A proton of mass 'm' moving with a speed v (`lt lt c`, velocity of light in vaccuum) 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)`

Answer

B

Explanation

`:.` The time period of the revolving charge is, `T= (2pi R)/(v)= (2pi)/(v) xx (mv)/(qB) or T= (2pi m)/(qB)` (Using `r= (mv)/(qB)`)
Here T is independent of v. So there is no change in the time period of proton if speed is increased to `sqrt2v`.