A rectangular loop of sides 10 cm and 5 cm with a cut is stationary between the pole pieces of an electromagnet.
The magnetic field of the magnet is normal to the loop. The current feeding the electromagnet is reduced, so that the field decreased from its initial value of 0.2 T at the rate of `0.02Omega`. If the cut is joined and the loop has a resistance of `2.0Omega`, then the power dissipated by the loop as heat is

Options

A. 5 nW
B. 4 nW
C. 3 nW
D. 2 nW

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the dimensions of the rectangular loop
The dimensions of the rectangular loop are given as:
- Length = 10 cm = 0.1 m
- Breadth = 5 cm = 0.05 m

### Step 2: Calculate the area of the loop
The area \( A \) of the rectangular loop can be calculated using the formula:
\[ A = \text{Length} \times \text{Breadth} \]
Substituting the values:
\[ A = 0.1 \, \text{m} \times 0.05 \, \text{m} = 0.005 \, \text{m}^2 = 50 \times 10^{-4} \, \text{m}^2 \]

### Step 3: Determine the rate of change of the magnetic field
The magnetic field \( B \) is decreasing at a rate of \( \frac{dB}{dt} = -0.02 \, \text{T/s} \) (negative because it is decreasing).

### Step 4: Calculate the induced EMF
The induced electromotive force (EMF) \( \mathcal{E} \) can be calculated using Faraday's law of electromagnetic induction:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
Where \( \Phi = B \cdot A \). Since the area is constant, we can express the change in flux as:
\[ \mathcal{E} = -A \frac{dB}{dt} \]
Substituting the values:
\[ \mathcal{E} = -50 \times 10^{-4} \, \text{m}^2 \times (-0.02 \, \text{T/s}) = 50 \times 10^{-4} \times 0.02 = 10 \times 10^{-4} \, \text{V} = 0.001 \, \text{V} = 10^{-3} \, \text{V} \]

### Step 5: Calculate the power dissipated in the loop
The power \( P \) dissipated in the loop can be calculated using the formula:
\[ P = \frac{\mathcal{E}^2}{R} \]
Where \( R \) is the resistance of the loop, given as \( R = 2 \, \Omega \).
Substituting the values:
\[ P = \frac{(10^{-3} \, \text{V})^2}{2 \, \Omega} = \frac{10^{-6} \, \text{V}^2}{2} = 5 \times 10^{-7} \, \text{W} = 0.5 \, \text{mW} = 500 \, \text{µW} \]

### Final Answer
The power dissipated by the loop as heat is \( 500 \, \mu W \) or \( 0.5 \, mW \).