For moving coil galvanometer,the deflection in the coil is `0.05rad` when a current of `10mA` is passed through it.If the torsional constant of suspension wire is `4.0times10^(-5)Nm` rad `^(-1)`,the magnetic field is `0.01T` and the number of turns in the coil is `200`,the area of each turn (in `cm^(2)` ) is :

Options

A. `2.0`
B. `1.0`
C. `1.5`
D. `0.5`

Answer

To solve the problem, we need to find the area of each turn of the coil in a moving coil galvanometer given the following parameters:

- Deflection (θ) = 0.05 rad
- Current (I) = 10 mA = 10 × 10^(-3) A
- Torsional constant (k) = 4.0 × 10^(-5) Nm/rad
- Magnetic field (B) = 0.01 T
- Number of turns (n) = 200

### Step-by-step Solution:

1. **Understand the relationship between torque and magnetic field:**
The torque (τ) experienced by the coil due to the magnetic field is given by:
\[ \tau = n \cdot I \cdot A \cdot B \cdot \sin(\theta) \]
where:
- \( n \) = number of turns
- \( I \) = current
- \( A \) = area of each turn
- \( B \) = magnetic field
- \( \theta \) = angle of deflection

For maximum deflection, we can assume \( \sin(\theta) \approx 1 \) when θ is small.

2. **Set the torque due to the spring equal to the torque due to the magnetic field:**
The torque due to the spring is given by:
\[ \tau = k \cdot \theta \]
Setting the two expressions for torque equal gives:
\[ k \cdot \theta = n \cdot I \cdot A \cdot B \]

3. **Rearranging the equation to solve for the area (A):**
\[ A = \frac{k \cdot \theta}{n \cdot I \cdot B} \]

4. **Substituting the known values into the equation:**
\[ A = \frac{(4.0 \times 10^{-5} \, \text{Nm/rad}) \cdot (0.05 \, \text{rad})}{200 \cdot (10 \times 10^{-3} \, \text{A}) \cdot (0.01 \, \text{T})} \]

5. **Calculating the numerator:**
\[ \text{Numerator} = 4.0 \times 10^{-5} \cdot 0.05 = 2.0 \times 10^{-6} \, \text{Nm} \]

6. **Calculating the denominator:**
\[ \text{Denominator} = 200 \cdot 10 \times 10^{-3} \cdot 0.01 = 200 \cdot 0.01 \cdot 10^{-2} = 2.0 \times 10^{-2} \, \text{N} \]

7. **Calculating area (A):**
\[ A = \frac{2.0 \times 10^{-6}}{2.0 \times 10^{-2}} = 1.0 \times 10^{-4} \, \text{m}^2 \]

8. **Converting area from m² to cm²:**
\[ A = 1.0 \times 10^{-4} \, \text{m}^2 = 1.0 \times 10^{-4} \times 10^4 \, \text{cm}^2 = 1.0 \, \text{cm}^2 \]

### Final Answer:
The area of each turn is \( 1.0 \, \text{cm}^2 \).