Shown in the figure is a circular loop of radius, r and resistance R. A variable magnetic field of induction `B=e^(-t)` is established inside the coil. If the key (K) is closed at t=0 , the electrical power developed at the instant is equal to

Options

A. `(pir^2)/R`
B. `(10r^3)/R`
C. `(pi^2r^4R)/5`
D. `(pi^2r^4)/(4R)`

Answer

To solve the problem, we need to find the electrical power developed at the instant when the key K is closed at \( t = 0 \). We will follow these steps:

### Step 1: Determine the Magnetic Flux
The magnetic field \( B \) is given as:
\[ B = e^{-t} \]
The area \( A \) of the circular loop with radius \( r \) is:
\[ A = \pi r^2 \]
Thus, the magnetic flux \( \Phi \) through the loop is given by:
\[ \Phi = B \cdot A = B \cdot \pi r^2 = e^{-t} \cdot \pi r^2 \]

### Step 2: Find the Induced EMF
According to Faraday's law of electromagnetic induction, the induced electromotive force (EMF) \( E \) is given by the negative rate of change of magnetic flux:
\[ E = -\frac{d\Phi}{dt} \]
Substituting the expression for \( \Phi \):
\[ E = -\frac{d}{dt}(e^{-t} \cdot \pi r^2) \]
Using the product rule, we differentiate:
\[ E = -\pi r^2 \cdot \frac{d}{dt}(e^{-t}) = -\pi r^2 \cdot (-e^{-t}) = \pi r^2 e^{-t} \]

### Step 3: Calculate the Power Developed
The electrical power \( P \) developed in the circuit is given by:
\[ P = \frac{E^2}{R} \]
Substituting the expression for \( E \):
\[ P = \frac{(\pi r^2 e^{-t})^2}{R} = \frac{\pi^2 r^4 e^{-2t}}{R} \]

### Step 4: Evaluate at \( t = 0 \)
Now we evaluate the power at \( t = 0 \):
\[ P(t=0) = \frac{\pi^2 r^4 e^{0}}{R} = \frac{\pi^2 r^4}{R} \]

### Final Answer
Thus, the electrical power developed at the instant when the key is closed is:
\[ P = \frac{\pi^2 r^4}{R} \]

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