Question

A charge moving in a magnetic field B with velocity v right angle to the magnetic field .what is that value of time period (A) 2πm/qb (B) 2π/mqb
(C) 2πm/qb (D) 2πq/mb

Answer 1

Given info : a charge moving in a magnetic field B with velocity v right angle to the magnetic field.

To find : time period of charge particle.

solution : when velocity of charge particle is perpendicular on magnetic field, particle moves in circular path.

at an instant,

Centripetal force = magnetic force

⇒mv²/r = Bqv

⇒r = mv/qB .....(1)

Angular frequency, ω = v/r

From equation (1) we get, ω = v/(mv/qB)

⇒ω = 2π/T = qB/m

⇒T = 2πm/qB

Therefore time period of particle is 2πm/qB. So option (A) is correct choice.