Question

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

Answer 1

Correct Answer - c
Using, impulse=change in linear momentum, we have
`intFdt=mv or int (iBl)dt=mv`
or `Blq=mv (as intidt=q)`
`:. q=(mv)/(Bl)`