A rectangular coil of 100 turns and size `0.1mxx0.05m` is placed perpendicular to a magnetic field of 0.1 T. If the field drops to 0.05 T in 0.05 s, the magnitude of the emf induced in the coil is

Options

A. 0.2 V
B. 0.3 V
C. 0.5 V
D. 0.6 V

Explanation

To solve the problem of finding the induced EMF in a rectangular coil when the magnetic field changes, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Number of turns in the coil, \( N = 100 \)
- Initial magnetic field, \( B_i = 0.1 \, \text{T} \)
- Final magnetic field, \( B_f = 0.05 \, \text{T} \)
- Time duration for the change in magnetic field, \( \Delta t = 0.05 \, \text{s} \)
- Dimensions of the coil: Length \( l = 0.1 \, \text{m} \) and Width \( w = 0.05 \, \text{m} \)

2. **Calculate the Change in Magnetic Field:**
\[ \Delta B = B_i - B_f = 0.1 \, \text{T} - 0.05 \, \text{T} = 0.05 \, \text{T} \]

3. **Calculate the Area of the Coil:**
\[ A = l \times w = 0.1 \, \text{m} \times 0.05 \, \text{m} = 0.005 \, \text{m}^2 = 5 \times 10^{-3} \, \text{m}^2 \]

4. **Calculate the Change in Magnetic Flux (\( \Delta \Phi \)):**
The change in magnetic flux is given by:
\[ \Delta \Phi = A \times \Delta B = 5 \times 10^{-3} \, \text{m}^2 \times 0.05 \, \text{T} = 2.5 \times 10^{-4} \, \text{Wb} \]

5. **Calculate the Induced EMF (\( \epsilon \)):**
The induced EMF can be calculated using Faraday's law of electromagnetic induction:
\[ \epsilon = -N \frac{\Delta \Phi}{\Delta t} \]
Substituting the values:
\[ \epsilon = -100 \times \frac{2.5 \times 10^{-4} \, \text{Wb}}{0.05 \, \text{s}} = -100 \times 5 \times 10^{-3} = -0.5 \, \text{V} \]
The magnitude of the induced EMF is:
\[ |\epsilon| = 0.5 \, \text{V} \]

### Final Answer:
The magnitude of the induced EMF in the coil is \( 0.5 \, \text{V} \).

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