A conducting circular loop is placed in a uniform magnetic field, `B=0.025T` with its plane perpendicular to the loop. The radius of the loop is made to shrink at a constant rate of `1 mms^(-1)`. The induced emf when the radius is `2 cm` is

Options

A. `2pimuV`
B. `pimuV`
C. `(pi)/(2)muV`
D. `2muV`

Explanation

To solve the problem step by step, we will calculate the induced EMF in the conducting circular loop based on the given parameters.

### Step 1: Identify the given values
- Magnetic field strength, \( B = 0.025 \, \text{T} \)
- Rate of change of radius, \( \frac{dr}{dt} = -1 \, \text{mm/s} = -1 \times 10^{-3} \, \text{m/s} \) (negative because the radius is shrinking)
- Radius of the loop at the moment of interest, \( r = 2 \, \text{cm} = 2 \times 10^{-2} \, \text{m} \)

### Step 2: Calculate the magnetic flux \( \Phi \)
The magnetic flux through the loop is given by the formula:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
Since the plane of the loop is perpendicular to the magnetic field, \( \theta = 0^\circ \) and \( \cos(0) = 1 \). The area \( A \) of the circular loop is:
\[ A = \pi r^2 \]
Thus, the magnetic flux becomes:
\[ \Phi = B \cdot \pi r^2 \]

### Step 3: Differentiate the magnetic flux to find induced EMF
According to Faraday's law of electromagnetic induction, the induced EMF \( E \) is given by:
\[ E = -\frac{d\Phi}{dt} \]
Substituting the expression for magnetic flux:
\[ E = -\frac{d}{dt}(B \cdot \pi r^2) \]
Since \( B \) is constant, we can take it out of the differentiation:
\[ E = -B \cdot \pi \frac{d}{dt}(r^2) \]
Using the chain rule:
\[ \frac{d}{dt}(r^2) = 2r \frac{dr}{dt} \]
Thus, the induced EMF becomes:
\[ E = -B \cdot \pi \cdot 2r \frac{dr}{dt} \]

### Step 4: Substitute the known values
Now, substituting the known values into the equation:
\[ E = -0.025 \cdot \pi \cdot 2 \cdot (2 \times 10^{-2}) \cdot (-1 \times 10^{-3}) \]
Calculating this step-by-step:
1. Calculate \( 2 \cdot (2 \times 10^{-2}) = 4 \times 10^{-2} \)
2. Now substitute:
\[ E = 0.025 \cdot \pi \cdot (4 \times 10^{-2}) \cdot (1 \times 10^{-3}) \]
3. Simplifying further:
\[ E = 0.025 \cdot \pi \cdot 4 \times 10^{-5} \]
4. Calculate \( 0.025 \cdot 4 = 0.1 \):
\[ E = 0.1 \cdot \pi \times 10^{-5} \]
5. Therefore:
\[ E = \pi \times 10^{-6} \, \text{V} \]

### Final Result
The induced EMF when the radius is 2 cm is:
\[ E = \pi \times 10^{-6} \, \text{V} \quad \text{(or } \mu V \text{)} \]