A conducting circular loop is placed in a uniform magnetic field $0.04 \mathrm{~T}$ with its plane…
A conducting circular loop is placed in a uniform magnetic field $0.04 \mathrm{~T}$ with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at $2 \mathrm{mms}^{-1}$. The induced emf in the loop when the radius is $2 \mathrm{~cm}$ is
$3.2 \pi \mu \mathrm{V}$
$4.8 \pi \mu \mathrm{V}$
$0.8 \pi \mu \mathrm{V}$
$1.6 \pi \mu \mathrm{V}$
Solution
Key Idea According to Faraday's second law of electromagnetic induction the induced emf is given by rate of change of magnetic flux linked with the circuit.
Here, $B=0.04 \mathrm{~T}$ and $\frac{-\mathrm{dr}}{\mathrm{dt}}=2 \mathrm{mms}^{-1}$
Induced emf, $\mathrm{e}=\frac{-\mathrm{d} \phi}{\mathrm{dt}}=\frac{-\mathrm{BdA}}{\mathrm{dt}}=-\mathrm{B} \frac{\mathrm{d}\left(\pi \mathrm{r}^2\right)}{\mathrm{dt}}$
$=-\mathrm{B} \pi 2 \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}$
Now, if $\mathrm{r}=2 \mathrm{~cm}$
$\begin{aligned}
\mathrm{e} & =-0.04 \times \pi \times 2 \times 2 \times 10^{-2} \times 2 \times 10^{-3} \\
& =3.2 \pi \mu \mathrm{V}
\end{aligned}$