A long solenoid is carrying a current $I=I_0 \sin (\omega t)$, having $N$ turns per unit length and radius…
A long solenoid is carrying a current $I=I_0 \sin (\omega t)$, having $N$ turns per unit length and radius $R$. A square loop is placed inside the solenoid with its plane perpendicular to the solenoid axis, and its corners touching the solenoid. Now the emf induced in the square coil.
$\mu_0 N I_0 R^2 \sin (\omega t)$
$2 \mu_0 N l_0 R^2 \sin (\omega t)$
$2 \mu_0 N I_0 \dot{R}^2 \omega \cos (\omega t)$
$\mu_0 N I_0 R^2 \pi \omega \cos (\omega t)$
Solution
Given, $I=I_0 \sin \omega t$
Magnetic field inside the solenoid, $B=\mu_0 N I$
$
\Rightarrow \quad B=\mu_0 N I_0 \sin \omega t
$
When side of square loop be $l$, then
$
\begin{array}{rlrl}
& & 4 l & =2 \pi R \\
\Rightarrow \quad & l & =\frac{\pi R}{2}
\end{array}
$
Area of square loop, $A=l^2$
$
=\left(\frac{\pi R}{2}\right)^2=\frac{\pi^2 R^2}{4}
$
Magnetic flux linked with square loop,
$
\begin{aligned}
\phi & =B A \\
& =\mu_0 N I_0 \sin \omega t \times \frac{\pi^2 R^2}{4} \\
\phi & =\frac{\mu_0 \pi^2 R^2 N I_0 \sin \omega t}{4}
\end{aligned}
$
Induced emf in the square coil, $e=\frac{d \phi}{d t}$
$
\begin{aligned}
& =\frac{d}{d t} \frac{\mu_0 \pi^2 R^2 N I_0 \sin \omega t}{4} \\
& =\frac{\mu_0 \pi^2 R^2 N I_0 \omega \cos \omega t}{4} \quad\left(\because \frac{\pi^2}{4}=246\right) \\
& =2,46 \mu_0 R^2 N I_0 \omega \cos \omega t
\end{aligned}
$
Which is closest to the value given in option (c)