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 $A B$ when $A B$ 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 \phi}{d t} $
For circuit Loop, $ \frac{d \phi}{d t}-i R-L \frac{d i}{d t}=0 $
$ \frac{d \phi}{d t}-\frac{d q}{d t} R-L \frac{d i}{d t}=0 $
$ \Rightarrow d q=\frac{d \phi-L d i}{R} $
$ \Delta q=\frac{\Delta \phi-L \Delta I}{R} $
$ =\frac{1}{R}\left[\left(\phi_{1}-\phi_{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] $