$A$ magnetic field in a certain region is given by $\vec B = B_0 \cos(\omega t) \hat k$. $A$ coil of radius $a$ with resistance $R$ is placed in the $xy$-plane with its centre at the origin in the magnetic field (see 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 $t = \frac{3\pi}{2\omega}$.

(N/A) The magnetic flux linked with the coil is given by $\phi = \vec B \cdot \vec A = B A \cos(0^\circ) = B A$.
Substituting $B = B_0 \cos(\omega t)$ and $A = \pi a^2$,we get $\phi = B_0 \pi a^2 \cos(\omega t)$.
The induced electromotive force (emf) is $\varepsilon = -\frac{d\phi}{dt} = -\frac{d}{dt} (B_0 \pi a^2 \cos(\omega t)) = B_0 \pi a^2 \omega \sin(\omega t)$.
The induced current is $I = \frac{\varepsilon}{R} = \frac{B_0 \pi a^2 \omega}{R} \sin(\omega t)$.
$1$. At $t = \frac{\pi}{2\omega}$:
$I = \frac{B_0 \pi a^2 \omega}{R} \sin(\omega \cdot \frac{\pi}{2\omega}) = \frac{B_0 \pi a^2 \omega}{R}$.
Since the magnetic flux is decreasing,by Lenz's law,the induced current flows in the anticlockwise direction. At $(a, 0, 0)$,the current is along $+\hat j$.
$2$. At $t = \frac{\pi}{\omega}$:
$I = \frac{B_0 \pi a^2 \omega}{R} \sin(\omega \cdot \frac{\pi}{\omega}) = 0$.
$3$. At $t = \frac{3\pi}{2\omega}$:
$I = \frac{B_0 \pi a^2 \omega}{R} \sin(\omega \cdot \frac{3\pi}{2\omega}) = -\frac{B_0 \pi a^2 \omega}{R}$.
The negative sign indicates the current flows in the clockwise direction. At $(a, 0, 0)$,the current is along $-\hat j$.