A circular coil of radius 5 cm, number of turns 200 carries a current of 2A.
The magnetic field at the centre of coil is:

Options

A. `5.027xx10^(-3)T`
B. `5xx10^(-2)T`
C. `5xx10^(-4)T`
D. None of these

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 (\(4\pi \times 10^{-7} \, \text{T m/A}\)),
- \(n\) 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. **Convert the radius from centimeters to meters:**
\[ r = 5 \, \text{cm} = 5 \times 10^{-2} \, \text{m} \]

2. **Identify the values given in the problem:**
- Number of turns, \(N = 200\)
- Current, \(I = 2 \, \text{A}\)

3. **Calculate the number of turns per unit length:**
Since the coil is circular, the number of turns per unit length \(n\) is simply the total number of turns \(N\):
\[ n = N = 200 \]

4. **Substitute the values into the formula:**
\[ B = \frac{(4\pi \times 10^{-7}) \times 200 \times 2}{2 \times (5 \times 10^{-2})} \]

5. **Simplify the expression:**
- The \(2\) in the numerator and denominator cancels out:
\[ B = \frac{(4\pi \times 10^{-7}) \times 200}{(5 \times 10^{-2})} \]

6. **Calculate the numerator:**
\[ 4\pi \times 200 = 800\pi \]

7. **Now substitute back into the equation:**
\[ B = \frac{800\pi \times 10^{-7}}{5 \times 10^{-2}} = \frac{800\pi}{5} \times 10^{-5} \]

8. **Calculate \( \frac{800}{5} \):**
\[ \frac{800}{5} = 160 \]

9. **Thus, we have:**
\[ B = 160\pi \times 10^{-5} \, \text{T} \]

10. **Now calculate \(160\pi\):**
\[ 160\pi \approx 160 \times 3.14 = 502.4 \]
Therefore,
\[ B \approx 502.4 \times 10^{-5} \, \text{T} = 5.024 \times 10^{-3} \, \text{T} \]

### Final Answer:
The magnetic field at the center of the coil is approximately:
\[ B \approx 5.024 \times 10^{-3} \, \text{T} \]