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. $\frac{v\,Bd}{R}e^{-t/ c}$
• B. $\frac{v\,Bd}{R}e^{-t/ Rc}$
• C. $\frac{v\,Bd}{R}e^{ Rtc}$
• D. $\frac{v\,Bd}{R}e^{t/ Rc}$

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 $q=vBdc+Ae^{-t /Rc}$ (where $A$ is a constant) $...\left(i\right) $
At $t= 0 ,q = 0 $
$\therefore A-vBdC$
From $\left(i\right)$ $q =vBdC\left[1-e^{-t /RC}\right]$
$I=\frac{dq}{dt}=vBdC\times\frac{1}{RC}e^{-t /RC}$
$=\frac{vBd}{R} e^{-t /RC}$