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

Options

A. `0.5xx10^(-5)T`
B. `1.25xx10^(-4)T`
C. `3xx10^(-5)T`
D. `4xx10^(-5)T`

Answer

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

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

Where:
- \( B \) is the magnetic field intensity 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 per unit length (in this case, total turns since we are considering the entire coil),
- \( I \) 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**:
\[ B = \frac{4\pi \times 10^{-7} \, \text{T m/A} \cdot 50 \cdot 2}{2 \cdot 0.5} \]

3. **Simplify the equation**:
- The denominator \( 2 \cdot 0.5 = 1 \).
- Thus, the equation simplifies to:
\[ B = 4\pi \times 10^{-7} \cdot 50 \cdot 2 \]

4. **Calculate the values**:
- First, calculate \( 50 \cdot 2 = 100 \).
- Now substitute back:
\[ B = 4\pi \times 10^{-7} \cdot 100 \]
- This gives:
\[ B = 400\pi \times 10^{-7} \, \text{T} \]

5. **Calculate \( 400\pi \)**:
- Using \( \pi \approx 3.14 \):
\[ 400\pi \approx 400 \cdot 3.14 = 1256 \]
- Therefore:
\[ B \approx 1256 \times 10^{-7} \, \text{T} = 12.56 \times 10^{-5} \, \text{T} \]

6. **Convert to standard form**:
- This can also be expressed as:
\[ B = 1.256 \times 10^{-4} \, \text{T} \]

### Final Answer:
The magnetic field intensity at the center of the coil is approximately \( 1.256 \times 10^{-4} \, \text{T} \) or \( 12.56 \times 10^{-5} \, \text{T} \).