“Motion of a charged particle class 12 ncert
notes “ will be very beneficial for the students who are engaged in the
preparation of upcoming board exam and
competitive exam.
In this topic,
the following terms will be illustrated.
Contents :
* Motion of a charged
particle in a uniform electric field
* Motion of charged
particle in uniform magnetic field
* Lorentz force
Let Us Start
Motion of a charged particle in a uniform electric and magnetic field | Class 12 Physics Notes
Motion of a charged particle in a uniform electric field
* Consider that a charged particle of charge +q and mass m enters at right angles to a uniform
electric field of strength E with velocity v along OX –axis.
* The electric field is along OY-axis and acts over a
horizontal distance x.
* Since the electric field is along OY-axis, so only vertical force will act along OY .
* No horizontal force acts on the charged particle entering
the field, Therefore, the horizontal velocity v of the charged particle
remains the same throughout the journey.
* Vertical Force on the charged particle, Fy = qE
* Acceleration of the charged particle, along OY , a
= qE/m
* Time taken to traverse the field, t =
x / v
In same time t
, vertical distance covered by particle,
Where ,
* This is the equation of a parabola.
Therefore, inside the electric field, the charged particle follows a parabolic
path OA.
* As the charged particle leaves the electric field at A, it
follows a straight line path AB tangent to path OA at A.
Important point :
(i) When a charged particle moves perpendicular to a
uniform electric field, it follows a parabolic path in the electric field.
(ii) When a charged particle of charge +q is at rest
or moving in the direction of electric field, it will experience a force qE in
the direction of E. As a result, charged particle is accelerated in the
direction of E.
(iii) If the charged particle of charge +q is moving
opposite to the direction of E, it will experience a retarding force -q E and
hence the motion of the particle is retarded.
Motion of charged particle in uniform magnetic field
When a charged particle of charge +q and mass m moving
with velocity v enters a uniform magnetic field B, it experiences
a magnetic force F, given by
F = q (v x B)
* The magnetic force F acts at right angles to the
plane containing v and B.
* The magnitude of this force is F = qvB sinθ , where θ is angle between v and B .
Case 1- When the initial
velocity of the particle is parallel to the magnetic field
* In this case, θ = 00
So , F = qvB
sin0 = 0
* Thus in this case, the magnetic field does not exert any
force on the charged particle. Therefore , the charged particle will continue
to move parallel to the field.
Case 2 - When the initial velocity is anti parallel to the magnetic field
In this case , θ= 1800
So, F = qvB
sin1800 = 0
Therefore , the particle will continue to move in the
original direction.
Case 3- When the initial velocity of the
particle is perpendicular to the magnetic field.
In this case , θ= 900
So, F = qvB
sin900 = qvB
* In this condition, force acting on charge will be maximum.
So, Fm = qvB
* The direction of force is perpendicular to direction of
velocity.
Therefore, Force cannot change the speed of the
charged particle; it only changes the direction of motion of the charged
particle.
* As a result, the charged particle moves in a circle of
radius r perpendicular to the field.
(a) To find radius of circular path
* Centripetal force required to maintain circular motion = qvB
Thus, qvB =
mv2 / r
where , r= radius of circular path
Thus, r = mv2
/ qvB
-So, r
= mv / qB
* for given mass,
charge and magnetic field , r α
v
* it means, fast particle moves in large circle and slow
particle moves in small circle.
(b) To find
time period of circular motion
Time period ,
T= 2πr / v
T = ( 2π /v ) x ( mv / qB )
So , T=
2πm / qB
( c ) Frequency
of revolution,
f = 1/T
= 1 / 2πm / qB
so, f = qB / 2πm
(d) Angular
frequency,
ω =
2π / T
So, ω
= qB / m
* For constant uniform magnetic field, Time
period , frequency and angular frequency
of a charge do not depend on
speed (v ) and radius ( r).
* All charge particles with same q/m value and moving in a uniform
magnetic field will have same value of T
, f and ω .
Case 4- When the initial velocity of the charged particle
makes an angle θ to the direction of magnetic field.
* Suppose the charged particle moving with velocity v enters
in a uniform magnetic field B
making an angle θ to the direction of the magnetic field
.
* The velocity v can be resolved into two perpendicular
components
(a) v1
= v cosθ , acting
in the direction of the field.
(b) v2 = v sinθ , acting perpendicular to the direction of the
field.
* The component ( v2
= v sinθ
) moves the charged particle in a circular path perpendicular to the
field.
* The component ( v1 = v cosθ ) moves charge in the direction of the magnetic
field.
* Thus, the charged
particle covers circular path as well as linear path simultaneously .
* In other words , circular path advance in direction of
magnetic field with speed v1
= v cosθ . i.e the charged
particle follows a helical path .
* Parameters of Helical motion,
(a) Radius of
helical path
r =
mvsinθ / qB
(b) Time period
of circular motion,
T =
2πm / qB
(c) Frequency ,
f = qB
/ 2m
(d) Angular
Frequency,
ω = qB / m
Pitch of helix:
It is the linear distance covered by the charged
particle when it completes one circular revolution i.e. it is the linear
distance covered by the charged particle during time T.
So , Pitch of
helix,
d = velocity component along magnetic field x
Time period
ie. d = v1
x T
or, d = v cosθ. ( 2πm / qB)
Lorentz force
The Net force experienced by a moving charged particle
in a region of electric and magnetic
fields is called Lorentz force.
* Suppose a charge +q moves with velocity v in region of
both electric field E and magnetic field B.
* Electric field exert a force on charged particle. This force
acts in the direction of E and does not depend on the velocity of the
charged particle.
Electric force,
Fe = qE
* Magnetic field exert
a force on moving charged particle . This force acts perpendicular to the plane
containing v and B and depends on the velocity of the charged
particle.
Magnetic force,
Fm = q ( vx
B )
* The total force experienced by the charged particle due to
both electric and magnetic fields
F = Fe + Fm
( Vector sum of electric and
magnetic force )
Thus, F = qE +q ( V x B )
Or , F = q [ E + ( V x B ) ]
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