1. Magnetic field at the center of the large coil

For a circular loop of radius $R$ carrying current $I$, the magnetic field at its center is

$B=\frac{\mu_0 I}{2R}$

Since the smaller coil is much smaller than the larger one, we can take this field to be approximately uniform over the area of the small coil.

2. Area of the smaller coil

If the smaller coil has radius $r$, its area is

$A=\pi r^2$

3. Flux through the smaller coil

Initially both coils are in the same plane, so the magnetic field is along the normal to the plane of the coils. The smaller coil rotates about a diameter with angular speed $\omega$, so after time $t$ its normal makes an angle

$\theta=\omega t$

with the magnetic field.

Hence the magnetic flux through the smaller coil is

$\Phi = BA\cos\theta$

$\Phi = \frac{\mu_0 I}{2R}\,\pi r^2 \cos(\omega t)$

4. Induced emf

By Faraday's law,

$e = -\frac{d\Phi}{dt}$

So,

$e = -\frac{d}{dt}\left(\frac{\mu_0 I}{2R}\pi r^2 \cos(\omega t)\right)$

$e = \frac{\mu_0 I}{2R}\pi r^2 \omega \sin(\omega t)$

Thus the induced emf is

$\boxed{e=\frac{\mu_0 I}{2R}\,\omega\pi r^2 \sin(\omega t)}$

5. Option matching

This matches:

$\boxed{\text{A}}$