A coil of area `500cm^(2)` and having `1000` turns is held perpendicular to a uniform field of `0.4` gauss. The coil is turned through `180^(@)` in `1//10sec` . Calculate the average induced e.m.f.

Options

A. 0.02 V
B. 0.04 V
C. 1.4 V
D. 0.08 V

Explanation

To solve the problem step by step, we will follow the principles of electromagnetic induction and apply the relevant formulas.

### Step 1: Convert the given values to SI units
1. **Area of the coil (A)**:
- Given: \(500 \, \text{cm}^2\)
- Conversion: \(500 \, \text{cm}^2 = 500 \times 10^{-4} \, \text{m}^2 = 5 \times 10^{-2} \, \text{m}^2\)

2. **Magnetic field (B)**:
- Given: \(0.4 \, \text{gauss}\)
- Conversion: \(0.4 \, \text{gauss} = 0.4 \times 10^{-4} \, \text{tesla} = 4 \times 10^{-5} \, \text{T}\)

3. **Number of turns (n)**:
- Given: \(1000\)

4. **Time (Δt)**:
- Given: \(1/10 \, \text{sec} = 0.1 \, \text{sec}\)

### Step 2: Calculate the change in magnetic flux (ΔΦ)
1. **Initial magnetic flux (Φ_initial)** when the coil is perpendicular to the magnetic field:
\[ \Phi_{\text{initial}} = n \cdot B \cdot A \]
\[ \Phi_{\text{initial}} = 1000 \cdot (4 \times 10^{-5}) \cdot (5 \times 10^{-2}) = 2 \times 10^{-5} \, \text{Wb} \]

2. **Final magnetic flux (Φ_final)** when the coil is turned 180 degrees:
\[ \Phi_{\text{final}} = n \cdot B \cdot A \cdot (-1) = -n \cdot B \cdot A \]
\[ \Phi_{\text{final}} = -2 \times 10^{-5} \, \text{Wb} \]

3. **Change in magnetic flux (ΔΦ)**:
\[ \Delta \Phi = \Phi_{\text{final}} - \Phi_{\text{initial}} = -2 \times 10^{-5} - 2 \times 10^{-5} = -4 \times 10^{-5} \, \text{Wb} \]

### Step 3: Calculate the average induced EMF (E)
1. **Using the formula for average induced EMF**:
\[ E = -\frac{\Delta \Phi}{\Delta t} \]
\[ E = -\frac{-4 \times 10^{-5}}{0.1} = \frac{4 \times 10^{-5}}{0.1} = 4 \times 10^{-4} \, \text{V} = 0.04 \, \text{V} \]

### Final Answer
The average induced EMF is \(0.04 \, \text{V}\).

---