A thin non conducting ring of mass `m`, radius a carrying a charge `q` can rotate freely about its own axis which is vertical. At the initial moment, the ring was at rest in horizontal position and no magnetic field was present. At instant `t=0`, a uniform magnetic field is switched on which is vertically downward and increases with time according to the law `B=B_0t`. Neglecting magnetism induced due to rotational motion of ring.
The magnitude of induced emf of the closed surface of ring will be

Options

A. `pia^2B_0`
B. `2a^2B_0`
C. zero
D. `1/2pia^2B_0`

Explanation

To solve the problem, we need to determine the induced electromotive force (emf) in the thin non-conducting ring when a magnetic field is applied. We will use Faraday's law of electromagnetic induction, which states that the induced emf is equal to the negative rate of change of magnetic flux through the loop.

### Step-by-step Solution:

1. **Identify the Magnetic Field**:
The magnetic field \( B \) is given by the equation \( B = B_0 t \), where \( B_0 \) is a constant and \( t \) is the time. This means that the magnetic field is increasing linearly with time.

2. **Determine the Area of the Ring**:
The area \( A \) of the ring can be calculated using the formula for the area of a circle:
\[ A = \pi a^2 \]
where \( a \) is the radius of the ring.

3. **Calculate the Magnetic Flux**:
The magnetic flux \( \Phi \) through the ring is given by the product of the magnetic field and the area:
\[ \Phi = B \cdot A = (B_0 t) \cdot (\pi a^2) = \pi a^2 B_0 t \]

4. **Find the Rate of Change of Magnetic Flux**:
To find the induced emf, we need to take the derivative of the magnetic flux with respect to time \( t \):
\[ \frac{d\Phi}{dt} = \frac{d}{dt}(\pi a^2 B_0 t) = \pi a^2 B_0 \]

5. **Apply Faraday's Law**:
According to Faraday's law, the induced emf \( \mathcal{E} \) is equal to the negative rate of change of magnetic flux:
\[ \mathcal{E} = -\frac{d\Phi}{dt} = -(\pi a^2 B_0) \]
Since we are interested in the magnitude of the induced emf, we can drop the negative sign:
\[ \mathcal{E} = \pi a^2 B_0 \]

### Final Answer:
The magnitude of the induced emf of the closed surface of the ring is:
\[ \mathcal{E} = \pi a^2 B_0 \]