A field of strenght `5 xx 10^(4)//pi` ampere turns`//`meter acts at right angles to the coil of `50` turns of area`10^(-2) m^(2)`. The coil is removed from the field in `0.1` second. Then the induced `e.m.f` in the coil is

Options

A. `0.1 V`
B. `0.2 V`
C. `1.96 V`
D. `0.98 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’s a step-by-step solution:

### Step 1: Identify the given values
- Magnetic field strength (H) = \( \frac{5 \times 10^4}{\pi} \) A/m
- Number of turns (N) = 50
- Area of the coil (A) = \( 10^{-2} \) m²
- Time (t) = 0.1 s

### Step 2: Calculate the magnetic flux density (B)
The magnetic flux density (B) can be calculated using the formula:
\[ B = \mu_0 H \]
where \( \mu_0 \) (the permeability of free space) is approximately \( 4\pi \times 10^{-7} \) T·m/A.

Substituting the values:
\[ B = (4\pi \times 10^{-7}) \times \left(\frac{5 \times 10^4}{\pi}\right) \]
\[ B = 4 \times 5 \times 10^{-7} \times 10^4 \]
\[ B = 20 \times 10^{-3} \]
\[ B = 2 \times 10^{-2} \text{ T} \]

### Step 3: Calculate the initial magnetic flux (Φ_initial)
The magnetic flux (Φ) through the coil is given by:
\[ \Phi = N \times B \times A \]
Substituting the values:
\[ \Phi_{\text{initial}} = 50 \times (2 \times 10^{-2}) \times (10^{-2}) \]
\[ \Phi_{\text{initial}} = 50 \times 2 \times 10^{-4} \]
\[ \Phi_{\text{initial}} = 10^{-2} \text{ Wb} \]

### Step 4: Calculate the final magnetic flux (Φ_final)
When the coil is removed from the magnetic field, the final magnetic flux is:
\[ \Phi_{\text{final}} = 0 \text{ Wb} \]

### Step 5: Calculate the change in magnetic flux (ΔΦ)
The change in magnetic flux is:
\[ \Delta \Phi = \Phi_{\text{final}} - \Phi_{\text{initial}} \]
\[ \Delta \Phi = 0 - 10^{-2} \]
\[ \Delta \Phi = -10^{-2} \text{ Wb} \]

### Step 6: Calculate the induced e.m.f (ε)
The induced e.m.f can be calculated using Faraday's law of electromagnetic induction:
\[ \epsilon = -\frac{\Delta \Phi}{\Delta t} \]
Substituting the values:
\[ \epsilon = -\frac{-10^{-2}}{0.1} \]
\[ \epsilon = \frac{10^{-2}}{0.1} \]
\[ \epsilon = 10^{-1} \]
\[ \epsilon = 0.1 \text{ V} \]

### Final Answer
The induced e.m.f in the coil is **0.1 V**.

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