Question

A wire of mass $m$ and length $l$ can slide freely on a pair of smooth, vertical rails as shown in the figure. A magnetic field $B$ exists in the region in the direction perpendicular to the plane of the rails. The rails are connected at the top end by a capacitor of capacitance $C$. Find the acceleration of the wire, neglecting any electric resistance.

• A. $\frac{mg}{2m+C{B}^2{l}^2}$
• B. $\frac{2mg}{m+C{B}^2{l}^2}$
• C. $\frac{mg}{m-C{B}^2{l}^2}$
• D. $\frac{mg}{m+C{B}^2{l}^2}$

Answer 1

The correct option is D $\frac{mg}{m+C{B}^2{l}^2}$

Let, at any instant, the velocity of the rod is $v$.

Emf induced across the rod or capacitor is,

$E=Bvl$

Charge on capacitor is,

$q=CE=C\times Bvl=CBvl$

Current flowing through the circuit,

$i=\frac{dq}{dt}=\frac{d}{dt}\left(CBvl\right)$

$=CBl\frac{dv}{dt}=CBla$

Where, $a$ is the acceleration of the rod.

And the direction of current in the circuit will be in anticlockwise direction (from Lenz's law).


Force of magnetic field and gravity on the rod are opposite to each other. So,

$mg-ilB=ma$

$mg-\left(CBla\right)lB=ma$

$\Rightarrow mg-C{B}^2{l}^{2}a=ma$

$\therefore a=\frac{mg}{m+C{B}^2{l}^2}$

Hence, $\left(\text{D}\right)$ is the correct answer.