A small circular loop of wire of radius $$a$$ is located at the centre of a much larger circular wire loop of radius $$b$$. The two loops are in the same plane. The outer loop of radius $$b$$ carries an alternating current $$I = I_0 \cos(\omega t)$$. The emf induced in the smaller inner loop is nearly:

We begin by recalling the magnetic field at the centre of a single circular loop of radius $$R$$ carrying a current $$I$$. The well-known expression is

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

In the present problem the outer loop has radius $$b$$ and carries an alternating current $$I = I_0 \cos(\omega t)$$. Substituting $$R = b$$ and the given time-dependent current into the formula, the magnetic field produced by the large loop at its own centre is

$$B(t)=\frac{\mu_0 I_0}{2b}\,\cos(\omega t).$$

The small loop of radius $$a$$ is placed exactly at this centre and lies in the same plane, so all lines of the magnetic field pass perpendicularly through the small loop. Therefore the field can be considered uniform over the area of the small loop, provided $$a \ll b$$ (which is stated by “a much larger loop”).

The area of the inner loop is

$$A = \pi a^2.$$

The magnetic flux $$\Phi$$ through the small loop equals the product of the field and the area (because the field is perpendicular to the plane of the loop and uniform over it):

$$\Phi(t) = B(t) \, A = \left(\frac{\mu_0 I_0}{2b}\,\cos(\omega t)\right)\,(\pi a^2).$$

So we have

$$\Phi(t) = \frac{\pi \mu_0 I_0 a^2}{2b}\,\cos(\omega t).$$

Now, according to Faraday’s law of electromagnetic induction, the induced emf $$\mathcal{E}$$ in the small loop is

$$\mathcal{E} = -\frac{d\Phi}{dt}.$$

Taking the time derivative of the flux, we differentiate term by term. The constant factors $$\frac{\pi \mu_0 I_0 a^2}{2b}$$ remain in front, while the derivative of $$\cos(\omega t)$$ is $$-\,\omega\sin(\omega t)$$. Thus

$$\mathcal{E} = -\left(\frac{\pi \mu_0 I_0 a^2}{2b}\right)\left(-\,\omega\sin(\omega t)\right).$$

The two minus signs cancel, leaving

$$\mathcal{E} = \frac{\pi \mu_0 I_0 a^2}{2b}\,\omega\,\sin(\omega t).$$

This emf is the magnitude of the induced voltage in the small loop; its sign (direction) is given by Lenz’s law, but the options list only the magnitude and its time dependence. Comparing our expression with the given alternatives, we see it matches exactly with Option D:

$$\mathcal{E}(t)=\frac{\pi \mu_0 I_0}{2}\cdot\frac{a^2}{b}\,\omega\,\sin(\omega t).$$

Hence, the correct answer is Option D.