A circular coil of radius 10 cm having 100 turns carries a current of 3.2 A. The magnetic field at the center of the coil is

Options

A. `2.01 xx 10^(-3)T`
B. `5.64 xx 10^(-3)T`
C. `2.64 xx 10^(-3)T`
D. `5.64 xx 10^(-3)T`

Explanation

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

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

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 of the coil,
- \( I \) is the current flowing through the coil,
- \( r \) is the radius of the coil.

### Step-by-step Solution:

1. **Identify the given values**:
- Radius of the coil, \( r = 10 \, \text{cm} = 10 \times 10^{-2} \, \text{m} = 0.1 \, \text{m} \)
- Number of turns, \( n = 100 \)
- Current, \( I = 3.2 \, \text{A} \)

2. **Substitute the values into the formula**:
\[ B = \frac{\mu_0 n I}{2r} \]
Substituting the known values:
\[ B = \frac{(4\pi \times 10^{-7}) \times 100 \times 3.2}{2 \times 0.1} \]

3. **Calculate the denominator**:
\[ 2r = 2 \times 0.1 = 0.2 \, \text{m} \]

4. **Calculate the numerator**:
\[ \mu_0 n I = (4\pi \times 10^{-7}) \times 100 \times 3.2 \]
First, calculate \( 100 \times 3.2 = 320 \):
\[ \mu_0 n I = 4\pi \times 10^{-7} \times 320 \]

5. **Calculate \( 4\pi \times 320 \)**:
\[ 4\pi \approx 12.5664 \quad \Rightarrow \quad 12.5664 \times 320 \approx 4021.238 \]
So,
\[ \mu_0 n I \approx 4021.238 \times 10^{-7} \, \text{T m/A} \]

6. **Now substitute back to find \( B \)**:
\[ B = \frac{4021.238 \times 10^{-7}}{0.2} \]
\[ B \approx 20106.19 \times 10^{-7} \, \text{T} \]
\[ B \approx 2.010619 \times 10^{-3} \, \text{T} \]

7. **Final result**:
\[ B \approx 2.01 \times 10^{-3} \, \text{T} \]

### Conclusion:
The magnetic field at the center of the coil is approximately \( 2.01 \times 10^{-3} \, \text{T} \).