of 8.0 A and 5.0 A in the

same direction are separated by a distance of

4.0 cm. Estimate the force on a 10 cm section

of wire A.

Soln. : 8 ,aI A 5 ,bI A 24 4 10d cm m

2 110 10 10 10 ,l cm m m F = (?),

Force on 10 cm section of wire A is,

0

2

a bI I lF

d

7 1

2

4 10 8 5 10

2 4 10

52 10 N

4.8 A closely wound solenoid 80 cm long has 5

layers of windings of 400 turns each. The

diameter of the solenoid is 1.8 cm. If the

current carried is 8.0 A, estimate the

magnitude of B inside the solenoid near its

centre.

Soln. : Here, 280 80 10l cm m , 8I A

Each layers of winding of 400 turns and there are 5

layers so total number of turns N = 5 400 = 2000

Now, 00

NIB nI

l

7

2

4 10 2000 8

80 10

3

2

8 10

2.5 10

T

T

4.9 A square coil of side 10 cm consists of 20 turns

and carries a current of 12 A. The coil is

suspended vertically and the normal to the

plane of the coil makes an angle of 30º with the

direction of a uniform horizontal magnetic

field of magnitude 0.80 T. What is the

magnitude of torque experienced by the coil?

Soln. : Here, A = 10 10 = 2100 cm

4 2 2 2100 10 10m m

N = 20, I = 12 A, = 30, B = 0.80 T

Magnitude of torque

sinθNAIB

220 10 12 0.80 sin30 0.96 Nm

I N

W E

S

r

I

N

W E

S


4.10 Two moving coil meters,

1M and 2

M have the

following particulars:

1 1

R = 10Ω, N = 30,

-3 2

1 1A = 3.6×10 m , B = 0.25 T

2 2

R = 14Ω, N = 42,

-3 2

2 2A = 1.8×10 m , B = 0.50 T

(The spring constants are identical for the two

meters). Determine the ratio of (a) current

sensitivity and (b) voltage sensitivity of M2 and

M1.

Soln. : Here,

3 2

1 1 110 , 30, 3.6 10 ,R N A m

1 0.25B T3 2

2 2 214 , 42, 1.8 10 ,R N A m

2 0.50 .B T

(a) Current sensitivity NAB

I k

2 2 2 2 2

1 1 11

1

Current sensitivity

Current sensitivity

N A B kI

k N A B

I

3

3

42 1.8 10 0.50

30 3.6 10 0.25

1.4

(b) Voltage sensitivity NAB

IR kR

2 2 2 2 1

2 1 1 11

Voltage sensitivity

Voltage sensitivity

N A B kR

kR N A B

3

3

42 1.8 10 0.50 10

30 3.6 10 0.25 14

10

1.414

1 ( Ratio of current sensitivity)

4.11 In a chamber, a uniform magnetic field of 6.5

G (1 G = 10–4 T) is maintained. An electron is

shot into the field with a speed of 4.8 × 106

ms–1 normal to the field. Explain why the path

of the electron is a circle. Determine the radius

of the circular orbit.

( -19 -31e = 1.6×10 C, = 9.1×10 kg

em )

Soln. : Here, 46.5 6.5 10B G T

6 194.8 10 / , 1.6 10v m s e C

319.1 10 ,em kg r = (?)

The magnetic force act normal to the direction of

motion thus provide the necessary centripetal force to

follow the circular path.

31 6

19 4

9.1 10 4.8 10

1.6 10 6.5 10

em vr

qB

24.2 10 4.2m cm

4.12 In Exercise 4.11 obtain the frequency of

revolution of the electron in its circular orbit.

Does the answer depend on the speed of the

electron? Explain.

Soln. : Frequency of revolution of the electron

19 4

31

1.6 10 6.5 10

2 2 3.14 9.1 10e

qB

m

618.2 10 Hz 18.2 MHz

4.13 (a) A circular coil of 30 turns and radius 8.0 cm

carrying a current of 6.0 A is suspended

vertically in a uniform horizontal magnetic

field of magnitude 1.0 T. The field lines make

an angle of 60° with the normal of the coil.

Calculate the magnitude of the counter torque

that must be applied to prevent the coil from

turning.

(b) Would your answer change, if the circular

coil in (a) were replaced by a planar coil of

some irregular shape that encloses the same

area? (All other particulars are also

unaltered.)

Soln. : (a) Here, N = 30, R = 8.0 cm = 8 10-2 m

I = 6 A, B = 1.0 T, = 60

Torque on coil sinNAIB

2 sinN R IB

2

230 3.14 8 10 6 1 sin 60

3.13 Nm

(b) No, the answer is unchanged because the formula

sinNAIB is true for a planar loop of any

shape.

Additional Exercises

4.14 Two concentric circular coils X and Y of radii

16 cm and 10 cm, respectively, lie in the same

vertical plane containing the north to south

direction. Coil X has 20 turns and carries a

current of 16 A; coil Y has 25 turns and carries

a current of 18 A. The sense of the current in

X is anticlockwise, and clockwise in Y, for an

observer looking at the coils facing west. Give


the magnitude and direction of the net

magnetic field due to the coils at their centre.

Soln. : For coil X :

2

1 116 16 10 , 20R cm m N , 1 16I A

Magnetic field at the centre of coil : X

0 1 1

1

12

N IB

R

7

2

4 10 20 16

2 16 10

44 10 T

Since the direction of current coil : X is anticlockwise,

so magnetic field is towards East.

For coil Y :

2

2 210 10 10 , 25R cm N , 2 18I A

Magnetic field at the centre of coil : Y

7

0 2 22 2

2

4 10 25 18

2 2 10 10

N IB

R

49 10 T

Since the direction of current in coil : Y is clockwise,

so magnetic field is towards west.

Net magnetic field

4 4

2 1 9 10 4 10B B B

4 35 10 1.6 10 ;T T towards west.

4.15 A magnetic field of 100 G (1 G = 10–4 T) is

required which is uniform in a region of linear

dimension about 10 cm and area of cross-

section about 10–3 m2. The maximum current-

carrying capacity of a given coil of wire is 15 A

and the number of turns per unit length that

can be wound round a core is at most 1000

turns m–1. Suggest some appropriate design

particulars of a solenoid for the required

purpose. Assume the core is not

ferromagnetic.

Soln. : Here, 4 2100 100 10 10B G T T

115 , 1000 turns I A n m

3 210 , ?A m N

Magnetic field inside a solenoid,

0B nI

2

7

0

107961 8000

4 10

BnI

We may take I = 10 A then n = 800

If we take length of solenoid l = 50 cm then

210 7961

50 10

N

398N turns 400 turns .

So length about 50 cm, radius about 4 cm, number of

turns about 400 and current about 10 A. These

particulars are not unique. Some adjustment with

limits is possible.

4.16 For a circular coil of radius R and N turns

carrying current I, the magnitude of the

magnetic field at a point on its axis at a

distance x from its centre is given by,

2

0

3/22 2

μ IR NB =

2 + Rx

(a) Show that this reduces to the familiar

result for field at the centre of the coil.

(b) Consider two parallel co-axial circular

coils of equal radius R, and number of turns

N, carrying equal currents in the same

direction, and separated by a distance R. Show

that the field on the axis around the mid-point

between the coils is uniform over a distance

that is small as compared to R, and is given by,

,0μ NIB = 0.72

R approximately.

[Such an arrangement to produce a nearly

uniform magnetic field over a small region is

known as Helmholtz coils.]

Soln. : (a) Magnetic field due to current carrying coil of

radius R having N turns at a distance x from centre is,

2

0

3/22 22

NIRB

x R

At the centre of the coil i.e. x = 0 so,

2

0

3/22

2

0

3

0

2 0

2

2

NIRB

R

NIR

R

NI

R

Observer

Coil : Y Coil : X


(b)

Consider two parallel co-axial circular coils of equal

radius R, and number of turns N, carrying equal

currents in the same direction and separated by a

distance R. Also consider a small region of length 2d

about the mid point O between the two coils.

Magnetic field at point P due to coil 1

2

0

1 3/22

222

NIRB

RR d

(along - PO2)

and at point P due to coil 2,

2

0

2 3/22

222

NIRB

RR d

(along - PO2)

Net magnetic field at point P will be,

1 2B B B

2

0

3/22

2 2

1

2

4

NIR

RR d Rd

3/2

22 2

1

4

RR d Rd

2

0

3/2 3/22

2

1 1

2 5 5

4 4

INR

RR Rd Rd

(neglecting 2d because d R )

2

0

3/2 3/2 3/2

2

1 1

5 4 42 1 1

4 5 5

NIR

d dR

R R

3/2 3/2 3/2

0 4 4 41 1

2 5 5 5

NI d d

R R R

(Using Binomial expansion i.e. (1+x)n = 1+nx

neglecting higher power terms of d/R)

3/2

04 6 6

1 12 5 5 5

d dNI

R R R

3/2

0 42

2 5

NI

R

3/2

04

5

NI

R

00.72 NIB

R

At Q point same magnetic field be possible.

4.17 A toroid has a core (non-ferromagnetic) of

inner radius 25 cm and outer radius 26 cm,

around which 3500 turns of a wire are wound.

If the current in the wire is 11 A, what is the

magnetic field (a) outside the toroid, (b) inside

the core of the toroid, and (c) in the empty

space surrounded by the toroid.

Soln. : 1r inner radius = 25 cm

2r outer radius = 26 cm

1 2

2

r rr

average radius

25 26

2

25.5 cm225.5 10 m

N = 3500, I = 11 A

(a) The field outside toroid is zero.

(b) The field inside the core of the toroid

00

2

NIB nI

r

7

2

4 10 3500 11

2 25.5 10

23 10 T

(c) Magnetic field in the empty space surrounded by

the toroid is zero.

4.18 Answer the following questions:

(a) A magnetic field that varies in magnitude

from point to point but has a constant

direction (east to west) is set up in a chamber.

A charged particle enters the chamber and

travels undeflected along a straight path with

Coil - 1 Coil - 2

Q P Q P


constant speed. What can you say about the

initial velocity of the particle?

(b) A charged particle enters an environment

of a strong and non-uniform magnetic field

varying from point to point both in magnitude

and direction, and comes out of it following a

complicated trajectory. Would its final speed

equal the initial speed if it suffered no

collisions with the environment?

(c) An electron travelling west to east enters a

chamber having a uniform electrostatic field

in north to south direction. Specify the

direction in which a uniform magnetic field

should be set up to prevent the electron from

deflecting from its straight line path.

Soln. : (a) Initial velocity v is either parallel or antiparallel to

B .

If v is parallel to B then sin 0 0F qvB and if

v is antiparallel to B then sin 180 0F qvB

(b) Yes, because magnetic force can change the

direction of v , not its magnitude.

(c) According to fleming’s left hand rule, the

magnetic field must act in the vertically downward

direction.

4.19 An electron emitted by a heated cathode and

accelerated through a potential difference of

2.0 KV, enters a region with uniform magnetic

field of 0.15 T. Determine the trajectory of the

electron if the field (a) is transverse to its initial

velocity, (b) makes an angle of 30º with the

initial velocity.

Soln. : Here, 320 2 10 , 0.15V KV V B T

An electron accelerated by a p.d. 2.0 KV

21

2em v eV

19 3

31

2 2 1.6 10 2 10

9.1 10e

eVv

m

72.65 10 /m s

(a) When magnetic field transverse to the initial

velocity

2

sin90mv

qvBr

31 7

19

9.1 10 2.65 10

1.6 10 0.15B

mvr

q

310 1 .m mm

(b) When magnetic field makes an angle of 30 to the

initial velocity .v

2mv

qv Br

sin 30mv mv

rq B q B

31 7

19

9.1 10 2.65 10 0.5

1.6 10 0.15

550.2 10 0.50 .m mm

4.20 A magnetic field set up using Helmholtz coils

(described in Exercise 4.16) is uniform in a

small region and has a magnitude of 0.75 T. In

the same region, a uniform electrostatic field is

maintained in a direction normal to the

common axis of the coils. A narrow beam of

(single species) charged particles all

accelerated through 15 KV enters this region

in a direction perpendicular to both the axis of

the coils and the electrostatic field. If the beam

remains undeflected when the electrostatic

field is 9.0 × 10–5 Vm–1, make a simple guess as

to what the beam contains. Why is the answer

not unique?

Soln. : Here, B = 0.75 T, V = 15 KV = 15 103 V

E = 9 105 V/m

Speed of charged particles

5

59 1012 10 /

0.75

Ev m s

B

The beam is accelerated through 15 KV then,

21

2mv qV

2 5 2

3

(12 10 )

2 2 15 10

q v

m V

74.8 10 /C kg

Now for deutrons,

charge of deutrone

= mass of deutrone 2

e

m

19

27

1.6 10

2 1.67 10

74.79 10

74.8 10 /C kg

Deuterion ions or deutrons; the answer is not unique

because only the ratio of charge to mass is

determined, other possible answers are ,He Li

etc.


Here, the particles may be deutrons, each of which

contains one proton and one neutron.

4.21 A straight horizontal conducting rod of length

0.45 m and mass 60 g is suspended by two

vertical wires at its ends. A current of 5.0 A is

set up in the rod through the wires.

(a) What magnetic field should be set up

normal to the conductor in order that the

tension in the wires is zero?

(b) What will be the total tension in the wires

if the direction of current is reversed keeping

the magnetic field same as before?

(Ignore the mass of the wires.) g = 9.8 m s–2.

Soln. : Here, 0.45 ,l m 5.0I A

360 60 10m g kg

(a)

Tension in the wires will be zero when the weight of

the rod is balanced by the upward force 1BIl F of

the magnetic field.

1 2F F

BIl mg

360 10 9.8

5 0.45

mgB

Il

0.26 T

According to Fleming’s left hand rule, the magnetic

field should be applied normally into the plane of

paper so as to exert an upward magnetic force on the

rod.

(b) If the direction of current is reversed, the magnetic

force will act in downward direction. So, total tension

T BIl mg mg mg

2T mg

32 60 10 9.8T 1.176 N

4.22 The wires which connect the battery of an

automobile to its starting motor carry a

current of 300 A (for a short time). What is the

force per unit length between the wires if they

are 70 cm long and 1.5 cm apart? Is the force

attractive or repulsive?

Soln. : Here, 1 2 300 ,I I A

270 70 10l cm m

21.5 1.5 10d cm m

f force per unit length 0 1 2

2

I I

d

7

2

4 10 300 300

2 1.5 10

1.2 /N m (repulsive)

Now, per unit length force = 1.2 N

So, for 270 10l m length force = (?)

21.2 70 10F

0.84N

4.23 A uniform magnetic field of 1.5 T exists in a

cylindrical region of radius 10.0 cm, its

direction parallel to the axis along east to west.

A wire carrying current of 7.0 A in the north

to south direction passes through this region.

What is the magnitude and direction of the

force on the wire if,

(a) the wire intersects the axis,

(b) the wire is turned from N-S to northeast-

northwest direction,

(c) the wire in the N-S direction is lowered

from the axis by a distance of 6.0 cm?

Soln. : Here, 1.5 ,B T 7I A

(a)

Length of wire of cylindrical region = diameter PQ of

cylindrical region

20 0.2l cm m

The magnetic field is in the direction East to West in

the cylindrical region.

Force on wire sin90F BIl l B

BIl

1.5 7 0.2 2.1 N

According to Fleming’s left hand rule, this force acts

in the vertically downward direction.

(b) When the wire turns from NS to NE and NW

direction, it makes angle with magnetic field.

Suppose length of magnetic field is 'l then

T

I I

F1

F2

l

N

S

ʘ West

East

Q P

I 10 cm


sin'

l

l

'sin

ll

So, force on wire

' 'sin sinsin

lF BIl BI

1.5 7 0.2 2.1BIl N

Force is in the vertically downward direction.

(c)

Now, the wire is lowered from the axis by 6 cm then

length of wire in cylindrical region is

" 10 6 16l cm 216 10 m

Force 2" " 1.5 7 16 10F BIl

2168 10 1.68 N

Force is in vertically downward.

4.24 A uniform magnetic field of 3000 G is

established along the positive z-direction. A

rectangular loop of sides 10 cm and 5 cm

carries a current of 12 A. What is the torque

on the loop in the different cases shown in

Fig.? What is the force on each case? Which

case corresponds to stable equilibrium?

Soln. : Here, 3000B G

43000 10 0.3T T

12I A , 10l cm , 5b cm

Area of loop 210 5 50A lb cm

4 250 10 m

Magnetic moment m I A

4 212 50 10 0.06 Am

Using right hand rule to various current loops, one can

decide the direction of magnetic moment.

(a) 2 ˆˆ0.06 , 0.3m i Am B k T

m B

ˆˆ0.06 0.3i k 2 ˆ1.8 10 j Nm

(b) 2 ˆˆ0.06 , 0.3m i Am B k T

m B 2 ˆ1.8 10 j Nm

(c) 2 ˆˆ0.06 , 0.3m j Am B k T

m B ˆˆ0.06 0.3j k

2 ˆ1.8 10 i Nm

(d) Dipole moment is at on 150 with +x direction so,

90 i.e. angle between m and B

2sin 1.8 10 sin90mB

21.8 10 Nm

(e) 2ˆ0.06 ,m k Am ˆ0.3B k T

m B ˆ ˆ0.06 0.3k k 0 Nm

(f) 2ˆ ˆ0.06 , 0.3m k Am B k T

m B ˆ ˆ0.06 0.3k k 0 Nm

4.25 A circular coil of 20 turns and radius 10 cm is

placed in a uniform magnetic field of 0.10 T

normal to the plane of the coil. If the current

in the coil is 5.0 A, what is the

(a) total torque on the coil,

(b) total force on the coil,

N

S

E W

θ

θ

10

cm

I

N

S

10 cm

10 cm

6 cm


(c) average force on each electron in the coil

due to the magnetic field?

(The coil is made of copper wire of cross-

sectional area 10–5 m2, and the free electron

density in copper is given to be about

1029 m–3.)

Soln. : N 2 120, 10 10 10 10R cm m m

0.10 ,B T 5I A

(a) Torque,

sinNAIB

0 0Nm

(b) Magnetic force on

the opposite arms of

coil are equal and

opposite and act in the

same plane; hence the

total force on the coil

is zero.

(c) Force on each

electron,

dF qv B

qBI

nAq dI nAqv

29 5

0.10 5

10 10

BI

nA

255 10 N (This is only magnetic force)

4.26 A solenoid 60 cm long and of radius 4.0 cm has

3 layers of windings of 300 turns each. A 2.0

cm long wire of mass 2.5 g lies inside the

solenoid (near its centre) normal to its axis;

both the wire and the axis of the solenoid are

in the horizontal plane. The wire is connected

through two leads parallel to the axis of the

solenoid to an external battery which supplies

a current of 6.0 A in the wire. What value of

current (with appropriate sense of circulation)

in the windings of the solenoid can support the

weight of the wire? g = 9.8 m s–2.

Soln. : Length of solenoid 2

1 60 60 10 .l cm m

24 4 10r cm m

Layers 3, 300, (?)N I

Length of wire 2

2 2 2 10l cm m

Mass of wire 2.5m g32.5 10 kg

Current for solenoid 1 ?I

Current for wire 2 6I A

Due to current in the windings of the solenoid which

can support the weight of the wire. And the magnetic

force on the wire equals the weight of the wire.

Now, 2

300 31500

60 10n

turns / m

Magnetic field in the solenoid is 0 1B nI

and magnetic force act on wire,

1 2 2F BI l and weight force of wire 2F mg

So, 1 2F F

2 2BI l mg

0 1 2 2nI I l mg

3

1 7 2

0 2 2

2.5 10 9.8

4 10 1500 6 2 10

mgI

nI l

3

9

24.5 10108.36

226080 10A

4.27 A galvanometer coil has a resistance of 12 Ω

and the metre shows full scale deflection for a

current of 3 mA. How will you convert the

metre into a voltmeter of range 0 to 18 V?

Soln. : 33 3 10 ,I mA A 18 voltV

Resistance of galvanometer 12GR

Series resistance R = (?)

G

VR R

I

3

1812 6000 12

3 10

5988

We can connect this resistance in series with given

galvanometer, one can get a voltmeter of given range.

4.28 A galvanometer coil has a resistance of 15 Ω

and the metre shows full scale deflection for a

current of 4 mA. How will you convert the

metre into an ammeter of range 0 to 6 A?

Soln. : Resistance of galvanometer 15GR

34 4 10 , 6gI mA A I A

Resistance of shunt (?)sr

3

3

4 10 15

6 4 10

Gg

s

g

I Rr

I I

3 360 10 60 10 1000

4 59966

1000

0.01 10 m

We can connect shunt in parallel with given

galvanometer so one can get an ammeter of given

range.

I

I


Solutions to NCERT Exemplar

Multiple Choice Questions (MCQ I)

4.1 Two charged particles traverse identical

helical paths in a completely opposite sense in

a uniform magnetic field ˆ0

B = B k.

(a)They have equal z-components of momenta.

(b)They must have equal charges.

(c)They necessarily represent a particle-

antiparticle pair.

(d)The charge to mass ratio satisfy :

01 2

e e+ = .

m m

Soln. : (d) The charge to mass ratio satisfy :

1 2

0e e

m m

Here, e/m ratio of two particles is same and charges

on them are of opposite.

4.2 Biot-Savart law indicates that the moving

electrons (velocity v ) produce a magnetic field

B such that

(a) B v.

(b) B || v.

(c) it obeys inverse cube law.

(d) it is along the line joining the electron and

point of observation.

Soln. : (a) B v

For moving charge v is in the direction to the tangent

while magnetic is either going into the plane or

coming out from plane i.e. .B v

4.3 A current carrying circular loop of radius R is

placed in the x-y plane with centre at the

origin. Half of the loop with x > 0 is now bent

so that it now lies in the y-z plane.

(a) The magnitude of magnetic moment now

diminishes.

(b) The magnetic moment does not change.

(c) The magnitude of B at (0.0.z), z >> R

increases.

(d) The magnitude of B at (0.0.z), z >> R is

unchanged.

Soln. : (a) The magnitude of magnetic moment now

diminishes.

Magnetic moment in xy plane is = M (along z

direction)

When half of the current loop is bent in y-z plane the

magnetic moment due to half current loop in xy plane

become 1

2

MM (along z direction) and magnetic

moment due to half current loop in y-z plane i.e.

22

MM (along x- direction)

So resultant magnetic moment 1 2'M M M

2 2

2 2

1 2'4 4

M MM M M

2

2 2

M MM

i.e. magnetic moment diminishes.

4.4 An electron is projected with uniform velocity

along the axis of a current carrying long

solenoid. Which of the following is true?

(a) The electron will be accelerated along the

axis.

(b) The electron path will be circular about the

axis.

(c) The electron will experience a force at 45°

to the axis and hence execute a helical path.

(d) The electron will continue to move with

uniform velocity along the axis of the solenoid.

Soln. : (d) The electron will continue to move with uniform

velocity along the axis of the solenoid.

When electron is projected in a uniform magnetic

field then magnetic force on it is,

sin 0 0,F qvB so the electron will continue to

move along the axis of the solenoid.

4.5 In a cyclotron, a charged particle

(a) undergoes acceleration all the time.

(b) speeds up between the dees because of the

magnetic field.

(c) speeds up in a dee.

(d) slows down within a dee and speeds up

between dees.

Soln. : (a) undergoes acceleration all the time.

4.6 A circular current loop of magnetic moment

M is in an arbitrary orientation in an external

magnetic field B . The work done to rotate the

loop by 30° about an axis perpendicular to its

plane is

(a) MB (b) MB

32

(c) MB

2 (d) zero

Soln. : (d) zero


Rotation of loop about an axis perpendicular to its

plane does not change the angle between magnetic

moment and magnetic field. Hence, no work is done.

Multiple Choice Questions (MCQ II)

4.7 The gyro-magnetic ratio of an electron in an

H-atom, according to Bohr model, is

(a) independent of which orbit it is in.

(b) negative.

(c) positive.

(d) increases with the quantum number n.

Soln. : (a), (b)

For H-atom, gyro-magnetic ratio of an electron l

l

2

2

1

2e r

m r

2

e

m

It is negative and independent of the orbit of the

electron.

4.8 Consider a wire carrying a steady current, I

placed in a uniform magnetic field B

perpendicular to its length. Consider the

charges inside the wire. It is known that

magnetic forces do no work. This implies that,

(a) motion of charges inside the conductor is

unaffected by B since they do not absorb

energy.

(b) some charges inside the wire move to the

surface as a result of B .

(c) if the wire moves under the influence of

B , no work is done by the force.

(d) if the wire moves under the influence of

B , no work is done by the magnetic force on

the ions, assumed fixed within the wire.

Soln. : (b), (d)

Due to F q v B force, some charges inside the

wire move to the surface of wire.

When wire moves under the influence of magnetic

field, then displacement of the ions is perpendicular

to the magnetic field. So, 90 hence

cos90 0.W Fd

4.9 Two identical current carrying coaxial loops,

carry current I in an opposite sense. A simple

amperian loop passes through both of them

once. Calling the loop as C,

(a) 0B dl = m2μ I

(b) the value of B dl is independent of sense

of C.

(c) there may be a point on C where B and dl

are perpendicular.

(d) B vanishes everywhere on C.

Soln. : (b), (c)

According to Ampere circuital law,

0 0,B dl I I

so it is independent of sense of C.

There will be a point on loop C, at the axis of two

loops where .B dl

4.10 A cubical region of space is filled with some

uniform electric and magnetic fields. An

electron enters the cube across one of its faces

with velocity v and a positron enters via

opposite face with velocity - v . At this instant,

(a) the electric forces on both the particles

cause identical accelerations.

(b) the magnetic forces on both the particles

cause equal accelerations.

(c) both particles gain or loose energy at the

same rate.

(d) the motion of the centre of mass (CM) is

determined by B alone.

Soln. : (b), (c), (d)

Force on electron and positron in given electric field,

1F eE and 2 ,F eE i.e. both accelerates in

opposite directions.

Force on electron and positron in given magnetic

field,

2F e v B and acceleration

2

2

e v BFa

m m

…… (1)

While 2

'F e v B and acceleration

22

''

e v BFa

m m

…… (2)

Hence 2 2

'a a

Now speed of both the particles

2 eBRmv

evB vR m

So, K.E. = 21

2mV


2 2 2 2 2 2

2

1

2 2

e B R e B Rm

m m

Net electrical force on both particle

0eF eE eE while net magnetic force on

both particle mF e v B e v B

2e v B

So, the motion of the CM of both particles is

determined by B alone.

4.11 A charged particle would continue to move

with a constant velocity in a region wherein,

(a) E = 0, B≠ 0. (b) E≠ 0, B ≠ 0.

(c) E ≠ 0, B = 0. (d) E = 0, B = 0.

Soln. : (a), (b), (d)

When 0, 0,E B so both electrical and magnetic

force on particle will be zero, if v is parallel or

antiparallel to magnetic field, and then do not change

velocity when entering in magnetic field.

When 0, 0,E B by balancing both forces,

velocity of charged particle remains constant.

When 0, 0,E B charge particle passes with

constant velocity.

Very Short Questions Answers (VSA)

4.12 Verify that the cyclotron frequency ω = eB/m

has the correct dimensions of [T]–1.

Soln. : For a charge particle moving perpendicular to the

magnetic field, 2mv

qvBR

v qB

R m v R

So, v

R

0 1 1

0 0 1

0 1 0

M LTM L T

M LT

4.13 Show that a force that does no work must be a

velocity dependent force.

Soln. : Work done dW F ds F v dt

When 0dW i.e. 0F v dt

0F v

F must be velocity dependent which implies that

angle between F and v is 90. If v changes

(direction) then F (directions) should also change so

that above condition is satisfied.

4.14 The magnetic force depends on v which

depends on the inertial frame of reference.

Does then the magnetic force differ from

inertial frame to frame? Is it reasonable that

the net acceleration has a different value in

different frames of reference?

Soln. : Magnetic force is, F q v B

Magnetic force is frame dependent. Net acceleration

arising from this is however frame independent for

inertial frames.

4.15 Describe the motion of a charged particle in a

cyclotron if the frequency of the radio

frequency (rf) field were doubled.

Soln. : Particle will accelerate and decelerate alternatively.

So the radius of path in the dee’s will remain

unchanged.

4.16 Two long wires carrying current I1 and I2 are

arranged as shown in Fig. 4.1. The one

carrying current I1 is along is the x-axis. The

other carrying current I2 is along a line

parallel to the y-axis given by x = 0 and z = d.

Find the force exerted at O2 because of the

wire along the x-axis.

Soln. : We have F Il B

Here, ˆB B j and ˆl l j

ˆ ˆ 0F Il j B j

So at O2, the magnetic field due to I1 is along y-axis.

The second wire is along the y – axis and hence force

is zero.

Short Questions Answers (SA)

4.17 A current carrying loop consists of 3 identical

quarter circles of radius R, lying in the positive

quadrants of the x-y, y-z and z-x planes with

their centres at the origin, joined together.

Find the direction and magnitude of B at the

origin.


Soln. : Magnetic field due to quarter circle of radius R with

current I is, 01

4 2

IB

R

For x-y plane : 0

1

1 ˆ4 2

IB k

R

For y-z plane : 02

1 ˆ4 2

IB i

R

For z-x plane : 03

1 ˆ4 2

IB j

R

Net magnetic field at origin

1 2 3B B B B 01 ˆˆ ˆ

4 2

Ii j k

R

4.18 A charged particle of charge e and mass m is

moving in an electric field E and magnetic

field B . Construct dimensionless quantities

and quantities of dimension [T ]–1.

Soln. : No dimensionless quantity can be obtain from given

data. From given, 2mv

evBr

mr

eBr

0 0 1eBM L T

m

4.19 An electron enters with a velocity 0ν = ν i into

a cubical region (faces parallel to coordinate

planes) in which there are uniform electric and

magnetic fields. The orbit of the electron is

found to spiral down inside the cube in plane

parallel to the x-y plane. Suggest a

configuration of fields E and B that can lead

to it.

Soln. : The spiral path of the electron in a plane parallel to

the x-y plane and velocity is 0ˆ,v v i electric field is

given by 0ˆE E i . Magnetic field should be

perpendicular to x-y plane i.e. along z-axis. So

0ˆB B k (here

0 00, 0E B ). Hence from

F q v B electron revolves in the x-y plane.

4.20 Do magnetic forces obey Newton’s third law.

Verify for two current elements ˆ1

dl = dli

located at the origin and ˆ2

dl = dlj located at

(0, R, 0). Both carry current I. Soln. :

Force due to 2dl on

1dl is zero. Force due to 1dl on

2dl is non-zero. Hence, the magnetic forces do not

obey Newton’s third law.

4.21 A multirange voltmeter can be constructed by

using a galvanometer circuit as shown in Fig.

4.2. We want to construct a voltmeter that can

measure 2V, 20V and 200V using a

galvanometer of resistance 10Ω and that

produces maximum deflection for current of 1

mA. Find R1, R2 and R3 that have to be used.

Soln. : 31 10gI mA A , 10GR

Find 1 2,R R and

3R Using : G

g

VR R

I

At A : 1 3

210 2000

10R

1 2000 10

1990 1.99

R

K

(0,R,0)

(0,0,0)

A B C

Ig


At B :

1 2 3

2010 20000

10R R

2 20000 2000

18000 1.8

R

K

At C : 1 2 3 3

20010

10R R R

200000

3 200000 20000R

180000 180 K

4.22 A long straight wire carrying current of 25A

rests on a table as shown in Fig. 4.3. Another

wire PQ of length

1m, mass 2.5 g

carries the same

current but in the

opposite direction.

The wire PQ is

free to slide up

and down. To what height will PQ rise?

Soln. : Magnetic field due to long wire carrying current 25 A

rests on a table on small wire is,

0

2

IB

h

Now, magnetic force on small wire

2

0sin2

I lF BIl BIl

h

…… (1)

Force applied on PQ balance the weight of small

current carrying wire, F mg …… (2)

From (1) and (2),

2

0

2

I lmg

h

272

0

3

4 10 25 1

2 2 2.5 10 9.8

I lh

mg

451 10 0.51cm

Long Questions Answers (LA)

4.23 A 100 turn rectangular coil ABCD (in XY

plane) is hung from

one arm of a balance

(Fig. 4.4). A mass

500g is added to the

other arm to balance

the weight of the coil.

A current 4.9 A

passes through the

coil and a constant

magnetic field of 0.2 T acting inward (in xz

plane) is switched on such that only arm CD of

length 1 cm lies in the field. How much

additional mass ‘m’ must be added to regain

the balance ?

Soln. : 500 0.5m gm kg , 0.2B T

When magnetic field is off, 0. Taking the

separation of each hung from midpoint bl r

then, coilmgl W l

0.5 9.8 4.9coilW mg N

Now when magnetic field is switched on then, mass

m is added in a pan to balance the beam. So,

' 0

' sin90coilmgl m gl W l BIL l

'm gl BIL l

2

'

0.2 4.9 10

9.8

BILm

g

310 kg 1 gm

4.24 A rectangular conducting loop consists of two

wires on two opposite sides of length l joined

together by rods of length d. The wires are

each of the same material but with cross-

sections differing by a factor of 2. The thicker

wire has a resistance R and the rods are of low

resistance, which in turn are connected to a

constant voltage source V0. The loop is placed

in uniform a magnetic field B at 45° to its

plane. Find τ , the torque exerted by the

magnetic field on the loop about an axis

through the centres of rods. Soln. :


Consider the thicker wire has resistance = R while

other wire has resistance = 2R. The wires are each of

the same material but with cross-sections differing by

a factor of 2.

Wire I : 01 1

VF Bi l Bl

R and

01 1

2 2 2 2

V BlddF

R

Wire II : 02 2

2

VF Bi l Bl

R and

02 2

2 2 4 2

V BlddF

R

Net torque, 1 2 0 0

2 2 4 2

V Bld V Bld

R R

0 0

4 2 4 2

V Bld V BA

R R

( A =ld = area of loop.)

4.25 An electron and a positron are released from

(0, 0, 0) and (0, 0, 1.5R) respectively, in a

uniform magnetic field ˆ0

B = B i , each with an

equal momentum of magnitude BR= .p e

Under what conditions on the direction of

momentum will the orbits be non-intersecting

circles ? Soln. :

As ˆB Bi is along the x-axis, for a circular orbit the

momentum of the two particles are in the y-z plane.

Let 1p and

2p be the momentum of the electron and

positron respectively. Both of them define a circle of

radius R. Then small define circles of opposite sense.

Let 1P make an angle with the y axis 2P must make

the same angle. The centres of the respective circles

must be perpendicular to the momentum and at a

distance R. Let the centre of the electron be at Ce and

that of the positron at CP.

The co-ordinates of Ce is,

0, sin , coseC R R

The co-ordinates of CP is,

3

0, sin , cos2

PC R R R

The circles of the two small not overlap, if the

distance between two centres are greater than 2R.

Let d be the distance between CP and Ce then

2

22 32 sin 2 cos

2d R R R

2 2 2 2 2 294 sin 6 cos 4 cos

4R R R R

2 2 294 6 cos

4R R R

Since d has to be greater than 2R,

i.e. 2 24d R

So, 2 2 2 294 6 cos 4

4R R R R

2 296 cos

4R R

96cos

4

Or 3

cos8

4.26 A uniform conducting wire of length 12a and

resistance R is wound up as a current carrying

coil in the shape of (i) an equilateral triangle of

side a; (ii) a square of sides a and, (iii) a regular

hexagon of sides a. The coil is connected to a

voltage source V0. Find the magnetic moment

of the coils in each case. Soln. :

Ce

a

a

a

a a

a

a

a a

a

a a

a


(i) For equilateral triangle :

Side = a, n = 4 as the total wire of length = 12 a

magnetic moment of coils 12

43

an

a

1m nIA

234

4I a

23Ia 1

2A b h

(ii) For square :


12

34

an

a

magnetic moment 2m nIA 23I a

(iii) For regular hexagon :


12

26

an

a

magnetic moment

3m nIA 26 3

24

I a

( area of hexagon = 6 area of equilateral triangle)

23 3Ia

4.27 Consider a circular current-carrying loop of

radius R in the x-y plane with centre at origin.

Consider the line integral L

-L

L = B dl

taken along z-axis.

(a) Show that L monotonically increases

with L.

(b) Use an appropriate Amperian loop to show

that ℑ 0= μ I, where I is the current in

the wire.

(c) Verify directly the above result.

(d) Suppose we replace the circular coil by a

square coil of sides R carrying the same

current I. What can you say about ℑ (L)

and ℑ (∞)?

Soln. :

(a) B (Z) points in the same direction on z-axis and

hence ℑ (L) is a monotonically increasing function of

L.

Since cos0B dl Bdl Bdl as B and dl

along the same direction.

(b) ℑ (L) + contribution from large distance on

contour 0C I as L

Contribution from large distance 0 (as 3

1B

r)

0I

(c) Magnetic field due to circular coil of radius R is,

(here along z axis)

2

0

3/22 22

z

IRB

Z R

2

0

3/22 22

z

IRB dz dz

R Z

Let tanz R 2secdz R d

So,

/2 220

3/22 2 2

/2

sec2 tan

z

IRB dz R d

R R

/2 /2

0 0

/2 /2

1cos

2 sec 2

I Id d

002

2

II

(d) B (Z) square < B (Z) circular coil

ℑ (L) square < ℑ (L) circular coil

But by using arguments as in (b)

square circular coil

4.28 A multirange current meter can be

constructed by using a galvanometer circuit as

shown in Fig. 4.5. We want a current meter

that can measure

10mA, 100mA and

1A using a

galvanometer of

resistance 10Ω and

that produces

maximum deflection

for current of 1mA.

Find S1, S2 and S3

that have to be used.

Soln. : 10 ,GR 31 10gI mA A

2

1 10 10I mA A , L + L


2 100I mA 3100 10 0.1A A

3 1I A

For 1 10I mA

1 2 3 1g G gI R S S S I I

3 2 3

1 2 310 10 10 10S S S

1 2 3

10

9S S S …… (1)

Similarly for 2 100 0.1I mA A

1 2 2 3g G gI R S I I S S

3 3

1 2 310 (10 ) 0.1 10S S S

1 2 310 99S S S …… (2)

Similarly for 3 1I A

1 2 3 3g G gI R S S I I S

3 3

1 2 310 10 10 1000 1S S S

1 2 310 999S S S …… (3)

By solving (1), (2) and (3) we get

1 2 31 , 0.1 , 0.01S S S

4.29 Five long wires A,

B, C, D and E, each

carrying current I are arranged to

form edges of a

pentagonal prism

as shown in Fig.

4.6. Each carries

current out of the

plane of paper.

(a) What will be magnetic induction at a point

on the axis O ? Axis is at a distance R from

each wire.

(b) What will be the field if current in one of

the wires (say A) is switched off ?

(c) What if current in one of the wire (say) A is

reversed ?

Soln. : (a) The five wires A, B, C, D and E be perpendicular

to the plane of paper at locations as shown in figure.

So magnetic field due to five wires will be

represented by different sides of a closed pentagon in

one order, lying in the plane of paper. So its value

become zero.

(b) Since total vector sum of all magnetic field at O is

zero, hence magnetic field due to one current carrying

wire is equal to the magnitude of four wires but

opposite in direction.

Hence, if current in one of the wires is switched off

then

0

2

IB

R

Perpendicular to AO towards left.

(c) If current in A is reversed then total magnetic field

at O = magnetic field due to wire A + magnetic field

due to B, C, D, E wires.

0 (Perpendicular to AO towards left)2

I

R R

0 (Perpendicular to AO towards left)2

I

R

0 (Perpendicular to AO towards left)I

R

G

S1 S2 S3

I1- Ig

I1 Ig

chapter - 4 moving charges and magnetism · tesla (t). i.e. “when the force acting on a unit...

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