A conducting metal circular-wire-loop of radius r is placed perpendicular to a magnetic field which varies with time as `B = B_0e^(-t//tau)`, where `B_0 and tau` are constants, at time = 0. If the resistance of the loop is R then the heat generated in the loop after a long time `(t to oo)` is :

Options

A. `(pi^(2)r^(4)B_(0)^(4))/(2tauR)`
B. `(pi^(2)r^(4)B_(0)^(2))/(2tauR)`
C. `(pi^(2)r^(4)B_(0)^(2)R)/(tau)`
D. `(pi^(2)r^(4)B_(0)^(2))/(tauR)`

Explanation

To solve the problem of calculating the heat generated in a conducting metal circular wire loop placed in a time-varying magnetic field, we can follow these steps:

### Step 1: Determine the Magnetic Flux
The magnetic field is given by \( B(t) = B_0 e^{-t/\tau} \). 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 \( r \), the area \( A \) is:
\[ A = \pi r^2 \]
Thus, the magnetic flux becomes:
\[ \Phi(t) = B_0 e^{-t/\tau} \cdot \pi r^2 \]

### Step 2: 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:
\[ \epsilon = -\frac{d}{dt} \left( B_0 \pi r^2 e^{-t/\tau} \right) = B_0 \pi r^2 \cdot \frac{1}{\tau} e^{-t/\tau} \]

### Step 3: Calculate the Heat Generated
The heat generated \( Q \) in the loop can be calculated using the formula:
\[ Q = \int_0^{\infty} \frac{\epsilon^2}{R} dt \]
Substituting the expression for \( \epsilon \):
\[ Q = \int_0^{\infty} \frac{(B_0 \pi r^2 \cdot e^{-t/\tau}/\tau)^2}{R} dt \]
This simplifies to:
\[ Q = \frac{B_0^2 \pi^2 r^4}{R \tau^2} \int_0^{\infty} e^{-2t/\tau} dt \]

### Step 4: Evaluate the Integral
The integral \( \int_0^{\infty} e^{-2t/\tau} dt \) can be evaluated as follows:
\[ \int_0^{\infty} e^{-kt} dt = \frac{1}{k} \quad \text{for } k > 0 \]
In our case, \( k = \frac{2}{\tau} \), so:
\[ \int_0^{\infty} e^{-2t/\tau} dt = \frac{\tau}{2} \]

### Step 5: Substitute Back to Find Q
Substituting this result back into the expression for \( Q \):
\[ Q = \frac{B_0^2 \pi^2 r^4}{R \tau^2} \cdot \frac{\tau}{2} = \frac{B_0^2 \pi^2 r^4}{2R \tau} \]

### Final Result
Thus, the heat generated in the loop after a long time is:
\[ Q = \frac{B_0^2 \pi^2 r^4}{2R \tau} \]

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