A conducting rod `PQ` of length `L = 1.0 m` is moving with a uniform speed `v = 2.0 m s^(-1)` in a uniform magnetic field `B = 4.0 T` directed into the plane of the paper.
A capacitor of capacity `C = 10muF` is connected as shown in , then

Options

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

Answer

A

Explanation

According to Fleming's right hand rule, P is at higher potential and Q is at lower potential. Therefore, A is positively charged and B is negatively charged.
Also, charge, Q = CV = C(Bvl)
`=10xx10^(-6)xx4xx2xx1=80muC`
`therefore" "q_(A)=80muC and q_(B)=-80muC`