Question

A metal bar $AB$ can slide on two parallel thick metallic rails separated by a distance $l$ . A resistance $R$ and an inductance $L$ are connected to the rails as shown in the figure. A long straight wire, carrying a constant current $I_0$ is placed in the plane of the rails as shown. The bar $AB$ is held at rest at a distance $x_0$ from the long wire. At $t=0$ , it is made to slide on the rails away from the wire.

It is observed that at time $t=T$ , the metal bar $AB$ is at a distance of $2x_0$ from the long wire and the resistance $R$ carries a current $i_1$ . The net charge that has flown through resistance $R$ from $t=0$ to $t=T$ is

• A. $\frac{1}{R}\left[\frac{{\mu }_{0}l}{2\pi }\ln (2)-L{i}_1\right]$
• B. $\frac{1}{R}\left[\frac{{\mu }_0{I}_{0}l}{\pi }\ln (2)-L{i}_1\right]$
• C. $\frac{1}{R}\left[\frac{{\mu }_0{I}_{0}l}{2\pi }\ln (2)-L{i}_1\right]$
• D. $\frac{1}{R}\left[\frac{{\mu }_0{I}_0}{2\pi }\ln (2)-L{i}_1\right]$

Answer 1

Answer: (C)

Let $v$ be velocity of $AB$ when $AB$ is at a distance $x$ from the wire,
$e_{AB}=B\cdot \ell \cdot v=\frac{{\mu }_0{I}_0}{2\pi x}I\cdot v=\frac{d\varphi }{dt}$
For circuit Loop, $\frac{d\varphi }{dt}-iR-L\frac{di}{dt}=0$
$\frac{d\varphi }{dt}-\frac{dq}{dt}R-L\frac{di}{dt}=0$
$\Rightarrow dq=\frac{d\varphi -Ldi}{R}$
$\Delta q=\frac{\Delta \varphi -L\Delta I}{R}$
$=\frac{1}{R}\left[\left({\varphi }_1-{\varphi }_2\right)-L{i}_1\right]$
$\Delta q=\frac{1}{R}\left[\frac{{\mu }_0\ell I_0}{2\pi }\ln (2)-L{i}_1\right]$