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
magnitude of charge flown through the loop from instant t = 0 to the instant when current reverses its direction.

Options

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

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the Area of the Loop
The area \( A \) of a circular loop with radius \( a \) is given by the formula:
\[ A = \pi a^2 \]

### Step 2: Calculate the Magnetic Flux
The magnetic field \( B \) varies with time as given by:
\[ B(t) = B_0 t (T - t) \]
The magnetic flux \( \Phi \) through the loop is given by:
\[ \Phi = B \cdot A = B_0 t (T - t) \cdot \pi a^2 \]
Thus,
\[ \Phi(t) = B_0 \pi a^2 t (T - t) \]

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

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

### Step 5: Determine the Charge Flowed
The charge \( Q \) that flows through the loop from \( t = 0 \) to the time when the current reverses direction can be found by integrating the current over time:
\[ Q = \int_0^{t_r} I \, dt \]
where \( t_r \) is the time when the current reverses direction. The current reverses when \( T - 2t = 0 \), which gives:
\[ t_r = \frac{T}{2} \]
Thus, we compute:
\[ Q = \int_0^{T/2} -\frac{B_0 \pi a^2 (T - 2t)}{R} \, dt \]

### Step 6: Evaluate the Integral
Now we evaluate the integral:
\[ Q = -\frac{B_0 \pi a^2}{R} \int_0^{T/2} (T - 2t) \, dt \]
Calculating the integral:
\[ \int (T - 2t) \, dt = Tt - t^2 \]
Evaluating from \( 0 \) to \( \frac{T}{2} \):
\[ = \left[ T\left(\frac{T}{2}\right) - \left(\frac{T}{2}\right)^2 \right] - \left[ 0 \right] = \frac{T^2}{2} - \frac{T^2}{4} = \frac{T^2}{4} \]
Thus, we have:
\[ Q = -\frac{B_0 \pi a^2}{R} \cdot \frac{T^2}{4} \]
Finally, the magnitude of the charge is:
\[ Q = \frac{B_0 \pi a^2 T^2}{4R} \]

### Final Result
The magnitude of the charge that flows through the loop from \( t = 0 \) to the instant when the current reverses its direction is:
\[ Q = \frac{B_0 \pi a^2 T^2}{4R} \]