A conducting rod of mass m and length l is placed over a smooth horizontal surface. A uniform magnetic field B is acting perpendicular to the rod. Charge q is suddenly passed through the rod and it acquires an initial velocity v on the surface, then q is equal to

Options

A. `(2mv)/(Bl)`
B. `(Bl)/(2mv)`
C. `(mv)/(Bl)`
D. `(Blv)/(2m)`

Answer

C

Explanation

Using, impulse = change in linear momentum, we have
`intF dt=mv` or `int(iBl)dt=mv` or `Blq=mv` (as `intidt=q):.q=(mv)/B`