A 800 turn coil of effective area `0.05 m^(2)` is kept perpendicular to a magnetic field `5xx10^(-5)` T. When the plane of the coil is rotated by `90^(@)` around any of its coplanar axis in 0.1 s, the emf induced in the coil will be:

Options

A. 0.02V
B. 2V
C. 0.2V
D. `2xx10^(-3)V`

Explanation

To solve the problem of finding the induced EMF in a coil when it is rotated in a magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Number of turns in the coil, \( N = 800 \)
- Effective area of the coil, \( A = 0.05 \, \text{m}^2 \)
- Magnetic field strength, \( B = 5 \times 10^{-5} \, \text{T} \)
- Time taken for rotation, \( \Delta t = 0.1 \, \text{s} \)

2. **Calculate the Initial Magnetic Flux (\( \Phi_i \)):**
The magnetic flux through the coil is given by:
\[ \Phi_i = N \cdot B \cdot A \]
Substituting the values:
\[ \Phi_i = 800 \cdot (5 \times 10^{-5}) \cdot (0.05) \]
\[ \Phi_i = 800 \cdot 5 \cdot 0.05 \times 10^{-5} \]
\[ \Phi_i = 800 \cdot 0.25 \times 10^{-5} = 200 \times 10^{-5} = 2 \times 10^{-3} \, \text{Wb} \]

3. **Calculate the Final Magnetic Flux (\( \Phi_f \)):**
When the coil is rotated by \( 90^\circ \), it becomes perpendicular to the magnetic field, and thus the magnetic flux becomes:
\[ \Phi_f = 0 \, \text{Wb} \]

4. **Calculate the Change in Magnetic Flux (\( \Delta \Phi \)):**
The change in magnetic flux is given by:
\[ \Delta \Phi = \Phi_f - \Phi_i \]
Substituting the values:
\[ \Delta \Phi = 0 - (2 \times 10^{-3}) = -2 \times 10^{-3} \, \text{Wb} \]

5. **Calculate the Induced EMF (\( \mathcal{E} \)):**
The induced EMF is calculated using Faraday's law of electromagnetic induction:
\[ \mathcal{E} = -\frac{\Delta \Phi}{\Delta t} \]
Substituting the values:
\[ \mathcal{E} = -\frac{-2 \times 10^{-3}}{0.1} \]
\[ \mathcal{E} = \frac{2 \times 10^{-3}}{0.1} = 0.02 \, \text{V} \]

### Final Answer:
The induced EMF in the coil is \( \mathcal{E} = 0.02 \, \text{V} \).

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