The problem involves calculating the induced current in a ring placed in a time-varying magnetic field. Let's break down the steps:
1. **Given Data:**
- The ring has an inner radius \(a\) and outer radius \(2a\).
- The resistivity of the material is \(\rho\).
- The thickness of the ring is \(h\).
- The magnetic field is \(B = kt\), where \(k\) is a constant, and the field is parallel to the axis of the ring.
2. **Induced EMF:**
According to Faraday's Law of Induction, the induced EMF in the ring is given by the rate of change of the magnetic flux through the ring:
\[
\mathcal{E} = -\frac{d\Phi}{dt}
\]
The magnetic flux \(\Phi\) is the integral of the magnetic field over the area of the ring. The area of the ring is the area of the outer circle minus the area of the inner circle:
\[
A = \pi (2a)^2 - \pi a^2 = \pi (4a^2 - a^2) = 3\pi a^2
\]
3. **Magnetic Flux:**
The flux \(\Phi\) is:
\[
\Phi = B \times A = (kt) \times 3\pi a^2
\]
So, the flux is:
\[
\Phi = 3\pi a^2 kt
\]
4. **Induced EMF:**
Taking the derivative of the flux with respect to time \(t\):
\[
\frac{d\Phi}{dt} = 3\pi a^2 k
\]
Therefore, the induced EMF is:
\[
\mathcal{E} = -3\pi a^2 k
\]
5. **Induced Current:**
The induced current \(I\) can be calculated using Ohm's Law:
\[
I = \frac{\mathcal{E}}{R}
\]
Where \(R\) is the resistance of the ring. The resistance of the ring is given by:
\[
R = \frac{\rho L}{A}
\]
Where \(L\) is the length of the ring and \(A\) is the cross-sectional area. The length of the ring is the circumference of the outer circle, which is \(2\pi(2a) = 4\pi a\). The cross-sectional area is \(h \times (2a - a) = h \times a\). Therefore:
\[
R = \frac{\rho \times 4\pi a}{h \times a} = \frac{4\pi \rho}{h}
\]
6. **Final Calculation:**
Using the formula for current:
\[
I = \frac{-3\pi a^2 k}{\frac{4\pi \rho}{h}} = \frac{3\pi a^2 k h}{4\pi \rho}
\]
Simplifying:
\[
I = \frac{3kh a^2}{4\rho}
\]
Therefore, the induced current in the ring is \(\frac{3kh a^2}{4\rho}\).
The correct answer is **(A)** \(\frac{3kh a^2}{4\rho}\).
