Question

A sliding rod AB of resistance R is shown in the 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 in the function of t, when switch S is closed at time t = 0 is

• A. $\left(\frac{vBd}{R}\right)e^{-t/C}$
• B. $\left(\frac{vBd}{R}\right)e^{-t/RC}$
• C. $\left(\frac{vBd}{R}\right)e^{RtC}$
• D. $\left(\frac{vBd}{R}\right)e^{t/RC}$

Answer 1

Answer: (B)

$\left(\frac{vBd}{R}\right)e^{-t/RC}$

Here magnetic field $\overrightarrow{B}$ 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)

At t = 0, q = 0
$\therefore A=-vBdC$
From $\left(i\right)q=vBdC[1-e^{-t/RC}]$

$I=\frac{dq}{dt}=vBdC\times \frac{1}{RC}{e}^{-t/RC}=\frac{vBd}{R}{e}^{-t/RC}$