A coil of area `0.1m^(2)` has 500 tums. After placing the coil in a magnetic field of strength `4xx10^(-4)Wb//m^(2)` it is rotated through `90^(@)` in 0.1 s. The average emf induced in the coil is

Options

A. 0.2 volt
B. 0.1 volt
C. 0.05 volt
D. 0.012 volt

Answer

To find the average induced electromotive force (emf) in the coil, we will follow these steps:

### Step 1: Calculate the initial magnetic flux (Φ_initial)
The magnetic flux (Φ) through a coil is given by the formula:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
Where:
- \(B\) is the magnetic field strength (in Weber/m²),
- \(A\) is the area of the coil (in m²),
- \(\theta\) is the angle between the magnetic field and the normal to the surface of the coil.

Given:
- \(B = 4 \times 10^{-4} \, \text{Wb/m}^2\)
- \(A = 0.1 \, \text{m}^2\)
- \(\theta_{\text{initial}} = 0^\circ\) (initially, the coil is aligned with the magnetic field)

Calculating the initial magnetic flux:
\[ \Phi_{\text{initial}} = B \cdot A \cdot \cos(0^\circ) = 4 \times 10^{-4} \cdot 0.1 \cdot 1 = 4 \times 10^{-5} \, \text{Wb} \]

### Step 2: Calculate the final magnetic flux (Φ_final)
After rotating the coil through \(90^\circ\), the angle becomes:
\(\theta_{\text{final}} = 90^\circ\)

Calculating the final magnetic flux:
\[ \Phi_{\text{final}} = B \cdot A \cdot \cos(90^\circ) = 4 \times 10^{-4} \cdot 0.1 \cdot 0 = 0 \, \text{Wb} \]

### Step 3: Calculate the change in magnetic flux (ΔΦ)
The change in magnetic flux is given by:
\[ \Delta \Phi = \Phi_{\text{final}} - \Phi_{\text{initial}} = 0 - 4 \times 10^{-5} = -4 \times 10^{-5} \, \text{Wb} \]

### Step 4: Calculate the average induced emf (ε)
The average induced emf can be calculated using Faraday's law of electromagnetic induction:
\[ \varepsilon = -\frac{\Delta \Phi}{\Delta t} \]
Where \(\Delta t\) is the time taken for the change in flux.

Given:
- \(\Delta t = 0.1 \, \text{s}\)

Calculating the average induced emf:
\[ \varepsilon = -\frac{-4 \times 10^{-5}}{0.1} = \frac{4 \times 10^{-5}}{0.1} = 4 \times 10^{-4} \, \text{V} \]

### Step 5: Considering the number of turns in the coil
Since the coil has 500 turns, the total induced emf is:
\[ \varepsilon_{\text{total}} = N \cdot \varepsilon = 500 \cdot 4 \times 10^{-4} = 2 \times 10^{-1} = 0.2 \, \text{V} \]

### Final Answer
The average induced emf in the coil is \(0.2 \, \text{V}\).

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