A coil of 100 turns having an average area of `100 cm^2` for each turn is held in a uniform field of 50 gauss, the direction of the field being at right angles to the plane of the coil. If the field is removed in 0.01 sec, then e.m.f. induced average in coil is

Options

A. 0.5 V
B. 10 V
C. 20 V
D. 50V

Explanation

To find the average induced electromotive force (e.m.f.) in the coil when the magnetic field is removed, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Number of turns in the coil, \( N = 100 \)
- Area of each turn, \( A = 100 \, \text{cm}^2 = 100 \times 10^{-4} \, \text{m}^2 = 10^{-2} \, \text{m}^2 \)
- Initial magnetic field, \( B_i = 50 \, \text{Gauss} = 50 \times 10^{-4} \, \text{Tesla} = 5 \times 10^{-3} \, \text{Tesla} \)
- Final magnetic field, \( B_f = 0 \, \text{Tesla} \)
- Time interval for the change, \( \Delta t = 0.01 \, \text{s} \)

2. **Calculate the Change in Magnetic Field (\( \Delta B \)):**
\[ \Delta B = B_f - B_i = 0 - 5 \times 10^{-3} = -5 \times 10^{-3} \, \text{Tesla} \]

3. **Use Faraday's Law of Electromagnetic Induction:**
The average induced e.m.f. (\( \mathcal{E} \)) is given by:
\[ \mathcal{E} = -N \frac{\Delta \Phi}{\Delta t} \]
where \( \Delta \Phi \) is the change in magnetic flux.

4. **Calculate the Change in Magnetic Flux (\( \Delta \Phi \)):**
The magnetic flux (\( \Phi \)) through one turn of the coil is given by:
\[ \Phi = B \cdot A \]
Therefore, the change in magnetic flux is:
\[ \Delta \Phi = A \cdot \Delta B = 10^{-2} \cdot (-5 \times 10^{-3}) = -5 \times 10^{-5} \, \text{Wb} \]

5. **Substitute Values into the e.m.f. Formula:**
\[ \mathcal{E} = -N \frac{\Delta \Phi}{\Delta t} = -100 \cdot \frac{-5 \times 10^{-5}}{0.01} \]

6. **Calculate the Induced e.m.f.:**
\[ \mathcal{E} = 100 \cdot \frac{5 \times 10^{-5}}{0.01} = 100 \cdot 5 \times 10^{-3} = 0.5 \, \text{V} \]

### Final Answer:
The average induced e.m.f. in the coil is \( 0.5 \, \text{V} \).

---