A conducting circular loop of radius a and resistance R is kept on a horizontal plane. A vertical time varying magnetic field B=2t is switched on at time t=0. Then

Options

A. power generated in the coil at any time t is constant
B. flow of charge passed through any section of coil is constant
C. total charge passed through any section between time `(t=0 "to" t =2 is ((4pia^(2))/(R))`.
D. All of the above

Answer

To solve the problem step by step, we will analyze the given information and apply relevant physics concepts.

### Step 1: Understand the Given Information
We have a conducting circular loop of radius \( a \) and resistance \( R \). A vertical time-varying magnetic field \( B = 2t \) is switched on at time \( t = 0 \).

### Step 2: Calculate the Magnetic Flux
The magnetic flux \( \Phi \) linked with the loop is given by the formula:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
Where:
- \( B \) is the magnetic field,
- \( A \) is the area of the loop,
- \( \theta \) is the angle between the magnetic field and the normal to the area.

Since the magnetic field is vertical and the area vector of the loop is also vertical, \( \theta = 0 \) degrees, and \( \cos(0) = 1 \).

The area \( A \) of the circular loop is:
\[ A = \pi a^2 \]
Thus, the magnetic flux becomes:
\[ \Phi = B \cdot A = (2t) \cdot (\pi a^2) = 2\pi a^2 t \]

### Step 3: Calculate the Induced EMF
According to Faraday's law of electromagnetic induction, the induced electromotive force (EMF) \( E \) in the loop is given by:
\[ E = -\frac{d\Phi}{dt} \]
Calculating the derivative:
\[ E = -\frac{d}{dt}(2\pi a^2 t) = -2\pi a^2 \]
Since we are interested in the magnitude, we have:
\[ E = 2\pi a^2 \]

### Step 4: Calculate the Current
Using Ohm's law, the current \( I \) flowing through the loop can be calculated as:
\[ I = \frac{E}{R} = \frac{2\pi a^2}{R} \]

### Step 5: Calculate the Power Generated
The power \( P \) generated in the loop can be calculated using the formula:
\[ P = I^2 R \]
Substituting the expression for current:
\[ P = \left(\frac{2\pi a^2}{R}\right)^2 R = \frac{4\pi^2 a^4}{R} \]
This shows that the power generated is constant since \( a \) and \( R \) are constants.

### Step 6: Calculate the Total Charge Passed
The total charge \( Q \) that passes through any section of the coil from \( t = 0 \) to \( t = 2 \) can be calculated as:
\[ Q = \int_0^2 I \, dt \]
Substituting the expression for current:
\[ Q = \int_0^2 \frac{2\pi a^2}{R} \, dt = \frac{2\pi a^2}{R} \int_0^2 dt = \frac{2\pi a^2}{R} \cdot (2 - 0) = \frac{4\pi a^2}{R} \]

### Conclusion
From the calculations:
1. The power generated in the coil is constant.
2. The flow of charge through any section of the coil is constant.
3. The total charge passed through any section between \( t = 0 \) to \( t = 2 \) is \( \frac{4\pi a^2}{R} \).

Thus, all options are correct, and the answer is:
**D: All of the above.**