Question

A conducting wire of length $\ell$ and mass $m$ can slide without friction on two parallel rails and is connected to capacitance $C$. Whole system lies in a magnetic field $B$ and a constant force $F$ is applied to the rod. Then

• A. the rod moves with constant velocity
• B. the rod moves with an acceleration of $\frac{F}{m+B^2{\ell }^{2}c}$
• C. there is constant charge on the capacitor
• D. charge on the capacitor increases with time
• E. answer required

Answer 1

Answer: (B), (D)

The correct options are
B the rod moves with an acceleration of $\frac{F}{m+B^2{\ell }^{2}c}$
D charge on the capacitor increases with time

Let at any instant, the rod is moving with velocity $v$ towards right due to force $F$and the acceleration of the wire be $a$ at that instant i.e $a=\frac{dv}{dt}$

Thus the EMF induced $E=Bvl$

$\therefore$ Charge flows across the loop $q=C\times E=CBvl$ .............(1)

The current flows in the loop in anti-clockwise direction $i=\frac{dq}{dt}=BlC\frac{dv}{dt}$

$⟹$ $i=BlCa$

The force $F^{\prime }=Bil$ acts towards left.

As $ma=F-Bil$

$ma=F-B\left(BlCa\right)l$

$⟹a=\frac{F}{m+B^2{l}^{2}C}$

As acceleration is constant, hence $v$ increases with time.

$\therefore$ Equation (1) implies that charge $q$ on the capacitor also increases with time.