A coil has `1,000` turns and `500 cm^(2)` as its area. The plane of the coil is placed at right angles to a magnetic induction field of `2 xx 10^(-5) web//m^(2)`. The coil is rotated through `180^(@)` in `0.2` second. The average emf induced in the coil, in milli volts, is :

Options

A. `5`
B. `10`
C. `15`
D. `20`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
- Number of turns in the coil, \( N = 1000 \)
- Area of the coil, \( A = 500 \, \text{cm}^2 = 500 \times 10^{-4} \, \text{m}^2 = 5 \times 10^{-2} \, \text{m}^2 \)
- Magnetic field strength, \( B = 2 \times 10^{-5} \, \text{Wb/m}^2 \)
- The coil is rotated through \( 180^\circ \) in \( t = 0.2 \, \text{s} \)

### Step 2: Calculate the initial magnetic flux
The magnetic flux \( \Phi \) through the coil is given by:
\[ \Phi = N \cdot B \cdot A \cdot \cos(\theta) \]
Initially, the coil is at a right angle to the magnetic field, so \( \theta = 0^\circ \) (the angle between the area vector and the magnetic field vector). Therefore, \( \cos(0^\circ) = 1 \).

Calculating the initial flux:
\[ \Phi_i = N \cdot B \cdot A \cdot \cos(0) = 1000 \cdot (2 \times 10^{-5}) \cdot (5 \times 10^{-2}) \cdot 1 \]
\[ \Phi_i = 1000 \cdot 2 \times 10^{-5} \cdot 5 \times 10^{-2} = 1000 \cdot 10^{-6} = 1 \times 10^{-3} \, \text{Wb} \]

### Step 3: Calculate the final magnetic flux
After rotating the coil through \( 180^\circ \), the angle becomes \( \theta = 180^\circ \). Thus, \( \cos(180^\circ) = -1 \).

Calculating the final flux:
\[ \Phi_f = N \cdot B \cdot A \cdot \cos(180) = 1000 \cdot (2 \times 10^{-5}) \cdot (5 \times 10^{-2}) \cdot (-1) \]
\[ \Phi_f = -1 \times 10^{-3} \, \text{Wb} \]

### Step 4: Calculate the change in magnetic flux
The change in magnetic flux \( \Delta \Phi \) is given by:
\[ \Delta \Phi = \Phi_f - \Phi_i = (-1 \times 10^{-3}) - (1 \times 10^{-3}) = -2 \times 10^{-3} \, \text{Wb} \]

### Step 5: Calculate the average induced emf
The average induced emf \( \mathcal{E} \) is given by:
\[ \mathcal{E} = -\frac{\Delta \Phi}{\Delta t} \]
Substituting the values:
\[ \mathcal{E} = -\frac{-2 \times 10^{-3}}{0.2} = \frac{2 \times 10^{-3}}{0.2} = 10^{-2} \, \text{V} = 10 \, \text{mV} \]

### Final Answer
The average emf induced in the coil is \( 10 \, \text{mV} \).

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