A galvanometer of number of turns 175 having `1cm^(2)` area through `1^(@)` when a current of `1mA` is passed. Find magnetic field if torsional constant of spring is `10^(-6)N-m`

Options

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

Answer

To solve the problem, we need to find the magnetic field \( B \) using the given parameters of the galvanometer. The relationship between the magnetic field, current, number of turns, area, torsional constant, and angular displacement is given by the formula:

\[ B \cdot I \cdot n \cdot A = C \cdot \theta \]

Where:
- \( B \) = magnetic field
- \( I \) = current in amperes
- \( n \) = number of turns
- \( A \) = area in square meters
- \( C \) = torsional constant in \( N \cdot m \)
- \( \theta \) = angular displacement in radians

### Step-by-step Solution:

1. **Identify the Given Values:**
- Number of turns, \( n = 175 \)
- Area, \( A = 1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2 \) (conversion from cm² to m²)
- Current, \( I = 1 \, \text{mA} = 1 \times 10^{-3} \, \text{A} \) (conversion from mA to A)
- Torsional constant, \( C = 10^{-6} \, \text{N} \cdot \text{m} \)
- Angular displacement, \( \theta = 1^\circ = \frac{\pi}{180} \, \text{radians} \) (conversion from degrees to radians)

2. **Substitute the Values into the Formula:**
Rearranging the formula to solve for \( B \):
\[ B = \frac{C \cdot \theta}{I \cdot n \cdot A} \]

Substitute the known values:
\[ B = \frac{10^{-6} \cdot \frac{\pi}{180}}{(1 \times 10^{-3}) \cdot 175 \cdot (1 \times 10^{-4})} \]

3. **Calculate the Denominator:**
Calculate \( I \cdot n \cdot A \):
\[ I \cdot n \cdot A = (1 \times 10^{-3}) \cdot 175 \cdot (1 \times 10^{-4}) = 1.75 \times 10^{-7} \]

4. **Calculate the Magnetic Field \( B \):**
Now plug this back into the equation for \( B \):
\[ B = \frac{10^{-6} \cdot \frac{\pi}{180}}{1.75 \times 10^{-7}} \]

Simplifying this gives:
\[ B = \frac{10^{-6} \cdot \pi}{180 \cdot 1.75 \times 10^{-7}} = \frac{\pi}{18 \cdot 1.75} \]

5. **Final Calculation:**
Calculate the numerical value:
\[ B \approx \frac{3.14}{31.5} \approx 0.001 \, \text{T} \]

Therefore, we can express this as:
\[ B \approx 10^{-3} \, \text{T} \]

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