Question

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

• A. $q_{A}=-80μC \, and \, q_{B}=+80μC$
• B. $q_{A}=+80 \, μC \, and \, q_{B}=-80 \, μC$
• C. $q_{A}=0=q_{B}$
• D. Charge stored in the capacitor increases expotentially with time

Answer 1

Answer: (B)

Motional emf across $PQ$
$V=Blv=4\left(1\right) \, \left(2\right)= \, 8 \, V$
This is the potential to which the capacitor is charged.
As $q=CV$
$\therefore \, \, q=\left(10 \times \left(10\right)^{- 6}\right)8=\left(10\right)^{- 5}C=80μC$
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}=+80μC \, \, \, and \, \, q_{B}=-80 \, μC \, $