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

Explanation

To find the average electromotive force (e.m.f.) induced in the coil, we can use Faraday's law of electromagnetic induction, which states that the induced e.m.f. (ε) in a coil is equal to the negative rate of change of magnetic flux through the coil.

### Step-by-Step Solution:

1. **Calculate the initial magnetic flux (Φ_initial)**:
The magnetic flux (Φ) through a coil is given by the formula:
\[ Φ = B \cdot A \cdot \cos(θ) \]
where:
- \(B\) = magnetic field strength = \(4 \times 10^{-5} \, Wb/m^2\)
- \(A\) = area of the coil = \(0.05 \, m^2\)
- \(θ\) = angle between the magnetic field and the normal to the coil surface = \(0°\) (since the coil is perpendicular to the field initially)

Thus,
\[ Φ_{\text{initial}} = B \cdot A \cdot \cos(0°) = 4 \times 10^{-5} \cdot 0.05 \cdot 1 = 2 \times 10^{-6} \, Wb \]

2. **Calculate the final magnetic flux (Φ_final)**:
After rotating the coil through \(90°\), the angle \(θ\) becomes \(90°\):
\[ Φ_{\text{final}} = B \cdot A \cdot \cos(90°) = 4 \times 10^{-5} \cdot 0.05 \cdot 0 = 0 \, Wb \]

3. **Calculate the change in magnetic flux (ΔΦ)**:
\[ ΔΦ = Φ_{\text{final}} - Φ_{\text{initial}} = 0 - 2 \times 10^{-6} = -2 \times 10^{-6} \, Wb \]

4. **Calculate the average e.m.f. (ε)**:
The average e.m.f. induced in the coil can be calculated using the formula:
\[ ε = -N \cdot \frac{ΔΦ}{Δt} \]
where:
- \(N\) = number of turns = \(800\)
- \(Δt\) = time taken for the change = \(0.1 \, s\)

Substituting the values:
\[ ε = -800 \cdot \frac{-2 \times 10^{-6}}{0.1} = 800 \cdot 2 \times 10^{-5} = 0.016 \, V \]

Thus, the average e.m.f. induced in the coil is \(0.016 \, V\) or \(16 \, mV\).