Magnitude of magnetic field (in SI units) at the centre of a hexagonal shape coil of side 10cm , 50 turns and carrying current I (Ampere) in units of `(mu_0I)/(pi)` is :

Options

A. `250 sqrt(3)`
B. `50sqrt(3)`
C. `500sqrt(3)`
D. `5sqrt(3)`

Explanation

To find the magnitude of the magnetic field at the center of a hexagonal coil with given parameters, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Side length of the hexagon, \( a = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Number of turns, \( N = 50 \)
- Current, \( I \, \text{(in Amperes)} \)

2. **Determine the Distance from the Center to a Side**:
- The distance \( R \) from the center of the hexagon to the midpoint of one of its sides can be calculated using the formula:
\[ R = \frac{a \cdot \sqrt{3}}{2} \]
- Substituting \( a = 0.1 \, \text{m} \):
\[ R = \frac{0.1 \cdot \sqrt{3}}{2} = 0.05\sqrt{3} \, \text{m} \]

3. **Calculate the Magnetic Field due to One Segment**:
- The magnetic field \( B \) at the center due to one segment of the hexagon can be calculated using the Biot-Savart Law:
\[ B = \frac{\mu_0 I}{4 \pi R} \left( \sin \alpha + \sin \beta \right) \]
- For a hexagon, each angle \( \alpha \) and \( \beta \) at the center is \( 30^\circ \):
\[ \sin 30^\circ = \frac{1}{2} \]
- Thus, we have:
\[ B = \frac{\mu_0 I}{4 \pi R} \left( \frac{1}{2} + \frac{1}{2} \right) = \frac{\mu_0 I}{4 \pi R} \]

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

5. **Substituting for \( R \)**:
- Substitute \( R = 0.05\sqrt{3} \):
\[ B_{\text{total}} = \frac{3 \mu_0 I}{2 \pi (0.05\sqrt{3})} = \frac{3 \mu_0 I}{0.1\pi\sqrt{3}} = \frac{30 \mu_0 I}{\pi\sqrt{3}} \]

6. **Expressing in the Required Units**:
- The problem asks for the answer in units of \( \frac{\mu_0 I}{\pi} \):
\[ B_{\text{total}} = 30 \cdot \frac{\mu_0 I}{\pi\sqrt{3}} \]
- To express this in the required form, we can factor out \( \frac{\mu_0 I}{\pi} \):
\[ B_{\text{total}} = \frac{30}{\sqrt{3}} \cdot \frac{\mu_0 I}{\pi} \]
- Simplifying \( \frac{30}{\sqrt{3}} = 10\sqrt{3} \):
\[ B_{\text{total}} = 10\sqrt{3} \cdot \frac{\mu_0 I}{\pi} \]

### Final Answer:
The magnitude of the magnetic field at the center of the hexagonal coil is:
\[ B_{\text{total}} = 10\sqrt{3} \cdot \frac{\mu_0 I}{\pi} \]