#### A magnetic field in a certain region is given by $B=B_{0}\cos \left ( \omega t \right )$ and a coil of radius a with resistance R is placed in the x-y plane with its centre at the origin in the magnetic field (figure). Find the magnitude and the direction of the current at (a, 0, 0) at$t=\frac{\pi }{2\omega },t=\frac{\pi }{\omega }\; and\; t=\frac{3\pi }{2\omega }$

$\phi _{m}=\vec{B}.\vec{A}=BA\; \cos \theta$

And as we know both $\vec{A}$ (area vector ) and $\vec{B}$  (magnetic field vector) are directed along z-axis. So, angle between them is 0.

So, $\cos \theta = 1 \left ( \because \theta =0 \right )$

$\Rightarrow \phi _{m}=BA$

Area of coil of radius  $a=\pi a ^{2}$

$\varepsilon =B_{0}(\pi a^{2})\; \cos \omega t$

By Faraday's law of electromagnetic induction, Magnitude of induced emf is given by

$\varepsilon =B_{0}(\pi a^{2})\; \omega \sin \omega t$

This causes flow of induced current, which is given by

$I=\frac{B_{0}(\pi a^{2})\; \omega \sin \omega t}{R}$

Now, the value of current at different instants,

(i) $t=\frac{\pi }{2\omega }$

$I=\frac{B_{0}(\pi a^{2})\omega }{R} along\; \hat{j}$

Because $\sin \omega t=\sin \left ( \omega \frac{\pi }{2\omega } \right )=\sin \frac{\pi }{2}=1$

(ii) $t=\frac{\pi }{\omega }$,$I=\frac{B(\pi a^{2})\omega }{R} =0$

because, $\sin \omega t=\sin \left ( \omega \frac{\pi }{2\omega } \right )=\sin \pi =0$

(iii) $t=\frac{3}{2}\frac{\pi }{\omega }$

$I = \frac{B(\pi a^{2})\omega }{R}\; along -\hat{j}$

$\sin \omega t=\sin \left ( \omega .\frac{3\pi }{2\omega } \right )=\sin \frac{3\pi }{2}=-1$