The magnetic flux through a stationary loop with resistance R varies during interval of time T as `phi` = at (T – t). The heat generated during this time neglecting the inductance of loop will be

Options

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

Answer

To find the heat generated in a stationary loop with resistance \( R \) when the magnetic flux \( \phi \) varies as \( \phi = a t (T - t) \), we can follow these steps:

### Step 1: Find the expression for EMF
The electromotive force (EMF) induced in the loop can be calculated using Faraday's law of electromagnetic induction, which states that:

\[ \text{EMF} = -\frac{d\phi}{dt} \]

### Step 2: Differentiate the magnetic flux
Given the magnetic flux \( \phi = a t (T - t) \), we need to differentiate this with respect to time \( t \):

\[ \frac{d\phi}{dt} = \frac{d}{dt}[a t (T - t)] \]

Using the product rule, we have:

\[ \frac{d\phi}{dt} = a \left( (T - t) + t \left(-1\right) \right) = a (T - 2t) \]

Thus, the EMF is:

\[ \text{EMF} = -\frac{d\phi}{dt} = -a (T - 2t) \]

### Step 3: Find the current in the loop
Using Ohm's law, the current \( I \) in the loop can be expressed as:

\[ I = \frac{\text{EMF}}{R} = \frac{-a (T - 2t)}{R} \]

### Step 4: Calculate the heat generated
The heat generated \( H \) in the resistor over the time interval \( T \) can be calculated using the formula:

\[ H = \int_0^T I^2 R \, dt \]

Substituting the expression for current \( I \):

\[ H = \int_0^T \left(\frac{-a (T - 2t)}{R}\right)^2 R \, dt \]

This simplifies to:

\[ H = \int_0^T \frac{a^2 (T - 2t)^2}{R} \, dt \]

### Step 5: Evaluate the integral
Now we need to evaluate the integral:

\[ H = \frac{a^2}{R} \int_0^T (T - 2t)^2 \, dt \]

Expanding \( (T - 2t)^2 \):

\[ (T - 2t)^2 = T^2 - 4Tt + 4t^2 \]

Now, we integrate each term separately:

\[ \int_0^T (T^2 - 4Tt + 4t^2) \, dt = \left[T^2 t - 2Tt^2 + \frac{4}{3}t^3\right]_0^T \]

Evaluating this from \( 0 \) to \( T \):

\[ = T^3 - 2T \cdot T^2 + \frac{4}{3}T^3 = T^3 - 2T^3 + \frac{4}{3}T^3 = \left(-T^3 + \frac{4}{3}T^3\right) = \frac{1}{3}T^3 \]

### Step 6: Substitute back to find \( H \)
Now substituting back into the equation for \( H \):

\[ H = \frac{a^2}{R} \cdot \frac{1}{3} T^3 = \frac{a^2 T^3}{3R} \]

Thus, the heat generated during the time interval \( T \) is:

\[ \boxed{H = \frac{a^2 T^3}{3R}} \]