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. `(vBd)/(R)e^(-t//C)`
B. `(vBd)/(R)e^(-t//RC)`
C. `(vBd)/(R)e^(tRC)`
D. `(vBd)/(R)e^(t//RC)`

Answer 1

Correct 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, he 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=(dq)/(dt)=(Bdv)/(R)-(q)/(RC)or (q)/(RC)+(dq)/(dt)=(Bdv)/(R)`
or `q=vBdC+A e^(-t//RC)" " ("where A is contant") .....(i)`
At `t=0,q=0`
`:. =-vBdC`
From (i) `q=vBdC[1-e^(-t//RC)]`
`I=(dq)/(dt)=vBdCx(1)/(RC)e^(-t//RC)=(vBd)/(R)`