A coil of effective area `2m^(2)` is placed at right angles to a uniform magnetic field of induction B. When the field reduces to ten percent of its original value in 0.6 sec, an e.m.f. Of 0.24 V is induced in it. The magnetic of magnetic induction (B)is

Options

A. `0.02 Wb//m^(2)`
B. `0.04 Wb//m^(2)`
C. `0.08 Wb//m^(2)`
D. `0.01 Wb//m^(2)`

Explanation

To solve the problem, we will use Faraday's law of electromagnetic induction, which states that the induced electromotive force (e.m.f.) in a coil is equal to the negative rate of change of magnetic flux through the coil.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Effective area of the coil, \( A = 2 \, \text{m}^2 \)
- Initial magnetic field, \( B \) (unknown)
- Final magnetic field after reduction, \( B_f = 0.1B \) (10% of the original)
- Time duration for the change, \( \Delta t = 0.6 \, \text{s} \)
- Induced e.m.f., \( \mathcal{E} = 0.24 \, \text{V} \)

2. **Calculate the Change in Magnetic Field:**
- The change in magnetic field, \( \Delta B = B_f - B = 0.1B - B = -0.9B \)

3. **Calculate the Change in Magnetic Flux:**
- The magnetic flux \( \Phi \) through the coil is given by:
\[ \Phi = B \cdot A \]
- Therefore, the change in magnetic flux \( \Delta \Phi \) is:
\[ \Delta \Phi = A \cdot \Delta B = A \cdot (-0.9B) = -0.9 \cdot 2 \cdot B = -1.8B \]

4. **Apply Faraday's Law:**
- According to Faraday's law, the induced e.m.f. is given by:
\[ \mathcal{E} = -\frac{\Delta \Phi}{\Delta t} \]
- Substituting the values we have:
\[ 0.24 = -\frac{-1.8B}{0.6} \]

5. **Solve for B:**
- Rearranging the equation:
\[ 0.24 = \frac{1.8B}{0.6} \]
- Multiply both sides by \( 0.6 \):
\[ 0.24 \times 0.6 = 1.8B \]
\[ 0.144 = 1.8B \]
- Now, divide both sides by \( 1.8 \):
\[ B = \frac{0.144}{1.8} = 0.08 \, \text{T} \]

### Final Answer:
The magnetic induction \( B \) is \( 0.08 \, \text{T} \).