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Answer» If a charge ‘q’ is moving with velocity v→ enters in uniform magnetic field B→, the magnetic force acting on the particle is given by,
FB = q→V →×B →
The direction of magnetic force is same as v→×B→if charge is positive and opposite if charge is negative.
Cases of Projection
Case I : If velocity of charge particle v→ is parallel to B→ then -
F = q V B Sin 0o ⇒ F = 0
Case II: If a charge enters a magnetic field at right angle to it, it moves in a circular path due to magnetic force which acts as a centripetal force.
Centripetal force is given by –
FB→ =
mV2r
Where,
m = Mass of the particle,
V = Linear velocity,
r = Radius of circular path.
As we know –
FB = qVB→
Put the value of FB →
∴ qVB =
mV2r.......(1) We can derive MULTIPLE relations from the above equation.
Velocity of charge
V =
qBrm
Observation: As magnetic field increases the velocity of charge increases.
Radius of circular path
r =
mVqB
Observation: To maintain the radius of circular path, as the magnetic field increases the velocity should also INCREASE.
Angular velocity
V =
qBrm
⇒ rω =
qBrm
∴ ω =
QBM
q/m is known as specific charge.
Observation: Angular velocity depends upon the specific charge and magnetic field.
Time period of rotation
ω =
2πT
∴ T =
2πmqB
Observation: As the magnetic field increases, the angular velocity increases and time period of rotation decreases.
Frequency
v =
1T
∴ v =
qB2πm
Observation: As the time period of rotation increases, the frequency of rotation decreases.
Kinetic ENERGY of the rotating charge
K.E =
12
mv2
∴ K.E =
q2B2r22m
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