Question

As a result of change in magnetic flux linked to the closed loop shown in figure, an emf $V$ volts is induced in the loop. The work done in taking a charge $Q$ coulomb once along the loop is,

• A. $QV$
• B. $2QV$
• C. $QV$
• D. $Zero$

Answer 1

The correct option is A $QV$

When a closed loop is placed in a varying magnetic field, then an electric field is induced in the closed loop, which is given by

$\oint \overrightarrow{E}\cdot \overrightarrow{dl}=-\frac{d{\varphi }_B}{dt}=E_{in}$

As per the question, $E_{in}=V$

Force on charge $Q$ due to induced electric field,

$\overrightarrow{F}=Q\overrightarrow{E}\Rightarrow \overrightarrow{E}=\frac{\overrightarrow{F}}{Q}$

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

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

$\Rightarrow W=VQ$

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