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 when switch S is closed at time t = 0 is

• A. $\frac{vBd}{R}e^{- t/C}$
• B. $\frac{vBd}{R}e^{- 1/RC}$
• C. $\frac{vBd}{R}e^{RtC}$
• D. $\frac{vBd}{R}e^{t/RC}$

Answer 1

Answer: (B)

$\frac{vBd}{R}e^{- 1/RC}$

Answer 2

Here magnetic field $\overset{\rightarrow}{B}$ is constant and is out of paper.

When the sliding rod AB moves with a velocity in the direction shown in 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 $q = vBdC + Qe^{- t/RC}$ (where A is a constant ….. (i)) At t = 0, q = 0

$\therefore A = - vBdc$

From (i) q = vBdC $\lbrack 1 - e^{- t/RC}\rbrack$

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

$= \frac{vBd}{R}e^{- t/RC}$