A conducting wire of length L is bent in the form of an equilateral triangle. If current I is made to pass through it, then calculate the magnetic field intensity at the centre of the triangle.

Options

A. `( 27 mu_(0)I)/(4 pi L) `
B. `(27 mu_(0)I)/(2 pi L)`
C. `(27 mu_(0)I)/(5 pi L)`
D. `(27 mu_(0) I)/(7 pi L)`

Explanation

To solve the problem of calculating the magnetic field intensity at the center of an equilateral triangle formed by a conducting wire of length L carrying a current I, we can follow these steps:

### Step 1: Determine the Length of Each Side of the Triangle
Since the wire is bent into the shape of an equilateral triangle, the length of each side (s) can be calculated as:
\[ s = \frac{L}{3} \]

### Step 2: Calculate the Distance from the Center to Each Side
For an equilateral triangle, the distance (R) from the center to any side can be derived using trigonometry. The height (h) of the triangle can be given by:
\[ h = \frac{\sqrt{3}}{2} s \]
The distance from the center to a side is:
\[ R = \frac{h}{3} = \frac{\sqrt{3}}{6} s \]
Substituting \( s = \frac{L}{3} \):
\[ R = \frac{\sqrt{3}}{6} \cdot \frac{L}{3} = \frac{L \sqrt{3}}{18} \]

### Step 3: Calculate the Magnetic Field Due to One Side
The magnetic field (B) at the center due to a straight current-carrying wire can be calculated using the formula:
\[ B = \frac{\mu_0 I}{4\pi R} \cdot \sin(\theta) \]
For an equilateral triangle, the angle \( \theta \) at the center for each side is \( 60^\circ \). Therefore:
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
Thus, the magnetic field due to one side is:
\[ B = \frac{\mu_0 I}{4\pi R} \cdot \frac{\sqrt{3}}{2} \]

### Step 4: Substitute the Value of R
Substituting \( R = \frac{L \sqrt{3}}{18} \) into the equation for B:
\[ B = \frac{\mu_0 I}{4\pi \left(\frac{L \sqrt{3}}{18}\right)} \cdot \frac{\sqrt{3}}{2} \]
This simplifies to:
\[ B = \frac{\mu_0 I \cdot 18}{4\pi L \sqrt{3}} \cdot \frac{\sqrt{3}}{2} = \frac{9 \mu_0 I}{2 \pi L} \]

### Step 5: Calculate the Total Magnetic Field at the Center
Since there are three sides contributing to the magnetic field at the center and all contribute equally and in the same direction (into the plane), the total magnetic field (B_net) is:
\[ B_{net} = 3B = 3 \cdot \frac{9 \mu_0 I}{2 \pi L} = \frac{27 \mu_0 I}{2 \pi L} \]

### Final Result
Thus, the magnetic field intensity at the center of the triangle is:
\[ B_{net} = \frac{27 \mu_0 I}{2 \pi L} \]