A particle of mass m and charge q is accelerated through a potential differece V to a velocity `vecv` towards south. The particle enters a region with both a magnetic field `vecB` (pointing eastwards) and electric field `vecE` (pointing downwards). The particle travels with a constant velocity through this region. The potential difference V through this region should be equal to

Options

A. `E//B`
B. `E//qB`
C. `2mE//qB`
D. `mE^2//2qB^2`

Answer

D

Explanation

`qV=1/2mv^2` or `v=sqrt(2qV//m)`
The charged particle will be moving with constant velocity in both, the magnetic field and electric field if, `qvB=qE` or `v=E//B`
or `sqrt(2qV//m)=E//B` or `V=(mE^2)/(2qB^2)`