The magnetic induction field at the centroid or an equilateral triangle of side 'I' and carrying a current 'i' is

Options

A. `(2sqrt2mu_(0)i)/(pil)`
B. `(9mu_(0)i)/(2pil)`
C. `(4mu_(0)i)/(pil)`
D. `(3sqrt3mu_(0)i)/(pil)`

Answer

To find the magnetic induction field at the centroid of an equilateral triangle of side 'l' carrying a current 'i', we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Geometry**:
- We have an equilateral triangle with each side of length 'l'.
- The centroid (O) of the triangle is the point where we want to calculate the magnetic field.

2. **Determine the Height of the Triangle**:
- The height (h) of an equilateral triangle can be calculated using the formula:
\[ h = \frac{\sqrt{3}}{2} l \]

3. **Calculate the Perpendicular Distance from the Centroid to a Side**:
- The perpendicular distance (r) from the centroid to any side of the triangle is given by:
\[ r = \frac{1}{3} h = \frac{1}{3} \left(\frac{\sqrt{3}}{2} l\right) = \frac{l}{2\sqrt{3}} \]

4. **Determine the Angles**:
- The angle subtended by the current-carrying side at the centroid is 60 degrees for each side.

5. **Apply Biot-Savart Law**:
- According to the Biot-Savart Law, the magnetic field (B) due to a straight current-carrying wire at a distance 'r' is given by:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{i}{r} \cdot (\sin \theta_1 + \sin \theta_2) \]
- For each side of the triangle, since both angles (θ1 and θ2) are 60 degrees:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{i}{r} \cdot ( \sin 60 + \sin 60) = \frac{\mu_0}{4\pi} \cdot \frac{i}{r} \cdot (2 \cdot \sin 60) \]

6. **Substituting Values**:
- Substitute \( r = \frac{l}{2\sqrt{3}} \) and \( \sin 60 = \frac{\sqrt{3}}{2} \):
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{i}{\frac{l}{2\sqrt{3}}} \cdot \left(2 \cdot \frac{\sqrt{3}}{2}\right) \]
- Simplifying gives:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{i \cdot 2\sqrt{3}}{l} \cdot \frac{\sqrt{3}}{2} = \frac{\mu_0 \cdot 3i}{4\pi l} \]

7. **Total Magnetic Field from All Sides**:
- Since there are three sides contributing to the magnetic field, the total magnetic field (B_total) at the centroid is:
\[ B_{total} = 3B = 3 \cdot \frac{\mu_0 \cdot 3i}{4\pi l} = \frac{9\mu_0 i}{4\pi l} \]

8. **Final Result**:
- The magnetic induction field at the centroid of the equilateral triangle is:
\[ B = \frac{9\mu_0 i}{4\pi l} \]