A coil having 50 turns and an area of `800 cm^(2)` is held with its plane perpendicular to uniform magnetic field of induction `5xx10^(-5)Wb//m^(2)`. If it is pulled out of the field in 2 second, then the emf induced in the coil will be

Options

A. `10^(-4)V`
B. `10^(-5)V`
C. `10^(-6)V`
D. `10^(-8)V`

Explanation

To solve the problem of finding the induced EMF in a coil being pulled out of a magnetic field, we will follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Number of turns (N) = 50
- Area of the coil (A) = 800 cm² = 800 × 10^(-4) m² = 0.08 m²
- Magnetic field induction (B) = 5 × 10^(-5) Wb/m²
- Time taken to pull out the coil (Δt) = 2 seconds

2. **Calculate Initial Magnetic Flux (Φ_initial):**
The magnetic flux (Φ) through the coil is given by the formula:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
Since the plane of the coil is perpendicular to the magnetic field, the angle (θ) = 0 degrees, and cos(0) = 1. Therefore:
\[ \Phi_{\text{initial}} = B \cdot A = (5 \times 10^{-5}) \cdot (0.08) = 4 \times 10^{-6} \text{ Wb} \]

3. **Calculate Final Magnetic Flux (Φ_final):**
After the coil is pulled out of the magnetic field, the magnetic field induction becomes zero:
\[ \Phi_{\text{final}} = 0 \text{ Wb} \]

4. **Calculate Change in Magnetic Flux (ΔΦ):**
The change in magnetic flux (ΔΦ) is given by:
\[ \Delta \Phi = \Phi_{\text{final}} - \Phi_{\text{initial}} = 0 - (4 \times 10^{-6}) = -4 \times 10^{-6} \text{ Wb} \]

5. **Calculate Induced EMF (ε):**
The induced EMF (ε) can be calculated using Faraday's law of electromagnetic induction:
\[ \varepsilon = -N \frac{\Delta \Phi}{\Delta t} \]
Substituting the values:
\[ \varepsilon = -50 \cdot \frac{-4 \times 10^{-6}}{2} = 50 \cdot \frac{4 \times 10^{-6}}{2} = 50 \cdot 2 \times 10^{-6} = 100 \times 10^{-6} \text{ V} = 10^{-4} \text{ V} \]

### Final Answer:
The induced EMF in the coil is \( \varepsilon = 10^{-4} \text{ V} \) or \( 100 \mu V \).

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