A coil of area `0.04 m^(2)` having 1000 turns is suspended perpendicular to a magnetic field of `5.0xx10^(-5) Wbm^(-2)`. It is rotated through `90^(@)` in 0.2 s. Calculate the average emf induced in it.

Answer

To calculate the average electromotive force (emf) induced in the coil, we can follow these steps:

### Step 1: Identify the given values
- Area of the coil, \( A = 0.04 \, m^2 \)
- Number of turns, \( N = 1000 \)
- Magnetic field strength, \( B = 5.0 \times 10^{-5} \, Wb/m^2 \)
- Initial angle, \( \theta_1 = 0^\circ \) (perpendicular to the magnetic field)
- Final angle, \( \theta_2 = 90^\circ \) (after rotation)
- Time taken for rotation, \( \Delta t = 0.2 \, s \)

### Step 2: Calculate the initial and final magnetic flux
The magnetic flux \( \Phi \) through the coil is given by the formula:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]

- **Initial flux** (\( \Phi_1 \)):
\[ \Phi_1 = B \cdot A \cdot \cos(\theta_1) = 5.0 \times 10^{-5} \cdot 0.04 \cdot \cos(0^\circ) = 5.0 \times 10^{-5} \cdot 0.04 \cdot 1 = 2.0 \times 10^{-6} \, Wb \]

- **Final flux** (\( \Phi_2 \)):
\[ \Phi_2 = B \cdot A \cdot \cos(\theta_2) = 5.0 \times 10^{-5} \cdot 0.04 \cdot \cos(90^\circ) = 5.0 \times 10^{-5} \cdot 0.04 \cdot 0 = 0 \, Wb \]

### Step 3: Calculate the change in magnetic flux
The change in magnetic flux \( \Delta \Phi \) is given by:
\[ \Delta \Phi = \Phi_2 - \Phi_1 = 0 - 2.0 \times 10^{-6} = -2.0 \times 10^{-6} \, Wb \]

### Step 4: Calculate the average emf induced
The average emf (\( E \)) induced in the coil can be calculated using Faraday's law of electromagnetic induction:
\[ E = -N \frac{\Delta \Phi}{\Delta t} \]
Substituting the values:
\[ E = -1000 \cdot \frac{-2.0 \times 10^{-6}}{0.2} = 1000 \cdot \frac{2.0 \times 10^{-6}}{0.2} \]
\[ E = 1000 \cdot 10^{-5} = 0.01 \, V \]

### Step 5: Conclusion
The average emf induced in the coil is:
\[ E = 0.01 \, V \]