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 an electric field on the circumference of the ring is

Options

A. `aB_0`
B. `2aB_0`
C. `1/2aB_0`
D. zero

Explanation

To find the magnitude of the electric field on the circumference of the ring, we can follow these steps:

### Step 1: Calculate the Magnetic Flux
The magnetic flux (Φ) through the ring can be calculated using the formula:
\[ \Phi = B \cdot A \]
where \(B\) is the magnetic field and \(A\) is the area of the ring. The area \(A\) of the ring is given by:
\[ A = \pi a^2 \]
At time \(t\), the magnetic field is given by:
\[ B = B_0 t \]
Thus, the magnetic flux becomes:
\[ \Phi = B_0 t \cdot \pi a^2 \]

### Step 2: Calculate the Induced EMF
According to Faraday's law of electromagnetic induction, the induced EMF (ε) is given by the negative rate of change of magnetic flux:
\[ \epsilon = -\frac{d\Phi}{dt} \]
Differentiating the expression for magnetic flux with respect to time \(t\):
\[ \epsilon = -\frac{d}{dt}(B_0 t \cdot \pi a^2) = -\pi a^2 B_0 \]
Since we are interested in the magnitude, we can write:
\[ \epsilon = \pi a^2 B_0 \]

### Step 3: Relate Induced EMF to Electric Field
The induced EMF (ε) is also related to the electric field (E) around the circumference of the ring. The relationship is given by:
\[ \epsilon = E \cdot L \]
where \(L\) is the circumference of the ring, given by:
\[ L = 2 \pi a \]
Substituting this into the equation:
\[ \epsilon = E \cdot (2 \pi a) \]

### Step 4: Solve for Electric Field
Now we can solve for the electric field \(E\):
\[ E = \frac{\epsilon}{2 \pi a} \]
Substituting the expression for ε:
\[ E = \frac{\pi a^2 B_0}{2 \pi a} = \frac{a B_0}{2} \]

### Final Result
Thus, the magnitude of the electric field on the circumference of the ring is:
\[ E = \frac{a B_0}{2} \]

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