A coil has 1,000 turns and `500 cm^2` as its area. The plane of the coil is placed at right angles to a magnetic induction field of `2xx10^(-5) wb//m^2` . The coil is rotated through `180^@` in 0.2 seconds. The average e.m.f induced in the coil, in milli-volts.

Options

A. 5
B. 10
C. 15
D. 20

Answer

To solve the problem step by step, we will use the formula for induced electromotive force (e.m.f) in a coil due to a change in magnetic flux.

### Step 1: Identify the given values
- Number of turns in the coil (N) = 1000 turns
- Area of the coil (A) = 500 cm² = 500 × 10^(-4) m² = 0.05 m²
- Magnetic field induction (B) = 2 × 10^(-5) Wb/m²
- Angle of rotation = 180°
- Time taken for rotation (Δt) = 0.2 seconds

### Step 2: Calculate the initial magnetic flux (Φ_initial)
The magnetic flux (Φ) through the coil is given by the formula:
\[ \Phi = N \cdot B \cdot A \cdot \cos(\theta) \]
Since the coil is initially placed at right angles to the magnetic field, the angle θ = 0°.
Thus,
\[ \Phi_{initial} = N \cdot B \cdot A \cdot \cos(0°) = N \cdot B \cdot A \]
Substituting the values:
\[ \Phi_{initial} = 1000 \cdot (2 \times 10^{-5}) \cdot (0.05) \]
\[ \Phi_{initial} = 1000 \cdot 2 \cdot 10^{-5} \cdot 0.05 \]
\[ \Phi_{initial} = 1000 \cdot 10^{-6} \]
\[ \Phi_{initial} = 1 \times 10^{-3} \, \text{Wb} \]

### Step 3: Calculate the final magnetic flux (Φ_final)
After rotating the coil by 180°, the angle θ becomes 180°. Thus,
\[ \Phi_{final} = N \cdot B \cdot A \cdot \cos(180°) \]
Since cos(180°) = -1,
\[ \Phi_{final} = N \cdot B \cdot A \cdot (-1) \]
\[ \Phi_{final} = - (1000 \cdot (2 \times 10^{-5}) \cdot (0.05)) \]
\[ \Phi_{final} = -1 \times 10^{-3} \, \text{Wb} \]

### Step 4: Calculate the change in magnetic flux (ΔΦ)
The change in magnetic flux (ΔΦ) is given by:
\[ \Delta \Phi = \Phi_{final} - \Phi_{initial} \]
Substituting the values:
\[ \Delta \Phi = -1 \times 10^{-3} - 1 \times 10^{-3} \]
\[ \Delta \Phi = -2 \times 10^{-3} \, \text{Wb} \]

### Step 5: Calculate the average induced e.m.f (E)
The average induced e.m.f (E) can be calculated using the formula:
\[ E = -\frac{\Delta \Phi}{\Delta t} \]
Substituting the values:
\[ E = -\frac{-2 \times 10^{-3}}{0.2} \]
\[ E = \frac{2 \times 10^{-3}}{0.2} \]
\[ E = 10 \times 10^{-3} \, \text{V} \]
\[ E = 10 \, \text{mV} \]

### Final Answer
The average e.m.f induced in the coil is **10 millivolts**.

---