1.

Motion of a charged particle in a uniform magnetic field byjus

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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