Question

A particle of mass m and charge q is moving in a circular path with constant speed V in a perpendicular uniform magnetic field B. Show that time T taken to complete one revolution is T=2m/qB

Answer 1

A particle of mass m and charge q is moving in a circular path with constant speed v in a perpendicular uniform magnetic field B as shown in figure .
force act on moving charged particle due to magnetic field B , Fm = Bqvsin90° = Bqv [because magnetic field and velocity of particle is perpendicular ]
now, at equilibrium ,
magnetic force, Fm = mv²/r
Bqv = mv²/r
r = mv/qB
now, time period , T = 2π/ω
also ω = v/r
ω = v/(mv/qB) = qB/m

now, Time period, T = 2π/(qB/m) = 2πm/qB
Hence, T = 2πm/qB

Answer 2

Let us assume the filed is directed into the page .
Then force experienced by the particle is F=qvB
We know that Force is always directed to velocity.
Hence particle moves along a circular path and the magnetic force on a charged particle acts like centripetal force.

Let r be the radius of circular path.
Centripetal Force=F=mV²/r
qvB=mV²/r
By solving we get :
r=mv/Bq
Time period=?
Time period= Distance/ speed
=2πr/v
=2πmv/Bqv
T=2πm/Bq