At the centre of a fixed large circular coil of radius R, a much smaller circular coil of radius r is placed. The two coils are concentric and are in the same plane. The larger coil carries a current I. The smaller coil is set to rotate with a constant angular velocity $$\omega$$ about an axis along their common diameter. Calculate the emf induced in the smaller coil after a time t of its start of rotation.

For a single circular coil of radius $$R$$ carrying a steady current $$I$$, the magnetic field at its centre (and in the neighbourhood of the centre) is given by the well-known expression

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

In our problem the large coil is fixed, so this field $$B$$ is constant in both magnitude and direction. Its direction is perpendicular to the plane of the coils, pointing along the common axis that passes through their centres.

The smaller coil of radius $$r$$ is placed at the same centre and in the same initial plane. The area enclosed by this secondary coil is

$$ A \;=\; \pi r^{2}. $$

Now the smaller coil starts rotating with a constant angular velocity $$\omega$$ about a diameter that lies in the plane of the coils. Because this diameter is in the plane, the axis of rotation is perpendicular to the magnetic field direction. Consequently, as the coil rotates, the angle between the magnetic field $$\vec B$$ and the area-vector (normal to the plane of the small coil) changes with time.

If we take $$t=0$$ as the instant when the area-vector of the smaller coil is parallel to $$\vec B$$, then after a time $$t$$ the angle between them becomes

$$ \theta = \omega t. $$

The magnetic flux linked with the small coil at this instant is, by definition,

$$ \Phi(t) \;=\; B A \cos \theta. $$

Substituting the explicit expressions for $$B$$, $$A$$ and $$\theta$$, we obtain

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

According to Faraday’s law of electromagnetic induction, the induced emf $$\mathcal E$$ in the small coil equals the negative time-rate of change of this flux, i.e.

$$ \mathcal E(t) \;=\; -\,\frac{d\Phi(t)}{dt}. $$

Differentiating the flux expression term by term, we have

$$ \frac{d\Phi(t)}{dt} = \Bigl(\frac{\mu_0 I}{2R}\Bigr)\,(\pi r^{2}) \,\frac{d}{dt}\bigl[\cos(\omega t)\bigr]. $$

The derivative of the cosine function is

$$ \frac{d}{dt}\bigl[\cos(\omega t)\bigr] = -\,\omega \sin(\omega t). $$

Substituting this result, we get

$$ \frac{d\Phi(t)}{dt} = -\,\Bigl(\frac{\mu_0 I}{2R}\Bigr)\,(\pi r^{2}) \,\omega \sin(\omega t). $$

Therefore the induced emf becomes

$$ \mathcal E(t) = -\Bigl[-\,\Bigl(\frac{\mu_0 I}{2R}\Bigr)\,(\pi r^{2}) \,\omega \sin(\omega t)\Bigr] = \Bigl(\frac{\mu_0 I}{2R}\Bigr)\,(\pi r^{2}) \,\omega \sin(\omega t). $$

Re-arranging the factors for clarity, we write

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

This expression matches Option C.

Hence, the correct answer is Option C.