A proton of mass m moving with a speed `v(ltltc`, velocity of light in vacuum) completes a circular orbit in time T in a uniform magnetic field. If the speed of the proton is increased to `sqrt2v`, what will be time needed to complete the circular orbit?

Options

A. `sqrt2T`
B. T
C. `(T)/(sqrt2)`
D. `(T)/(2)`

Explanation

The magnetic force on proton of mass m moving along a circular path in a uniform magnetic field,
`vecF_(m)=q(vecvxxvecB)`
`:.F_(m)=qvBsin90^(@)`
[B= magnetic field q=charge of proton]
`=qvB`
Since the proton is moving along a circular path of radius r,
`(mv^(2))/(r)=qvB:.v=(qBr)/(m)`
Now the time taken by the proton to make one complete revolution is `(2pir)/(v)`
Hence, `T=(2pir)/(v)or,T=(2pirxxm)/(qBr)or,T=(2pim)/(qB)`
Since T does not depend on the velocity of the proton, so if the velocity changes to `sqrt2v` then T will remain nuchanged.