Explanation

(a) The current flows through the coil 1 is $I_{1} = I_{0}\sin{\omega t}$
Where $I_{0}$ is the peak value of current
Magnetic flux linked with the coil 2 is
$\phi_{2} = MI_{1} = MI_{0}\sin{\omega t}$
Where $M$ is the mutual inductance between the two coils
The magnitude of induced emf in coil 2 is
$\left| \varepsilon_{2} \right| = \frac{d\phi_{2}}{dt} = \frac{d}{dt}\ \left( MI_{0}\sin{\omega t} \right) = MI_{0}\omega\cos{\omega t}$
$\therefore$ Peak value of voltage induced in the coil 2 is
$= MI_{0}\omega = 150 \times 10^{- 3} \times 2 \times 2\pi \times 50 = 30\pi\ V$