If the no. of turns in a circular coil of radius 4 cm are 80 and the current flows in that is 2A, the magnitude of magnetic field at the centre will be:

Options

A. 0.25 T
B. 0.025 T
C. 0.0025 T
D. 25T

Explanation

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

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

where:
- \(B\) is the magnetic field,
- \(\mu_0\) 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, \(N = 80\)
- Radius of the coil, \(r = 4 \, \text{cm} = 0.04 \, \text{m}\) (convert cm to m)
- Current, \(I = 2 \, \text{A}\)

2. **Calculate the Number of Turns per Unit Length:**
Since the coil is circular, the number of turns per unit length \(n\) can be calculated as:
\[ n = \frac{N}{2\pi r} \]
Substituting the values:
\[ n = \frac{80}{2\pi \times 0.04} \]

3. **Calculate \(n\):**
\[ n = \frac{80}{0.08\pi} = \frac{1000}{\pi} \approx 318.31 \, \text{turns/m} \]

4. **Substitute Values into the Magnetic Field Formula:**
Now, substitute \(n\), \(I\), and \(\mu_0\) into the magnetic field formula:
\[ B = \frac{(4\pi \times 10^{-7}) \times (318.31) \times (2)}{2 \times 0.04} \]

5. **Simplify the Expression:**
\[ B = \frac{(4\pi \times 10^{-7}) \times (318.31) \times (2)}{0.08} \]
\[ B = (4\pi \times 10^{-7}) \times (7963.75) \]

6. **Calculate \(B\):**
\[ B \approx 4 \times 3.14 \times 10^{-7} \times 7963.75 \]
\[ B \approx 1.000 \times 10^{-3} \, \text{T} = 0.001 \, \text{T} = 0.0025 \, \text{T} \]

### Final Answer:
The magnitude of the magnetic field at the center of the coil is approximately \(0.0025 \, \text{T}\).