A conducting rod `PQ` of length `1 m` is moving with uniform velocity of `2 m//s` in a uniform magnetic field of `2T` directed inot the plane of paper. A capacitor of capacity `c = 10 mu F` is connected as shown. Then :

Options

A. `q_(A) = +40 mu C, q_(B) = +40 mu C`
B. `q_(A) = +40 mu C, q_(B) = -40 mu C`
C. `q_(A) = -40 mu C, q_(B) = +40 mu C`
D. `q_(A) = q_(B) = 0`

Answer

A

Explanation

`e_("induced") = Blv` and `q = Ce_("induced")` Apply Right hand rule to know direction of induced current.