A condacting 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 `2 mm / s`. The indaced emf in the loop whicn the redius is `2 cm` is `y pi` microvolt, then find `y`.

Answer

To solve the problem, we need to determine the induced electromotive force (emf) in a conducting circular loop placed in a uniform magnetic field as its radius shrinks. Here’s the step-by-step solution:

### Step 1: Understand the Given Information
- Magnetic field strength, \( B = 0.04 \, \text{T} \)
- Rate of change of radius, \( \frac{dr}{dt} = -2 \, \text{mm/s} = -0.002 \, \text{m/s} \)
- Radius of the loop when calculating induced emf, \( r = 2 \, \text{cm} = 0.02 \, \text{m} \)

### Step 2: Calculate the Area of the Loop
The area \( A \) of a circular loop is given by the formula:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi (0.02)^2 = \pi (0.0004) = 0.0004\pi \, \text{m}^2 \]

### Step 3: Calculate the Magnetic Flux
The magnetic flux \( \Phi \) through the loop is given by:
\[ \Phi = B \cdot A \]
Substituting the values:
\[ \Phi = 0.04 \cdot 0.0004\pi = 0.000016\pi \, \text{Wb} \]

### Step 4: Find the Rate of Change of Magnetic Flux
To find the induced emf, we need to calculate the rate of change of magnetic flux \( \frac{d\Phi}{dt} \). Using the chain rule:
\[ \frac{d\Phi}{dt} = B \cdot \frac{dA}{dt} \]
Since \( A = \pi r^2 \), we differentiate with respect to time:
\[ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt} \]
Substituting \( r = 0.02 \, \text{m} \) and \( \frac{dr}{dt} = -0.002 \, \text{m/s} \):
\[ \frac{dA}{dt} = 2\pi (0.02)(-0.002) = -0.00008\pi \, \text{m}^2/s \]

### Step 5: Calculate the Induced EMF
Now substituting back to find \( \frac{d\Phi}{dt} \):
\[ \frac{d\Phi}{dt} = B \cdot \frac{dA}{dt} = 0.04 \cdot (-0.00008\pi) = -0.0000032\pi \, \text{V} \]
The induced emf \( \mathcal{E} \) is given by:
\[ \mathcal{E} = -\frac{d\Phi}{dt} = 0.0000032\pi \, \text{V} = 3.2 \times 10^{-6}\pi \, \text{V} \]

### Step 6: Convert to Microvolts
Since \( 1 \, \text{V} = 10^6 \, \mu\text{V} \):
\[ \mathcal{E} = 3.2\pi \, \mu\text{V} \]

### Step 7: Identify the Value of \( y \)
From the expression \( \mathcal{E} = y\pi \, \mu\text{V} \), we can see that:
\[ y = 3.2 \]

### Final Answer
Thus, the value of \( y \) is \( \boxed{3.2} \).

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