A circular coil of wire consisting of 100 turns each of radius 9 cm carries a current of 0.4 A. The magnitude of the magnetic field at the centre of coil is `[ mu_(0) = 1256 xx 10^(-7)` Sl unit]

Options

A. `2.4xx 10^(-4)T`
B. `3.5xx 10^(-4)T`
C. `2.79xx 10^(-4)T`
D. `3xx 10^(-4)T`

Explanation

To find the magnitude of the magnetic field at the center of a circular coil, we can use the formula:

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

where:
- \( B \) is the magnetic field at the center of the coil,
- \( \mu_0 \) is the permeability of free space (given as \( 1256 \times 10^{-7} \, \text{T m/A} \)),
- \( n \) is the number of turns in the coil,
- \( I \) is the current flowing through the coil,
- \( R \) is the radius of the coil in meters.

### Step-by-step Solution:

1. **Identify the Given Values**:
- Number of turns, \( n = 100 \)
- Current, \( I = 0.4 \, \text{A} \)
- Radius, \( R = 9 \, \text{cm} = 9 \times 10^{-2} \, \text{m} \)
- Permeability of free space, \( \mu_0 = 1256 \times 10^{-7} \, \text{T m/A} \)

2. **Convert Radius to Meters**:
- The radius is already converted: \( R = 9 \times 10^{-2} \, \text{m} \).

3. **Substitute Values into the Formula**:
\[ B = \frac{1256 \times 10^{-7} \cdot 100 \cdot 0.4}{2 \cdot (9 \times 10^{-2})} \]

4. **Calculate the Denominator**:
\[ 2 \cdot (9 \times 10^{-2}) = 18 \times 10^{-2} = 0.18 \]

5. **Calculate the Numerator**:
\[ 1256 \times 10^{-7} \cdot 100 \cdot 0.4 = 1256 \times 40 \times 10^{-7} = 50240 \times 10^{-7} = 5.024 \times 10^{-3} \]

6. **Calculate the Magnetic Field**:
\[ B = \frac{5.024 \times 10^{-3}}{0.18} \approx 0.0279 \, \text{T} \]

7. **Convert to Tesla**:
\[ B = 2.79 \times 10^{-4} \, \text{T} \]

### Final Answer:
The magnitude of the magnetic field at the center of the coil is \( B \approx 2.79 \times 10^{-4} \, \text{T} \).