The rectangular coil of a moving coil galvanometer has area of `8 xx 10^(-4) m^(2)` and 100 turns. The coil is kept in a radial horizontal magnetic field of` 0.2` T.
Calculate the restoring torque constant of hair spring connected to coil when current of `4 xx 10^(-5)` A passes through it and deflects the scale by `20^(@)` .

Answer

To solve the problem, we need to calculate the restoring torque constant (K) of the hair spring connected to the coil of the moving coil galvanometer. We will use the formula for the torque acting on the coil and the relationship between the torque and the deflection angle.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Area of the coil (A) = \(8 \times 10^{-4} \, m^2\)
- Number of turns (N) = 100
- Magnetic field (B) = 0.2 T
- Current (I) = \(4 \times 10^{-5} \, A\)
- Deflection angle (θ) = 20° (we will convert this to radians for calculations)

2. **Convert the Deflection Angle to Radians:**
- To convert degrees to radians, use the formula:
\[ \text{radians} = \frac{\pi}{180} \times \text{degrees} \]
- Thus,
\[ θ = \frac{\pi}{180} \times 20 = \frac{\pi}{9} \, \text{radians} \]

3. **Calculate the Torque (τ) Acting on the Coil:**
- The torque acting on the coil can be calculated using the formula:
\[ τ = N \cdot I \cdot A \cdot B \cdot \sin(θ) \]
- Since the angle θ is small, we can use sin(θ) ≈ θ in radians:
\[ τ = N \cdot I \cdot A \cdot B \cdot θ \]
- Substituting the values:
\[ τ = 100 \cdot (4 \times 10^{-5}) \cdot (8 \times 10^{-4}) \cdot (0.2) \cdot \left(\frac{\pi}{9}\right) \]

4. **Calculate τ:**
- First, calculate the product:
\[ τ = 100 \cdot 4 \cdot 8 \cdot 0.2 \cdot \frac{\pi}{9} \times 10^{-5} \times 10^{-4} \]
- Simplifying:
\[ τ = 100 \cdot 4 \cdot 8 \cdot 0.2 \cdot \frac{\pi}{9} \times 10^{-9} \]
- Calculating the numerical part:
\[ τ = \frac{640 \pi}{9} \times 10^{-9} \approx 2.24 \times 10^{-7} \, \text{N m} \]

5. **Calculate the Restoring Torque Constant (K):**
- The restoring torque constant is given by:
\[ K = \frac{τ}{θ} \]
- Substituting the values:
\[ K = \frac{2.24 \times 10^{-7}}{\frac{\pi}{9}} \approx 3.2 \times 10^{-8} \, \text{N m/degree} \]

### Final Answer:
The restoring torque constant (K) of the hair spring connected to the coil is approximately:
\[ K \approx 3.2 \times 10^{-8} \, \text{N m/degree} \]