Question

A circuit $\text{ABCD}$ is held perpendicular to the uniform magnetic field of $B=5\times {10}^{-2}\text{T}$ extending over the region $\text{PQRS}$ and directed into the plane of the paper. The circuit is moving out of the field at a uniform speed of $0.2{\text{ms}}^{-1}$ for $1.5\text{s}$. During this time, the current in the $5\Omega$ resistor is

• A. $0.6\text{mA}$ from $\text{B}$ to $\text{C}.$
• B. $0.9\text{mA}$ from $\text{C}$ to $\text{B}.$
• C. $0.6\text{mA}$ from $\text{C}$ to $\text{B}.$
• D. $0.9\text{mA}$ from $\text{B}$ to $\text{C}.$

Answer 1

The correct option is A $0.6\text{mA}$ from $\text{B}$ to $\text{C}.$

Since, the rod $\text{AD}$ is moving in the magnetic field, an emf will induce across it.

The induced emf across the rod will be equal to the potential difference across the resistor,

$E=Bv{l}_{AD}=5\times {10}^{-2}\times 0.2\times 0.3$

$=3\times {10}^{-3}\text{V}=3\text{mV}$

$\therefore$ The current flowing in the circuit is,

$i=\frac{E}{R}=\frac{3\times {10}^{-3}}{5}=0.6\text{mA}$

Area and flux are decreasing. So, according to the Lenz's law the current flows to increase the flux. Clearly, current should be clockwise.
Therefore, the current will flow from $\text{B}$ to $\text{C}$.

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