A charged particle moving in a uniform magnetic field penetrates a layer of lead and thereby loses one half of its kinetic energy. How does the time period of revolution of particle change.

Explanation

The radius of the circular path, `r=m v//qB`. If `E_K` is the kinetic energy of the particle, then
`E_K=1/2mv^2` or `v=sqrt(2E_K//m)`
`:.` radius, `r=(msqrt(2E_K//m))/(qB)=(sqrt(2mE_K))/(qB)`,
Time period `=(2pir)/(v)=(2pisqrt(2mE_K))/(qBsqrt(2K_K//m))=(2pim)/(qB)`
It means time period of revolution is independent of K.E. This shows that if KE is halved, there is no change in time period of revolution.