Question

A sliding rod $AB$ of resistance $R$ is shown in figure. Here magnetic field $B$ is constant and is out of the paper Parallel wires have no resistance and the rod is moving with constant velocity $v$. The current in the sliding rod $AB$ when switch $S$ is closed at time $t=0$ is

• A. $\frac{vBd}{R}{e}^{-t/c}$
• B. $\frac{vBd}{R}{e}^{-t/Rc}$
• C. $\frac{vBd}{R}{e}^{Rtc}$
• D. $\frac{vBd}{R}{e}^{t/Rc}$

Answer 1

Answer: (B)

Here magnetic field $\vec{B}$ is constant and is out of paper
When the sliding rod $AB$ moves with a velocity $v$ in the direction shown in the figure, the induced current in $AB$ is from $A$ to $B$.

As the switch $S$ is closed at time $t=0$ , the capacitor gets charged.
If $q$ is the charge on the capacitor, then
$I=\frac{dq}{dt}=\frac{Bdv}{R}-\frac{q}{RC}or\frac{q}{RC}+\frac{dq}{dt}=\frac{Bdv}{R}$
or $q=vBdc+A{e}^{-t/Rc}$ (where $A$ is a constant) $...\left(i\right)$
At $t=0,q=0$
$\therefore A-vBdC$
From $\left(i\right)$ $q=vBdC\left[1-e^{-t/RC}\right]$
$I=\frac{dq}{dt}=vBdC\times \frac{1}{RC}{e}^{-t/RC}$
$=\frac{vBd}{R}{e}^{-t/RC}$