A conduction circular loop is placed in a uniform magnetic field 0.02T with its place normal to the field . The radius of loop starts shrinking at a rate of `1.0 mm//sec`. Then induced emf in the loop at an instant when the radius is 2 cm :-

Options

A. `2.5 mu V`
B. `0.25 mu V`
C. `0.25 V`
D. `2.5 mV`

Answer

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 its radius shrinks. Let's go through the solution step by step.

### Step 1: Understand the Given Information
- Magnetic field strength (B) = 0.02 T (Tesla)
- Rate of change of radius (dr/dt) = -1 mm/s = -0.001 m/s (since the radius is shrinking)
- Radius at the instant we are considering (r) = 2 cm = 0.02 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 (Φ)
Magnetic flux (Φ) through the loop is given by:
\[ \Phi = B \cdot A \]
Since the area vector is normal to the magnetic field:
\[ \Phi = B \cdot \pi r^2 = 0.02 \cdot 0.0004\pi = 0.000008\pi \, \text{Wb} \]

### Step 4: Differentiate the Flux with Respect to Time
To find the induced emf (E), we use Faraday's law of electromagnetic induction:
\[ E = -\frac{d\Phi}{dt} \]
Using the product rule for differentiation:
\[ \frac{d\Phi}{dt} = B \cdot \frac{dA}{dt} \]
And since \( A = \pi r^2 \):
\[ \frac{dA}{dt} = \pi \cdot 2r \cdot \frac{dr}{dt} \]
Substituting this into the flux change:
\[ \frac{d\Phi}{dt} = B \cdot \pi \cdot 2r \cdot \frac{dr}{dt} \]

### Step 5: Substitute Known Values
Now substituting the known values:
- \( B = 0.02 \, \text{T} \)
- \( r = 0.02 \, \text{m} \)
- \( \frac{dr}{dt} = -0.001 \, \text{m/s} \)

So,
\[ \frac{d\Phi}{dt} = 0.02 \cdot \pi \cdot 2 \cdot 0.02 \cdot (-0.001) \]

### Step 6: Calculate the Induced emf
Now, substituting the values:
\[ \frac{d\Phi}{dt} = 0.02 \cdot \pi \cdot 2 \cdot 0.02 \cdot (-0.001) \]
\[ = -0.0000008\pi \, \text{Wb/s} \]

Thus, the induced emf is:
\[ E = -\left(-0.0000008\pi\right) = 0.0000008\pi \, \text{V} \]

### Step 7: Convert to Microvolts
To convert the induced emf to microvolts:
\[ E = 0.0000008\pi \, \text{V} \approx 0.00000251 \, \text{V} \]
\[ E \approx 2.51 \, \mu V \]

### Final Answer
The induced emf in the loop at the instant when the radius is 2 cm is approximately **2.51 microvolts**.

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