Find magnetic field at centre of hexagon wire of side a system carrying current 'I' at centre with 50 turn in multiple of `\mu_0I/\(a pi)`

Options

A. `500sqrt3`
B. `50sqrt3`
C. 50
D. 100

Answer

To find the magnetic field at the center of a hexagonal wire carrying current 'I' with 50 turns, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- A hexagon can be divided into 6 equilateral triangles. Each side of the hexagon is of length 'a'.
- The center of the hexagon is equidistant from all sides.

2. **Finding the Perpendicular Distance**:
- The distance from the center of the hexagon to the midpoint of one side (perpendicular distance) can be calculated using trigonometry.
- For an equilateral triangle, the height can be calculated as:
\[ r = a \cdot \cos(30^\circ) = a \cdot \frac{\sqrt{3}}{2} \]

3. **Magnetic Field Due to One Segment**:
- The magnetic field at a distance 'r' from a long straight wire carrying current 'I' is given by:
\[ B = \frac{\mu_0 I}{2\pi r} \]
- However, since we are considering the contribution from both sides of the wire segment, we need to account for the angles.
- The angle between the wire and the line joining the center to the wire is 30 degrees on either side.
- Therefore, the contribution from one segment can be expressed as:
\[ B_{\text{segment}} = \frac{\mu_0 I}{4\pi r} \cdot (\sin(30^\circ) + \sin(30^\circ)) = \frac{\mu_0 I}{4\pi r} \cdot (0.5 + 0.5) = \frac{\mu_0 I}{4\pi r} \]

4. **Total Magnetic Field from All Segments**:
- Since there are 6 segments in the hexagon, the total magnetic field at the center due to all segments is:
\[ B_{\text{total}} = 6 \cdot B_{\text{segment}} = 6 \cdot \frac{\mu_0 I}{4\pi r} = \frac{3\mu_0 I}{2\pi r} \]

5. **Substituting the Value of r**:
- Substitute \( r = \frac{\sqrt{3}a}{2} \):
\[ B_{\text{total}} = \frac{3\mu_0 I}{2\pi \cdot \frac{\sqrt{3}a}{2}} = \frac{3\mu_0 I}{\pi \sqrt{3}a} \]

6. **Considering the Number of Turns**:
- Since there are 50 turns, the total magnetic field becomes:
\[ B_{\text{final}} = 50 \cdot B_{\text{total}} = 50 \cdot \frac{3\mu_0 I}{\pi \sqrt{3}a} = \frac{150\mu_0 I}{\pi \sqrt{3}a} \]

7. **Expressing in Terms of Given Quantity**:
- The problem asks for the answer in multiples of \( \frac{\mu_0 I}{a \pi} \):
\[ B_{\text{final}} = 150 \cdot \frac{\mu_0 I}{\pi a \sqrt{3}} \]
- Thus, the multiple is:
\[ \text{Multiple} = \frac{150}{\sqrt{3}} = 50 \sqrt{3} \]

### Final Answer:
The magnetic field at the center of the hexagonal wire is \( 50 \sqrt{3} \cdot \frac{\mu_0 I}{a \pi} \).