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.
Find intantaneous power developed by electric force acting on the ring at `t=1`s

Options

A. `(2q^(2)B_(0)^(2)a^(2))/(14m)`
B. `(q^(2)B_(0)^(2)a^(2))/(8m)`
C. `(3q^(2)B_(0)^(2)a^(2))/(m)`
D. `(q^(2)B_(0)^(2)a^(2))/(4m)`

Explanation

To solve the problem step by step, we will follow the reasoning outlined in the video transcript while providing clear explanations for each step.

### Step 1: Understanding the System
We have a thin non-conducting ring of mass `m`, radius `a`, and charge `q`. The ring can rotate freely about its vertical axis. Initially, the ring is at rest in a horizontal position, and at time `t=0`, a uniform magnetic field `B` is switched on, which increases with time according to the law `B = B_0 t`.

### Step 2: Magnetic Flux Calculation
The magnetic field is directed vertically downward, and the area of the ring is given by:
\[ A = \pi a^2 \]
The magnetic flux `Φ` through the ring is given by:
\[ \Phi = B \cdot A = B \cdot \pi a^2 \]
Substituting `B = B_0 t`, we get:
\[ \Phi = B_0 t \cdot \pi a^2 \]

### Step 3: Induced EMF Calculation
According to Faraday's law of electromagnetic induction, the induced electromotive force (emf) `ε` is given by the rate of change of magnetic flux:
\[ \epsilon = -\frac{d\Phi}{dt} \]
Differentiating the flux with respect to time:
\[ \frac{d\Phi}{dt} = \frac{d}{dt}(B_0 t \cdot \pi a^2) = B_0 \pi a^2 \]
Thus, the induced emf is:
\[ \epsilon = -B_0 \pi a^2 \]

### Step 4: Electric Field Calculation
The induced electric field `E` around the ring can be found using the relationship between emf and electric field:
\[ \epsilon = \oint E \cdot dl \]
For a ring, the line integral around the ring is:
\[ \epsilon = E \cdot (2\pi a) \]
Setting these equal gives:
\[ E \cdot (2\pi a) = B_0 \pi a^2 \]
Solving for `E`:
\[ E = \frac{B_0 a}{2} \]

### Step 5: Electric Force Calculation
The electric force `F` acting on the charge `q` in the electric field `E` is given by:
\[ F = qE = q \cdot \frac{B_0 a}{2} = \frac{q B_0 a}{2} \]

### Step 6: Torque Calculation
The torque `τ` acting on the ring due to this force is given by:
\[ \tau = F \cdot r_{\perpendicular} \]
Since the force acts tangentially at the radius `a`, we have:
\[ \tau = F \cdot a = \left(\frac{q B_0 a}{2}\right) \cdot a = \frac{q B_0 a^2}{2} \]

### Step 7: Angular Velocity Calculation
The angular velocity `ω` of the ring can be determined from the relationship between torque and angular acceleration:
\[ \tau = I \alpha \]
where `I` is the moment of inertia of the ring, given by:
\[ I = ma^2 \]
Thus, we have:
\[ \alpha = \frac{\tau}{I} = \frac{\frac{q B_0 a^2}{2}}{ma^2} = \frac{q B_0}{2m} \]
Since the ring starts from rest, the angular velocity at time `t` is:
\[ \omega = \alpha t = \frac{q B_0}{2m} t \]
At `t = 1 s`, we have:
\[ \omega = \frac{q B_0}{2m} \]

### Step 8: Instantaneous Power Calculation
The instantaneous power `P` developed by the electric force acting on the ring is given by:
\[ P = \tau \cdot \omega \]
Substituting the expressions for `τ` and `ω`:
\[ P = \left(\frac{q B_0 a^2}{2}\right) \cdot \left(\frac{q B_0}{2m}\right) = \frac{q^2 B_0^2 a^2}{4m} \]

### Final Answer
Thus, the instantaneous power developed by the electric force acting on the ring at `t = 1 s` is:
\[ \boxed{\frac{q^2 B_0^2 a^2}{4m}} \]