The current I ampere is flowing in an equilateral triangle of side a the magnetic field induction at the centroid will be

Options

A. `(9mu_0I)/(2pia)`
B. `(5sqrt2mu_0I)/(3pia)`
C. `(3mu_0I)/(2pia)`
D. `(mu_0I)/(3sqrt3pia)`

Explanation

To find the magnetic field induction at the centroid of an equilateral triangle with current \( I \) flowing through its sides, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Geometry**:
- We have an equilateral triangle with each side of length \( a \).
- The centroid of the triangle is the point where all three medians intersect.

2. **Determine the Position of the Centroid**:
- The centroid divides each median in the ratio 2:1. The distance from the centroid to any vertex is \( \frac{2}{3} \) of the median length.
- The length of the median \( m \) of an equilateral triangle can be calculated using the formula:
\[ m = \frac{\sqrt{3}}{2} a \]
- Therefore, the distance \( R \) from the centroid to each vertex (which is also the perpendicular distance from the centroid to each side) is:
\[ R = \frac{2}{3} m = \frac{2}{3} \cdot \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{3} a \]

3. **Calculate the Magnetic Field Due to One Side**:
- The magnetic field \( B_0 \) at a distance \( R \) from a long straight conductor carrying current \( I \) is given by:
\[ B_0 = \frac{\mu_0 I}{2 \pi R} \]
- Substituting \( R \):
\[ B_0 = \frac{\mu_0 I}{2 \pi \left(\frac{\sqrt{3}}{3} a\right)} = \frac{3 \mu_0 I}{2 \pi \sqrt{3} a} \]

4. **Consider the Contributions from All Three Sides**:
- The magnetic field contributions from each side of the triangle at the centroid will have the same magnitude but different directions.
- Using the right-hand rule, we find that the magnetic fields due to each side will add up vectorially.
- The angle between the magnetic field due to one side and the line connecting the centroid to the vertex is \( 30^\circ \) (since the angles in an equilateral triangle are \( 60^\circ \)).
- Therefore, the total magnetic field \( B \) at the centroid is:
\[ B = 3 B_0 \sin(30^\circ) = 3 \cdot \frac{3 \mu_0 I}{2 \pi \sqrt{3} a} \cdot \frac{1}{2} = \frac{9 \mu_0 I}{4 \pi \sqrt{3} a} \]

5. **Final Result**:
- After simplifying, we find:
\[ B = \frac{9 \mu_0 I}{2 \pi a} \]

### Conclusion:
The magnetic field induction at the centroid of the equilateral triangle is given by:
\[ B = \frac{9 \mu_0 I}{2 \pi a} \]