A coil has an area of `0.05 m^(2)` and it has `800` turns. It is placed perpendicular in a magnitude field of strength `4xx10^(-5)Wb//m^(2)`, it is rotated through `90^(@)` in `0.1` sec. the average e.m.f. induced in the coil is

Options

A. 0.056 V
B. 0.046 V
C. 0.026 V
D. 0.016 V

Explanation

To find the average e.m.f. induced in the coil, we can follow these steps:

### Step 1: Identify the given parameters
- Area of the coil, \( A = 0.05 \, m^2 \)
- Number of turns, \( N = 800 \)
- Magnetic field strength, \( B = 4 \times 10^{-5} \, Wb/m^2 \)
- Initial angle, \( \theta_i = 0^\circ \)
- Final angle, \( \theta_f = 90^\circ \)
- Time taken for rotation, \( t = 0.1 \, s \)

### Step 2: Calculate the initial and final magnetic flux
The magnetic flux \( \Phi \) through the coil is given by the formula:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]

**Initial flux (\( \Phi_i \)) when \( \theta = 0^\circ \):**
\[ \Phi_i = B \cdot A \cdot \cos(0^\circ) = B \cdot A \cdot 1 = (4 \times 10^{-5}) \cdot (0.05) = 2 \times 10^{-6} \, Wb \]

**Final flux (\( \Phi_f \)) when \( \theta = 90^\circ \):**
\[ \Phi_f = B \cdot A \cdot \cos(90^\circ) = B \cdot A \cdot 0 = 0 \, Wb \]

### Step 3: Calculate the change in magnetic flux
\[ \Delta \Phi = \Phi_f - \Phi_i = 0 - (2 \times 10^{-6}) = -2 \times 10^{-6} \, Wb \]

### Step 4: Calculate the average e.m.f. induced
The average e.m.f. (\( \mathcal{E} \)) induced in the coil can be calculated using the formula:
\[ \mathcal{E} = -\frac{\Delta \Phi}{t} \]
Substituting the values:
\[ \mathcal{E} = -\frac{-2 \times 10^{-6}}{0.1} = \frac{2 \times 10^{-6}}{0.1} = 2 \times 10^{-5} \, V \]

### Step 5: Multiply by the number of turns
Since the coil has \( N = 800 \) turns, the total e.m.f. induced is:
\[ \mathcal{E}_{total} = N \cdot \mathcal{E} = 800 \cdot (2 \times 10^{-5}) = 0.016 \, V \]

### Final Answer
The average e.m.f. induced in the coil is \( 0.016 \, V \).

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