A stationary circular loop of radius a is located in a magnetic field which varies with time from `t = 0 to t = T` according to law `B = B_(0) t(T - t)`. If plane of loop is normal to the direction of field and resistance of the loop is R, calculate
amount of heat generated in the loop during this interval.

Options

A. `(pi^(2)a^(4)B_(0)^(2)T^(3))/(3R)`
B. `(pi^(2)a^(4)B_(0)^(2)T^(3))/(R)`
C. `(3pi^(2)a^(4)B_(0)^(2)T^(3))/(R)`
D. None of these

Answer

To solve the problem of calculating the amount of heat generated in a stationary circular loop of radius \( a \) located in a time-varying magnetic field, we will follow these steps:

### Step 1: Determine the Magnetic Field
The magnetic field is given by the equation:
\[ B(t) = B_0 t (T - t) \]
where \( B_0 \) is a constant, \( T \) is the total time, and \( t \) is the time variable.

### Step 2: Calculate the Magnetic Flux
The magnetic flux \( \Phi \) through the loop is given by:
\[ \Phi = B \cdot A \]
where \( A \) is the area of the loop. For a circular loop of radius \( a \), the area \( A \) is:
\[ A = \pi a^2 \]
Thus, the magnetic flux becomes:
\[ \Phi(t) = B(t) \cdot A = B_0 t (T - t) \cdot \pi a^2 = B_0 \pi a^2 t (T - t) \]

### Step 3: Calculate the Induced EMF
According to Faraday's law of electromagnetic induction, the induced electromotive force (emf) \( \epsilon \) is given by:
\[ \epsilon = -\frac{d\Phi}{dt} \]
Calculating the derivative of the magnetic flux:
\[ \frac{d\Phi}{dt} = B_0 \pi a^2 \left( T - 2t \right) \]
Thus, the induced emf is:
\[ \epsilon = -B_0 \pi a^2 (T - 2t) \]

### Step 4: Calculate the Induced Current
Using Ohm's law, the induced current \( I \) in the loop can be expressed as:
\[ I = \frac{\epsilon}{R} = -\frac{B_0 \pi a^2 (T - 2t)}{R} \]

### Step 5: Calculate the Power Dissipated
The power \( P \) dissipated in the loop is given by:
\[ P = I^2 R \]
Substituting for \( I \):
\[ P = \left(-\frac{B_0 \pi a^2 (T - 2t)}{R}\right)^2 R = \frac{B_0^2 \pi^2 a^4 (T - 2t)^2}{R} \]

### Step 6: Calculate the Total Heat Generated
The total heat \( Q \) generated in the loop over the time interval from \( 0 \) to \( T \) is given by:
\[ Q = \int_0^T P \, dt = \int_0^T \frac{B_0^2 \pi^2 a^4 (T - 2t)^2}{R} \, dt \]
Factoring out the constants:
\[ Q = \frac{B_0^2 \pi^2 a^4}{R} \int_0^T (T - 2t)^2 \, dt \]

### Step 7: Evaluate the Integral
Now we need to evaluate the integral:
\[ \int_0^T (T - 2t)^2 \, dt \]
Expanding the integrand:
\[ (T - 2t)^2 = T^2 - 4Tt + 4t^2 \]
Thus:
\[ \int_0^T (T - 2t)^2 \, dt = \int_0^T T^2 \, dt - 4T \int_0^T t \, dt + 4 \int_0^T t^2 \, dt \]
Calculating each integral:
1. \(\int_0^T T^2 \, dt = T^2 \cdot T = T^3\)
2. \(\int_0^T t \, dt = \frac{T^2}{2}\)
3. \(\int_0^T t^2 \, dt = \frac{T^3}{3}\)

Putting it all together:
\[ \int_0^T (T - 2t)^2 \, dt = T^3 - 4T \cdot \frac{T^2}{2} + 4 \cdot \frac{T^3}{3} = T^3 - 2T^3 + \frac{4T^3}{3} = \frac{T^3}{3} \]

### Step 8: Substitute Back to Find \( Q \)
Now substituting back into the equation for \( Q \):
\[ Q = \frac{B_0^2 \pi^2 a^4}{R} \cdot \frac{T^3}{3} = \frac{B_0^2 \pi^2 a^4 T^3}{3R} \]

### Final Answer
Thus, the amount of heat generated in the loop during the interval from \( t = 0 \) to \( t = T \) is:
\[ Q = \frac{B_0^2 \pi^2 a^4 T^3}{3R} \]