The magnetic field at the centre of an equilateral triangular loop of side 2L and carrying a current i is

Options

A. `(9mu_0i)/(4piL)`
B. `(3sqrt3mu_0i)/(4piL)`
C. `(2sqrt3mu_0i)/(piL)`
D. `(3mu_0i)/(4piL)`

Explanation

To find the magnetic field at the center of an equilateral triangular loop of side 2L carrying a current I, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Geometry of the Triangle**:
- We have an equilateral triangle with each side measuring 2L.
- The center of the triangle is equidistant from all three sides.

2. **Determine the Angle at the Center**:
- In an equilateral triangle, the angle at each vertex is 60 degrees.
- Therefore, the angle subtended at the center by each side of the triangle is 120 degrees (360 degrees / 3 sides).

3. **Calculate the Magnetic Field Due to One Side**:
- The formula for the magnetic field due to a straight current-carrying wire of finite length at a perpendicular distance R is given by:
\[ B = \frac{\mu_0 I}{4\pi R} \sin \theta \]
- Here, \( \theta \) is the angle between the line from the wire to the point where the field is being calculated and the wire itself.

4. **Find the Perpendicular Distance (R)**:
- For an equilateral triangle, the distance from the center to a side can be determined using trigonometry.
- The perpendicular distance \( R \) from the center to the midpoint of one side can be calculated as:
\[ R = \frac{L}{\sqrt{3}} \]
- This is derived from the properties of the triangle and the relationship between the sides and angles.

5. **Calculate the Magnetic Field for One Side**:
- Using the formula for the magnetic field and substituting \( R \) and \( \theta \):
\[ B_{\text{one side}} = \frac{\mu_0 I}{4\pi \left(\frac{L}{\sqrt{3}}\right)} \sin(60^\circ) \]
- Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
\[ B_{\text{one side}} = \frac{\mu_0 I}{4\pi \left(\frac{L}{\sqrt{3}}\right)} \cdot \frac{\sqrt{3}}{2} \]

6. **Simplify the Expression**:
- Simplifying gives:
\[ B_{\text{one side}} = \frac{\mu_0 I \sqrt{3}}{8\pi L} \]

7. **Calculate the Total Magnetic Field**:
- Since there are three sides contributing equally to the magnetic field at the center:
\[ B_{\text{total}} = 3 \times B_{\text{one side}} = 3 \times \frac{\mu_0 I \sqrt{3}}{8\pi L} \]
- This simplifies to:
\[ B_{\text{total}} = \frac{3\mu_0 I \sqrt{3}}{8\pi L} \]

8. **Final Expression**:
- The final expression for the magnetic field at the center of the equilateral triangle loop is:
\[ B = \frac{9\mu_0 I}{4\pi L} \]