Q.
A conducting rod $PQ\ of\ length\ L = 1.0$ m is moving with a uniform speed $v = 2.0\ ms^{- 1}$ in a uniform magnetic field = 4.0 T directed into the paper. A capacitor of capacity $C = 10\ \mu F$ is connected as shown in figure. Then,
A) $q_{A} = - 80\mu\ C\ and\ q_{B} = + 80\mu\ C$
B) $q_{A} = + 80\mu\ C\ and\ q_{B} = - 80\mu\ C$
C) $q_{A} = 0 = q_{B}$
D) Charge stored in the capacitor increases expotentially with time
Explanation
(b) Motional emf across $PQ$
$V = Blv = 4(1)\ (2) = \ 8\ $volt
This is the potential to which the capacitor is charged.
As $q = CV$
$\therefore\ \ q = \left( 10 \times 10^{- 6} \right)8 = 10^{- 5}C = 80\mu\ C$
As magnetic force on electron in the conducting rod $PQ$ is towards $Q,\ therefore,\ A$ is positively charged and $B$ is negatively charged
$ie,\ \ \ q_{A} = + 80\mu\ C\ \ \ and\ \ q_{B} = - 80\ \mu\ C\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.