Question

As a result of change in magnetic flux linked to the stationary closed loop placed perpendicular to the magnetic field as shown in the figure, an emf of magnitude $V$ is induced in the loop. The work done in moving a charge $Q$, once around the loop is -

• A. $QV$
• B. $2QV$
• C. $\frac{QV}{2}$
• D. Zero

Answer 1

The correct option is A $QV$

When a stationary closed loop is placed in a varying magnetic field, an electric field $\left(\overrightarrow{E}\right)$ is induced in it.

Also,

$\oint \overrightarrow{E}.\overrightarrow{dl}=-\frac{d{\varphi }_B}{dt}=E_{\text{induced}}$

Given, $E_{\text{induced}}=V...\left(i\right)$

The force on the charge $Q$, due to the induced electric field, is,

$\overrightarrow{F}=Q\overrightarrow{E}\Rightarrow \overrightarrow{E}=\frac{\overrightarrow{F}}{Q}...\left(ii\right)$

From equations $\left(i\right)$ and $\left(ii\right)$,

$\oint \frac{\overrightarrow{F}}{Q}.\overrightarrow{dl}=V$

$\Rightarrow \oint \overrightarrow{F}.\overrightarrow{dl}=QV$

$\therefore \text{Work done}=QV$

Hence, option $\left(A\right)$ is the correct answer.