If number of turns of `70 cm^(2)` coil is 200 and it is placed in a magnetic field of `0.8 Wb//m^(2)` which is perpendicular to the plane of coil and it is rotated through an angle `180^(@)` in `0.1` sec , then induced emf in coil :

Options

A. `11.2`V
B. `1.12`V
C. `22.4 V`
D. `2.24 V`

Answer

To solve the problem of induced EMF in a rotating coil placed in a magnetic field, we can follow these steps:

### Step 1: Understand the Given Data
- Number of turns (N) = 200
- Area of the coil (A) = 70 cm² = 70 × 10⁻⁴ m² (conversion to m²)
- Magnetic field strength (B) = 0.8 Wb/m²
- Angle of rotation (θ) = 180°
- Time taken for rotation (t) = 0.1 s

### Step 2: Calculate Initial and Final Magnetic Flux
The magnetic flux (Φ) through the coil is given by the formula:
\[ \Phi = N \cdot B \cdot A \cdot \cos(\theta) \]

**Initial Flux (Φ_initial)**:
- When the coil is perpendicular to the magnetic field, θ = 0°.
- Thus, \(\cos(0°) = 1\).
- Therefore, the initial flux is:
\[ \Phi_{\text{initial}} = N \cdot B \cdot A \cdot \cos(0°) = 200 \cdot 0.8 \cdot (70 \times 10^{-4}) \]
\[ \Phi_{\text{initial}} = 200 \cdot 0.8 \cdot 0.007 = 1.12 \text{ Wb} \]

**Final Flux (Φ_final)**:
- After rotating 180°, the angle becomes θ = 180°.
- Thus, \(\cos(180°) = -1\).
- Therefore, the final flux is:
\[ \Phi_{\text{final}} = N \cdot B \cdot A \cdot \cos(180°) = 200 \cdot 0.8 \cdot (70 \times 10^{-4}) \cdot (-1) \]
\[ \Phi_{\text{final}} = -1.12 \text{ Wb} \]

### Step 3: Calculate Change in Magnetic Flux
The change in magnetic flux (ΔΦ) is given by:
\[ \Delta \Phi = \Phi_{\text{final}} - \Phi_{\text{initial}} \]
\[ \Delta \Phi = -1.12 - 1.12 = -2.24 \text{ Wb} \]

### Step 4: Calculate Induced EMF
The induced EMF (ε) can be calculated using Faraday's law of electromagnetic induction:
\[ \epsilon = -\frac{\Delta \Phi}{\Delta t} \]
Substituting the values:
\[ \epsilon = -\frac{-2.24}{0.1} = \frac{2.24}{0.1} = 22.4 \text{ V} \]

### Final Answer
The induced EMF in the coil is **22.4 V**.

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