A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.2A. A uniform horizontal magnetic field of strength 2T is turned on. The axis of the coil is initially in the direction of the field . The coil rotates through an angle of `90^(@)` under the influence of the magnetic field. The moment of inertia of the coil is 0.1 kg `m^(2)`
What is the field at the centre of the coil?

Options

A. `2 xx 10 T`
B. `2 xx 10^(-3) T`
C. `2 xx 10^(-6) T`
D. `2 xx 10^(-9) T`

Answer

To find the magnetic field at the center of a closely wound circular coil, we can use the formula for the magnetic field due to a coil:

\[ B = \frac{\mu_0 \cdot n \cdot I}{2R} \]

Where:
- \( B \) is the magnetic field at the center of the coil,
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \),
- \( n \) is the number of turns per unit length,
- \( I \) is the current in amperes,
- \( R \) is the radius of the coil in meters.

### Step-by-step Solution:

1. **Identify the Given Values:**
- Number of turns, \( N = 100 \)
- Radius of the coil, \( R = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Current, \( I = 3.2 \, \text{A} \)
- Permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)

2. **Substitute the Values into the Formula:**
\[ B = \frac{\mu_0 \cdot N \cdot I}{2R} \]
\[ B = \frac{(4\pi \times 10^{-7}) \cdot 100 \cdot 3.2}{2 \cdot 0.1} \]

3. **Calculate the Magnetic Field:**
- First, calculate the numerator:
\[ 4\pi \times 10^{-7} \cdot 100 \cdot 3.2 = 4\pi \times 320 \times 10^{-7} \]
- Now, calculate \( 2R \):
\[ 2 \cdot 0.1 = 0.2 \]
- Now, substitute back into the equation:
\[ B = \frac{4\pi \times 320 \times 10^{-7}}{0.2} \]
- Simplifying further:
\[ B = 4\pi \times 1600 \times 10^{-7} \]
- Using \( \pi \approx 3.14 \):
\[ B \approx 4 \times 3.14 \times 1600 \times 10^{-7} \approx 2.01 \times 10^{-3} \, \text{T} \]

4. **Final Result:**
\[ B \approx 2.01 \, \text{mT} \text{ (milliTesla)} \]

### Conclusion:
The magnetic field at the center of the coil is approximately \( 2.01 \, \text{mT} \).