JEE Main 2019 — Electromagnetic Induction Question with Solution

It is given that, the magnetic field,

$B=0.4\mathrm{s}\mathrm{i}\mathrm{n}\left(50\pi t\right)$

and the area of the given circular conducting loop,

$A=3.5\times {10}^{-2} {\mathrm{m}}^2$

Magnetic flux,

$\varphi =BA ...\left(1\right)$

Then, induced emf,

$\epsilon =\left|\frac{d\varphi }{dt}\right| ...\left(2\right)$

When resistance in the circuit is $R$, the differential charge accumulated, based on current $i$,

$$\begin{gathered} dQ=idt=\frac{\epsilon }{R}dt=\frac{1}{R}d\varphi \\ \Rightarrow ∆Q=\frac{1}{R} \int d\phi =\frac{1}{R} ∆\varphi \end{gathered}$$

Substituting given values and using equation $\left(1\right)$ and $\left(2\right)$,
$$\begin{gathered} ∆Q=\frac{1}{10}\left[0.4\sin \left(50\pi \times 10\times {10}^{-3}\right)-0\right]\times 3.5\times {10}^{-2} \\ \Rightarrow ∆Q=\frac{0.4\times 3.5\times {10}^{-2}}{10}=1.4\times {10}^{-3} \mathrm{C} \end{gathered}$$

$\Rightarrow ∆Q=1.4 \mathrm{mC}$