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

Options

A. `(mu_(0)I)/(3sqrt(3)pia)`
B. `(3mu_(0)i)/(2pia)`
C. `(5sqrt(2)mu_(0)i)/(3pia)`
D. `(9mu_(0)i)/(2pia)`

Answer

To find the magnetic field induction at the centroid of an equilateral triangle with side length \( a \) carrying a current \( I \) in an anti-clockwise direction, we can follow these steps:

### Step 1: Understanding the Geometry
- We have an equilateral triangle with vertices A, B, and C, and we need to find the magnetic field at the centroid (point O).
- The centroid of an equilateral triangle divides each median in the ratio 2:1.

### Step 2: Calculate the Distance from the Centroid to Each Side
- The distance from the centroid to any side of the triangle can be calculated using trigonometry.
- For an equilateral triangle, the height \( h \) can be calculated as:
\[ h = \frac{\sqrt{3}}{2} a \]
- The distance \( D \) from the centroid to a side is:
\[ D = \frac{h}{3} = \frac{\sqrt{3}}{6} a \]

### Step 3: Magnetic Field Due to One Side
- The magnetic field \( B \) due to a straight current-carrying wire at a distance \( D \) is given by the formula:
\[ B = \frac{\mu_0 I}{4 \pi D} \]
- Since the current is flowing in an anti-clockwise direction, we can calculate the magnetic field due to one side of the triangle (let's say side AB):
\[ B_1 = \frac{\mu_0 I}{4 \pi \left(\frac{\sqrt{3}}{6} a\right)} = \frac{\mu_0 I \cdot 6}{4 \pi \sqrt{3} a} \]

### Step 4: Magnetic Field Due to All Three Sides
- The magnetic field contribution from each side of the triangle will be the same in magnitude and direction (outward from the plane of the triangle).
- Therefore, the total magnetic field \( B_{total} \) at the centroid is:
\[ B_{total} = 3B_1 = 3 \cdot \frac{\mu_0 I \cdot 6}{4 \pi \sqrt{3} a} = \frac{18 \mu_0 I}{4 \pi \sqrt{3} a} \]

### Step 5: Simplifying the Expression
- Simplifying the expression gives:
\[ B_{total} = \frac{9 \mu_0 I}{2 \pi \sqrt{3} a} \]

### Final Result
- The magnetic field induction at the centroid of the equilateral triangle is:
\[ B = \frac{9 \mu_0 I}{2 \pi \sqrt{3} a} \]