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=-40muC, q_B=+40 mu C`
D. `q_A=q_B=0`

Answer

B

Explanation

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