A square coil of 100 turns and side length 1 cm is rotating in a uniform magnetic field such that axis of rotation is perpendicular to magnetic field. If coil completes 1 rotation in 1ms.Find average EMF induced in `muV)` in half rotation. ( Magnetic field `( B) = 10^(-5) T ` ) .

Explanation

To find the average EMF induced in a square coil rotating in a magnetic field, we can follow these steps:

### Step 1: Identify the given parameters
- Number of turns (N) = 100
- Side length of the square coil (a) = 1 cm = 0.01 m
- Magnetic field (B) = \(10^{-5}\) T
- Time for one complete rotation (T) = 1 ms = \(10^{-3}\) s

### Step 2: Calculate the area of the square coil
The area (A) of the square coil can be calculated using the formula:
\[ A = a^2 \]
Substituting the value of \(a\):
\[ A = (0.01)^2 = 0.0001 \, m^2 = 10^{-4} \, m^2 \]

### Step 3: Calculate the angular velocity (ω)
The angular velocity (ω) can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \(T\):
\[ \omega = \frac{2\pi}{10^{-3}} = 2\pi \times 10^3 \, \text{rad/s} \]

### Step 4: Calculate the maximum EMF (E₀)
The maximum EMF (E₀) induced in the coil can be calculated using the formula:
\[ E_0 = N \cdot B \cdot A \cdot \omega \]
Substituting the values:
\[ E_0 = 100 \cdot (10^{-5}) \cdot (10^{-4}) \cdot (2\pi \times 10^3) \]
Calculating this step-by-step:
1. \(N \cdot B = 100 \cdot 10^{-5} = 10^{-3}\)
2. \(A \cdot \omega = 10^{-4} \cdot (2\pi \times 10^3) = 2\pi \times 10^{-1}\)
3. Now, \(E_0 = 10^{-3} \cdot 2\pi \times 10^{-1} = 2\pi \times 10^{-4}\)

### Step 5: Calculate the average EMF (E_avg) in half rotation
The average EMF over half a rotation can be calculated using:
\[ E_{avg} = \frac{2E_0}{\pi} \]
Substituting the value of \(E_0\):
\[ E_{avg} = \frac{2 \cdot (2\pi \times 10^{-4})}{\pi} = 4 \times 10^{-4} \, \text{V} \]
Converting to microvolts:
\[ E_{avg} = 4 \times 10^{2} \, \mu V = 400 \, \mu V \]

### Final Answer
The average EMF induced in the coil during half a rotation is **400 microvolts**.