A field of strength `8 xx 10^4//pi` ampere turns / meter acts at right angles to the coil of 500 turns of area `10^(-2) m^2` . The coil is removed from the field in 0.2 second. Then the induced e.m.f in the coil is

Options

A. 0.1 V
B. 0.2 V
C. 1.6 V
D. 2.5 V

Explanation

To solve the problem, we need to calculate the induced electromotive force (e.m.f) in a coil when it is removed from a magnetic field. Here are the steps to arrive at the solution:

### Step 1: Calculate the Magnetic Field Strength (B)
The magnetic field strength (B) can be calculated using the formula:
\[ B = \mu_0 H \]
where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space)
- \( H = \frac{8 \times 10^4}{\pi} \, \text{A/m} \)

Substituting the value of \( H \):
\[ B = 4\pi \times 10^{-7} \times \frac{8 \times 10^4}{\pi} \]
\[ B = 4 \times 8 \times 10^{-7} \times 10^4 \]
\[ B = 32 \times 10^{-3} \, \text{T} \]
Thus, the magnetic field strength \( B = 0.032 \, \text{T} \).

### Step 2: Determine the Magnetic Flux (Φ)
The magnetic flux (Φ) through the coil is given by:
\[ \Phi = n \cdot B \cdot A \]
where:
- \( n = 500 \) (number of turns)
- \( A = 10^{-2} \, \text{m}^2 \) (area of the coil)

Substituting the values:
\[ \Phi = 500 \cdot 0.032 \cdot 10^{-2} \]
\[ \Phi = 500 \cdot 0.032 \cdot 0.01 \]
\[ \Phi = 0.16 \, \text{Wb} \]

### Step 3: Calculate the Change in Magnetic Flux (ΔΦ)
Since the coil is removed from the magnetic field, the final magnetic flux is 0. Therefore, the change in magnetic flux (ΔΦ) is:
\[ \Delta \Phi = \Phi_{\text{final}} - \Phi_{\text{initial}} \]
\[ \Delta \Phi = 0 - 0.16 \]
\[ \Delta \Phi = -0.16 \, \text{Wb} \]

### Step 4: Calculate the Induced e.m.f (ε)
The induced e.m.f can be calculated using Faraday's law of electromagnetic induction:
\[ \epsilon = -\frac{d\Phi}{dt} \]
Given that the time taken to remove the coil from the field is \( dt = 0.2 \, \text{s} \):
\[ \epsilon = -\frac{-0.16}{0.2} \]
\[ \epsilon = \frac{0.16}{0.2} \]
\[ \epsilon = 0.8 \, \text{V} \]

### Final Answer
The induced e.m.f in the coil is \( 0.8 \, \text{V} \).

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