A moving coil galvanometer has a coil with 175 turns and area `1 cm^(2)`. It uses a torsion band of torsion constant `10^(-6) N-m//rad`. The coil is placed in a magnetic field B parallel to its plane. The coil deflects by `1^(@)` for a current of 1 mA. The value of B (in Tesla) is approximately.

Options

A. `10^(-4)`
B. `10^(-2)`
C. `10^(-1)`
D. `10^(-3)`

Answer

To solve the problem, we need to find the magnetic field \( B \) in a moving coil galvanometer given the following parameters:

- Number of turns, \( n = 175 \)
- Area of the coil, \( A = 1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2 \)
- Torsion constant, \( C = 10^{-6} \, \text{N-m/rad} \)
- Deflection angle, \( \theta = 1^\circ \)
- Current, \( I = 1 \, \text{mA} = 1 \times 10^{-3} \, \text{A} \)

### Step-by-Step Solution:

1. **Convert the angle from degrees to radians:**
\[ \theta = 1^\circ = \frac{\pi}{180} \, \text{radians} \]

2. **Write the equation for torque:**
The torque \( \tau \) on the coil due to the magnetic field is given by:
\[ \tau = C \cdot \theta \]
where \( C \) is the torsion constant and \( \theta \) is the deflection in radians.

3. **Write the expression for the torque due to the magnetic moment:**
The torque \( \tau \) on a current-carrying coil in a magnetic field is also given by:
\[ \tau = m \cdot B \]
where \( m \) is the magnetic moment, given by:
\[ m = n \cdot I \cdot A \]
Here, \( n \) is the number of turns, \( I \) is the current, and \( A \) is the area of the coil.

4. **Combine the equations:**
Setting the two expressions for torque equal to each other:
\[ C \cdot \theta = n \cdot I \cdot A \cdot B \]

5. **Rearranging to find \( B \):**
\[ B = \frac{C \cdot \theta}{n \cdot I \cdot A} \]

6. **Substituting the known values:**
- Convert \( \theta \) to radians:
\[ \theta = \frac{\pi}{180} \, \text{radians} \]
- Substitute \( C = 10^{-6} \, \text{N-m/rad} \), \( n = 175 \), \( I = 1 \times 10^{-3} \, \text{A} \), and \( A = 1 \times 10^{-4} \, \text{m}^2 \):
\[ B = \frac{10^{-6} \cdot \frac{\pi}{180}}{175 \cdot 1 \times 10^{-3} \cdot 1 \times 10^{-4}} \]

7. **Calculating the denominator:**
\[ 175 \cdot 1 \times 10^{-3} \cdot 1 \times 10^{-4} = 175 \times 10^{-7} = 1.75 \times 10^{-5} \]

8. **Calculating \( B \):**
\[ B = \frac{10^{-6} \cdot \frac{\pi}{180}}{1.75 \times 10^{-5}} = \frac{10^{-6} \cdot 3.14159}{180 \cdot 1.75 \times 10^{-5}} \]
\[ B \approx \frac{3.14159 \times 10^{-6}}{3.15 \times 10^{-5}} \approx 0.1 \, \text{T} \]

9. **Final approximation:**
The value of \( B \) is approximately \( 10^{-3} \, \text{T} \).

### Conclusion:
The approximate value of the magnetic field \( B \) is \( 10^{-3} \, \text{T} \).