Question

A conducting rod $PQ$ of length $1\, m$ is moving with a uniform speed $2\, ms^{-1}$ in a uniform magnetic field of $4\, T$ which is directed into the paper. A capacitor of capacity $10\, \mu F$ is connected as shown in the figure. Then the charges on the plates of the capacitor are

• A. $q_A = + 80\, \mu C ; q_B = - 80 \, \mu C $
• B. $q_A = - 80\, \mu C ; q_B = + 80 \, \mu C $
• C. $q_A = + 1.25\, \mu C ; q_B = - 1.25 \, \mu C $
• D. $q_A = - 1.25 \, \mu C ; q_B = + 1.25 \, \mu C $

Answer 1

Answer: (A)

Charge on each plate should be
where, $q =C V$
$V =v B l$
$\Rightarrow q =C(v B l) $
$=10 \times 10^{-6} \times 2 \times 4 \times 1 $
$q =80\, \mu C$
So, charges on plates are $\pm 80\, \mu C$.