A conducting circular loop is placed in a uniform magnetic field `0.04T` with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at `2mm//sec` . The induced emf in the loop when the radius is `2cm` is:
(1.) 3.2 π μ V
(2.) 4.8 π μ V
(3.) 0.8 π μ V
(4.) 1.6 π μ V

Options

A. `3.2 pi muV`
B. `4.8 pi muV`
C. `0.8pi muV`
D. `1.6pi muV`

Explanation

To solve the problem, we need to find the induced electromotive force (emf) in a conducting circular loop placed in a uniform magnetic field as the radius of the loop shrinks. We will use Faraday's law of electromagnetic induction.

### Step-by-Step Solution:

1. **Identify the given values:**
- Magnetic field strength, \( B = 0.04 \, T \)
- Radius of the loop when calculating, \( r = 2 \, cm = 2 \times 10^{-2} \, m \)
- Rate of change of radius, \( \frac{dr}{dt} = -2 \, mm/s = -2 \times 10^{-3} \, m/s \) (negative because the radius is shrinking)

2. **Calculate the area of the loop:**
- The area \( A \) of a circular loop is given by:
\[ A = \pi r^2 \]
- Substituting the value of \( r \):
\[ A = \pi (2 \times 10^{-2})^2 = \pi (4 \times 10^{-4}) = 4\pi \times 10^{-4} \, m^2 \]

3. **Use Faraday's law to find the induced emf:**
- The induced emf \( \varepsilon \) is given by:
\[ \varepsilon = -B \frac{dA}{dt} \]
- We can express \( \frac{dA}{dt} \) in terms of \( r \):
\[ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = \pi \cdot 2r \frac{dr}{dt} \]
- Substituting \( r = 2 \times 10^{-2} \, m \) and \( \frac{dr}{dt} = -2 \times 10^{-3} \, m/s \):
\[ \frac{dA}{dt} = \pi \cdot 2(2 \times 10^{-2}) \cdot (-2 \times 10^{-3}) = -8\pi \times 10^{-5} \, m^2/s \]

4. **Substituting into the emf formula:**
- Now substitute \( B \) and \( \frac{dA}{dt} \) into the emf equation:
\[ \varepsilon = -B \frac{dA}{dt} = -0.04 \cdot (-8\pi \times 10^{-5}) = 0.04 \cdot 8\pi \times 10^{-5} \]
- Calculating this gives:
\[ \varepsilon = 3.2 \times 10^{-6} \pi \, V \]

5. **Convert to microvolts:**
- Since \( 10^{-6} \) is equivalent to micro, we can express the final answer as:
\[ \varepsilon = 3.2 \pi \, \mu V \]

### Final Answer:
The induced emf in the loop when the radius is \( 2 \, cm \) is:
\[ \varepsilon = 3.2 \pi \, \mu V \]