A magnetic flux through a stationary loop with a resistance `R` varies during the time interval `tau` as `phi=at(tau-t)`. Find the amount of heat the generated in the loop during that time

Options

A. `(atau^2)/(2R)`
B. `(a^2tau^2)/(3R)`
C. `(2a^2tau^3)/(3R)`
D. `(atau)/(3R)`

Explanation

To solve the problem, we need to find the amount of heat generated in a loop with resistance \( R \) when the magnetic flux \( \phi \) varies with time as given by the equation \( \phi = a t (\tau - t) \).

### Step-by-Step Solution:

1. **Find the EMF (Electromotive Force):**
The EMF \( \mathcal{E} \) induced in the loop can be calculated using Faraday's law of electromagnetic induction, which states that the induced EMF is equal to the negative rate of change of magnetic flux:
\[ \mathcal{E} = -\frac{d\phi}{dt} \]
Given \( \phi = a t (\tau - t) \), we differentiate \( \phi \) with respect to \( t \):
\[ \frac{d\phi}{dt} = a(\tau - t) + at(-1) = a(\tau - 2t) \]
Therefore, the induced EMF is:
\[ \mathcal{E} = -a(\tau - 2t) \]

2. **Calculate the Current (I):**
Using Ohm's law, the current \( I \) flowing through the loop can be expressed as:
\[ I = \frac{\mathcal{E}}{R} = \frac{-a(\tau - 2t)}{R} \]

3. **Calculate the Heat Generated (H):**
The heat generated in the resistor over the time interval \( \tau \) can be calculated using the formula:
\[ H = \int_0^{\tau} I^2 R \, dt \]
Substituting for \( I \):
\[ H = \int_0^{\tau} \left(\frac{-a(\tau - 2t)}{R}\right)^2 R \, dt \]
Simplifying this gives:
\[ H = \frac{a^2}{R} \int_0^{\tau} (\tau - 2t)^2 \, dt \]

4. **Evaluate the Integral:**
We need to evaluate the integral \( \int_0^{\tau} (\tau - 2t)^2 \, dt \):
\[ (\tau - 2t)^2 = \tau^2 - 4\tau t + 4t^2 \]
Thus,
\[ \int_0^{\tau} (\tau - 2t)^2 \, dt = \int_0^{\tau} (\tau^2 - 4\tau t + 4t^2) \, dt \]
Evaluating this integral term by term:
- The first term: \( \int_0^{\tau} \tau^2 \, dt = \tau^2 \cdot t \big|_0^{\tau} = \tau^3 \)
- The second term: \( \int_0^{\tau} -4\tau t \, dt = -4\tau \cdot \frac{t^2}{2} \big|_0^{\tau} = -2\tau^3 \)
- The third term: \( \int_0^{\tau} 4t^2 \, dt = 4 \cdot \frac{t^3}{3} \big|_0^{\tau} = \frac{4\tau^3}{3} \)

Combining these results:
\[ \int_0^{\tau} (\tau - 2t)^2 \, dt = \tau^3 - 2\tau^3 + \frac{4\tau^3}{3} = \left(1 - 2 + \frac{4}{3}\right)\tau^3 = \frac{-2 + \frac{4}{3}}{1}\tau^3 = \frac{-6 + 4}{3}\tau^3 = \frac{-2}{3}\tau^3 \]

5. **Final Calculation of Heat (H):**
Substituting back into the equation for \( H \):
\[ H = \frac{a^2}{R} \cdot \frac{2}{3} \tau^3 = \frac{2a^2\tau^3}{3R} \]

### Final Answer:
The amount of heat generated in the loop during the time interval \( \tau \) is:
\[ H = \frac{2a^2\tau^3}{3R} \]