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 find the induced emf in the conducting circular loop when its radius is 2 cm, we can use Faraday's law of electromagnetic induction. The law states that the induced emf (ε) in a closed loop is equal to the negative rate of change of magnetic flux (Φ) through the loop.

### Step 1: Calculate the area of the loop
The area \( A \) of a circular loop is given by the formula:
\[ A = \pi r^2 \]
where \( r \) is the radius of the loop.

Given that the radius \( r = 2 \, \text{cm} = 0.02 \, \text{m} \):
\[ A = \pi (0.02)^2 = \pi (0.0004) \approx 0.00125664 \, \text{m}^2 \]

### Step 2: Calculate the magnetic flux
The magnetic flux \( \Phi \) through the loop is given by:
\[ \Phi = B \cdot A \]
where \( B = 0.025 \, \text{T} \) (the magnetic field strength).

Substituting the values:
\[ \Phi = 0.025 \cdot 0.00125664 \approx 3.1416 \times 10^{-5} \, \text{Wb} \]

### Step 3: Determine the rate of change of area
Since the radius is shrinking at a rate of \( \frac{dr}{dt} = -1 \, \text{mm/s} = -0.001 \, \text{m/s} \), we can find the rate of change of area \( \frac{dA}{dt} \) using the formula:
\[ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt} \]
Substituting \( r = 0.02 \, \text{m} \) and \( \frac{dr}{dt} = -0.001 \, \text{m/s} \):
\[ \frac{dA}{dt} = 2\pi (0.02)(-0.001) \approx -0.000125664 \, \text{m}^2/\text{s} \]

### Step 4: Calculate the induced emf
Using Faraday's law, the induced emf \( \varepsilon \) is given by:
\[ \varepsilon = -B \frac{dA}{dt} \]
Substituting the values:
\[ \varepsilon = -0.025 \cdot (-0.000125664) \approx 3.1416 \times 10^{-6} \, \text{V} = 3.14 \, \mu\text{V} \]

### Final Answer
The induced emf when the radius is 2 cm is approximately \( 3.14 \, \mu\text{V} \).

---