Magnetic field intensity at the centre of coil of `50` turns, radius `0.5m` and carrying a current of `2A` is

Options

A. `(sqrt(2)mu_(0)nI)/(pil)`
B. `(sqrt(2)mu_(0)nI)/(2pil)`
C. `(sqrt(2)mu_(0)nI)/(4pil)`
D. `(2mu_(0)nI)/(pil)`

Explanation

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

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

where:
- \( B \) is the magnetic field intensity 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. **Identify the given values:**
- Number of turns, \( N = 50 \)
- Radius of the coil, \( R = 0.5 \, \text{m} \)
- Current, \( I = 2 \, \text{A} \)

2. **Substitute the values into the formula:**
- We need to calculate \( B \) using the formula. Since the number of turns \( N \) is given, we can directly use it in the formula:
\[ B = \frac{\mu_0 N I}{2R} \]

3. **Substitute \( \mu_0 \), \( N \), \( I \), and \( R \) into the equation:**
\[ B = \frac{(4\pi \times 10^{-7} \, \text{T m/A}) \times 50 \times 2}{2 \times 0.5} \]

4. **Simplify the equation:**
- The denominator simplifies to \( 1 \):
\[ B = (4\pi \times 10^{-7} \, \text{T m/A}) \times 50 \times 2 \]

5. **Calculate the numerical value:**
- First calculate \( 50 \times 2 = 100 \):
\[ B = 4\pi \times 10^{-7} \times 100 \]
- This simplifies to:
\[ B = 4\pi \times 10^{-5} \, \text{T} \]
- Using \( \pi \approx 3.14 \):
\[ B \approx 4 \times 3.14 \times 10^{-5} \approx 1.256 \times 10^{-4} \, \text{T} \]

6. **Final answer:**
- The magnetic field intensity at the center of the coil is approximately:
\[ B \approx 1.25 \times 10^{-4} \, \text{T} \]

### Conclusion:
The magnetic field intensity at the center of the coil of 50 turns, radius 0.5 m, and carrying a current of 2 A is approximately \( 1.25 \times 10^{-4} \, \text{T} \).