Question

A solenoid of n turns per unit length is carrying sinusoidal current $i = i_0 \sin \omega t$. There is coil of area A, which rotates with constant angular velocity $\omega$ as shown. At t = 0, axis of coil coincides with axis of solenoid then flux of magnetic field through the coil at any instant of time is

Answer 1

$\frac{1}{2} \mu_0 n i_0 A \sin(2\omega t)$

Answer 2

The magnetic field inside the solenoid is $B(t) = \mu_0 n i(t) = \mu_0 n i_0 \sin \omega t$. The magnetic field is uniform and along the solenoid's axis. The coil rotates, and its area vector $\vec{A}(t)$ makes an angle $\theta = \omega t$ with the solenoid's axis at time $t$. The flux is $\Phi(t) = \vec{B} \cdot \vec{A} = B A \cos \theta$. Substituting the expressions for $B$ and $\theta$, we get $\Phi(t) = (\mu_0 n i_0 \sin \omega t) A \cos(\omega t)$. Using the identity $\sin(2\theta) = 2 \sin \theta \cos \theta$, we simplify this to $\Phi(t) = \frac{1}{2} \mu_0 n i_0 A \sin(2\omega t)$.