Question

A conducting rod is moving with a constant velocity $v$ over the parallel conducting rails, which are connected to the ends through a resistor $R$ and capacitor $C$ as shown in the figure. Magnetic field $B$ exist in the region perpendicular to the plane of paper.


Consider the following statements:

$\left(i\right)$ Current in loop $\text{AEFBA}$ is anticlockwise.

$\left(ii\right)$ Current in loop $\text{AEFBA}$ is clockwise.

$\left(iii\right)$ Current through the capacitor is zero.

$\left(iv\right)$ Energy stored in the capacitor is
$\frac{1}{2}C{B}^2{L}^2{v}^2$.

Which of the following options is correct?

• A. Statements $\left(i\right)$ and $\left(iii\right)$ are correct
• B. Statements $\left(ii\right)$ and $\left(iv\right)$ are correct
• C. Statements $\left(i\right)$, $\left(iii\right)$ and $\left(iv\right)$ are correct
• D. None of these statements are correct.

Answer 1

The correct option is C Statements $\left(i\right)$, $\left(iii\right)$ and $\left(iv\right)$ are correct

Emf induced across rod $AB$ is given by

$E=Bvl=BvL[\because l=L]$

The equivalent circuit can be drawn as,


Therefore, the current in the loop will flow in anticlockwise direction.

Here, the potential difference across the capacitor is constant, hence no current will flow through the capacitor.

Energy stored in the capacitor is given by,

$E=\frac{1}{2}C{E}^2=\frac{1}{2}C\times (BvL{)}^2=\frac{C{B}^2{V}^2{L}^2}{2}$

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

$\text{Note}$ : When the wire is moving with the constant velocity, no current will flow through the capacitor.