A conducting rod `PQ` of length `l=1.0m` is moving with a uniform speed `v2.0m//s` in a uniform magnetic field `B=4.0T` directed into the paper.
A capacitor of capacity `C=10muF` is connected as shown in figure. Then

Options

A. `q_(A)=+80muC` and `q_(B)=-80muC`
B. `q_(A)=-80 muC and q_(B)=+80 muC`
C. `q_(A)=0=q_(B)`
D. Charge stored in the capacitor increase exponentially with time

Answer

A

Explanation

`Q=CV=C(Bvl)=10xx10^(-6)xx4xx2xx1=80muC`
According to fleming's right hand hand rule, induced current flows from `Q` to `P`. Hence, `P` is at higher potential and `Q` is at lower potential. Therefore, `A` is positively charge and `B` is negatively charged.