class xii physics chapter- 4 moving charges and …
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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