Atomic Energy Education Society, Mumbai

CLASS XII PHYSICS

Chapter- 4

MOVING CHARGES AND MAGNETISM

Module - 1

By

Girish Kumar

PGT (Physics) AECS Narora

1) Lorentz magnetic force 2) Fleming left hand rule 3)Force on a current carrying conductor in a

uniform magnetic field 4)Motion of a charged particle in a magnetic

field 5)Velocity selector 6)Cyclotron

Moving Charges and Magentism

Content Module - 1

Lorentz Magnetic Force:

A current carrying conductor placed in a magnetic field experiences a force

which means that a moving charge in a magnetic field experiences force.

Fm = q (v x B)

or

F = (q v B sin θ) n

where θ is the angle between v and B

Special Cases:

i) If the charge is at rest, i.e. v = 0, then Fm = 0.

So, a stationary charge in a magnetic field does

not experience any force.

ii) If θ = 0°or 180°i.e. if the charge moves parallel

or anti-parallel to the direction of the magnetic

field, then Fm = 0.

iii) If θ = 90°i.e. if the charge moves perpendicular

to the magnetic field, then the force is

maximum.

Fm (max) = q v B

If the central finger, fore finger and thumb

of left hand are stretched mutually

perpendicular to each other and the

central finger points to current, fore finger

points to magnetic field, then thumb

points in the direction of motion (force)

on the current carrying conductor.

Fleming’s Left Hand Rule:

TIP:

Remember the phrase ‘e m f’ to represent

field and force in anticlockwise direction of the fingers of left hand.

Force on a moving charge in uniform Electric and Magnetic

Fields:

When a charge q moves with velocity v in region in which both electric

field E and magnetic field B exist, then the Lorentz force is

F = qE + q (v x B) or F = q (E + v x B)

Force on a current-carrying conductor in a uniform

Magnetic Field:

Force experienced by each electron in the conductor is f =

- e (vd x B)

If n be the number density of electrons, A

be the area of cross section of the

conductor, then no. of electrons in the

element dl is n A dl.

Force experienced by the electrons in dl is

dF = n A dl [ - e (vd x B)] = - n e A vd (dl X B)

where I = neAvd and -ve sign represents that

the direction of dl is opposite to that of vd) F = ∫ dF = ∫ I (dl x B)

= I (dl x B)

F = I (l x B) F = I l B sin θ

Motion of a charged particle in a magnetic field

As a magnetic field does not affect the

motion of a charged particle when

it is moving in the direction of

magnetic field.

Case1: - Circular path when the charged particle is moving perpendicular to

the field

When velocity of charged particle v is perpendicular to magnetic field B so

the force will be maximum( =qvB) and always directed perpendicular to

motion( and also to magnetic field); so the path will be circular ( with it's

plane perpendicular to the field) as in a circle velocity along tangent and

radius are always perpendicular to each other.

Here centripetal force is provided by the force qvB

mv2/r = qvB

r = mv/qB

Case 2:- Helical path when the charged particle is moving at

an angle to the field.( Other than 0 ,90 or 180) Consider a charged particle q entering a Uniform magnetic field B with velocity

v inclined at an angleθ with the direction of B,

The velocity v can be resolved into two rectangular components:

i)The v along the direction of the field i.e.,

along X-axis, vx = vcosθ The parallel

component remains unaffected by the

magnetic field and so the charged

particle continues to move along

the field with a speed of vcosθ

(ii) The component v perpendicular to the

direction of the field along Y- axis, vy= vsinθ

Due to this component of velocity, the charged particle experiences a force F=

qvB which acts perpendicular to both vsinθ and B. This force makes the particle

move along a circular path in Y-Z plane.

The radius of the circular path is r = mvsinθ/qB

The period of revolution is

T = 2πr/vsinθ = 2πmvsinθ /vsinθ qB = 2πm/qB

This a charged particle moving in a uniform

magnetic field has two concurrent motions:

a linear motion in the direction of B(along X-axis)

a circular motion In a plane perpendicular to B( in Y-Z plane)

Hence, the resultant path of the charged particle will be a helix, with it's axis along

direction of B.

Pitch of the Helix

It is the linear distance covered by charged particle in one rotation.

pitch =( vcosθ)T = vcosθ 2πm/Bq

Velocity selector A charge q moving with velocity v in the presence of both electric and magnetic fields

experiences a force F = q(E + vB)

F = FE + FB

Consider electric and magnetic fields are perpendicular

to each other and also perpendicular to the velocity of

the particle.

E = Ei , B = Bk, v = vj

Therefore F = q(E-vB)j

Thus, electric and magnetic forces are in opposite direction. If magnitude of electric

force and magnetic force are equal then charge will move in the field undeflected

qE = qvB

v = E/B

Thus, conditions can be used to select charged particles of particular velocity out of a

beam containing charges moving with different velocity.

The crossed field E and B therefore, serve as a velocity selector

Working: Imagining D1 is positive and D2 is negative, the + vely charged

particle kept at the centre and in the gap between the dees get accelerated

towards D2. Due to perpendicular magnetic field and according to Fleming’s

Left Hand Rule the charge gets deflected and describes semi-circular path.

When it is about to leave D2, D2 becomes + ve and D1 becomes – ve.

Therefore the particle is again accelerated into D1 where it continues to

describe the semi-circular path. The process continues till the charge

traverses through the whole space in the dees and finally it comes out with

very high speed through the window.

Theory:

The magnetic force experienced by the charge provides centripetal force

required to describe circular path.

(where m – mass of the charged particle,

q – charge, v – velocity on the path of

radius – r, B is magnetic field and 90°is the

angle b/n v and B)

Time taken inside the dee depends only on

the magnetic field and m/q ratio and not on

the speed of the charge or the radius of the

path.

If t is the time taken by the charge to describe the semi-circular path

inside the dee, then

If T is the time period of the high frequency oscillator, then for resonance,

If f is the frequency of the high frequency oscillator (Cyclotron

Frequency), then

B q f =

2πm

T = 2 t or 2πm

B q T =

mv2 / r = qvB sin 90°

B q r

m v =

π r

v or t =

B q

π m t =

Maximum Energy of the Particle:

Kinetic Energy of the charged particle is

The expressions for Time period and Cyclotron frequency only when

m remains constant. (Other quantities are already constant.)

If frequency is varied in synchronisation with the variation of mass of the

charged particle (by maintaining B as constant) to have resonance, then the

cyclotron is called synchro – cyclotron.

If magnetic field is varied in synchronisation with the variation of mass of

the charged particle (by maintaining f as constant) to have resonance, then

the cyclotron is called isochronous – cyclotron.

NOTE: Cyclotron can not be used for accelerating neutral particles. Electrons can

not be accelerated because they gain speed very quickly due to their lighter mass

and go out of phase with alternating e.m.f. and get lost within the dees.

But m varies with v according to

Einstein’s Relativistic Principle as per m =

m0

[1 – (v2 / c2)]½

Maximum Kinetic Energy of the charged particle is when r = R (radius of the D’s).

max K.E. = 1/2

B2 q2 R2

m

K.E. = ½ m v2 = ½ m ( B q r

)2

B2 q2 r2

m m = ½

Limitations of Cyclotron

(i) According to Einstein's special theory of relatively, the mass of a particle

increases with the increase in it's velocity as

At high velocities, the Cyclotron frequency will decrease due to increase in

mass. This will throw the particles out of resonance with the oscillating field.

That is because the, as ions reach the gap between the deed, the polarity of

the deed is not reversed at that instant. Consequently the ions are not

accelerated further.

(ii) Electrons cannot be accelerated in a cyclotron because a large increase in

energy increases their velocity to a very large extent. This throws the

electrons out of step with the oscillating field.

(iii) Neutrons being electrically neutral, cannot be accelerated in a Cyclotron.

Uses of Cyclotron

(i) The high energy particles produced in a Cyclotron are used to bombard

nuclei.

(ii) It is used to produce radioactive isotopes which are used in hospitals for

diagnosis and treatment.

m = [1 – (v2 / c2)]½

m0