Question

A conducting rod $PQ$ of length $l=1.0\text{m}$ is moving with a uniform speed $v=2\text{m/s}$ in a uniform magnetic field $B=4.0T$ directed into the paper. A capacitor of capacity $C=10\mu F$ is connected as shown in the figure. Then charge on plate $A$(in $\mu C$) is

Answer 1

According to Flemings right-hand rule induced current flows from $P$ to $Q$. Hence $P$ is at higher potential and $Q$ is at lower potential. Therefore $A$ is positively charged and $B$ is negatively charged.


Due to change in magnetic flux, potential difference generated in the loop is $V=Bvl$
Charge on capacitor is
$Q=CV=C\left(Bvl\right)=10\times {10}^{-6}\times 4\times 2\times 1=80\mu C$