Question

A conducting rod $PQ$ of length $L=1.0m$ is moving with a uniform speed $v=2.0m{s}^{-1}$ in a uniform magnetic field $=4.0T$ directed into the paper. A capacitor of capacity $C=10\mu F$ is connected as shown in the figure. Then,

• A. $q_A=-80\mu Cand{q}_B=+80\mu C$
• B. $q_A=+80\mu Cand{q}_B=-80\mu C$
• C. $q_A=0=q_B$
• D. Charge stored in the capacitor increases expotentially with time

Answer 1

Answer: (B)

Motional emf across $PQ$
$V=Blv=4\left(1\right)\left(2\right)=8V$
This is the potential to which the capacitor is charged.
As $q=CV$
$\therefore q=\left(10\times {\left(10\right)}^{-6}\right)8={\left(10\right)}^{-5}C=80\mu C$
As the magnetic force on an electron in the conducting rod $PQ$ is towards $Q,$ therefore, $A$ is positively charged and $B$ is negatively charged
$ie,q_A=+80\mu Cand{q}_B=-80\mu C$