A current of 2A is flowing in the sides of an equilateral triangle of side 9 cm. The magnetic field at the centroid of the triangle is

Options

A. `1.66xx10^(-6)T`
B. `1.22xx10^(-4)T`
C. `1.33xx10^(-6)T`
D. `1.44xx10^(-4)T`

Answer

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

### Step 1: Understand the Geometry
We have an equilateral triangle with each side of length \( A = 9 \, \text{cm} \). The centroid of the triangle is the point where the three medians intersect, and it is located at a distance of \( \frac{A}{\sqrt{3}} \) from each side.

### Step 2: Calculate the Distance from the Centroid to Each Side
The distance \( R \) from the centroid to each side of the triangle can be calculated as:
\[ R = \frac{A}{\sqrt{3}} = \frac{9 \, \text{cm}}{\sqrt{3}} = 3\sqrt{3} \, \text{cm} = 0.03\sqrt{3} \, \text{m} \]

### Step 3: Apply the Formula for the Magnetic Field Due to a Finite Wire
The magnetic field \( B \) at a distance \( R \) from a long straight wire carrying current \( I \) is given by:
\[ B = \frac{\mu_0 I}{4\pi R} \left( \sin \alpha + \sin \beta \right) \]
where \( \alpha \) and \( \beta \) are the angles subtended by the wire at the point where the magnetic field is being calculated.

For our equilateral triangle, the angles \( \alpha \) and \( \beta \) are both \( 60^\circ \):
\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \]

### Step 4: Calculate the Magnetic Field for One Side
Substituting the values into the formula:
\[ B_1 = \frac{\mu_0 I}{4\pi R} \left( \sin 60^\circ + \sin 60^\circ \right) = \frac{\mu_0 I}{4\pi R} \left( 2 \cdot \frac{\sqrt{3}}{2} \right) = \frac{\mu_0 I \sqrt{3}}{4\pi R} \]

### Step 5: Calculate the Total Magnetic Field at the Centroid
Since the current flows in all three sides of the triangle, the total magnetic field \( B_{\text{net}} \) at the centroid will be three times the magnetic field due to one side:
\[ B_{\text{net}} = 3B_1 = 3 \cdot \frac{\mu_0 I \sqrt{3}}{4\pi R} \]

### Step 6: Substitute the Values
Now, substituting \( I = 2 \, \text{A} \) and \( R = 0.03\sqrt{3} \, \text{m} \):
\[ B_{\text{net}} = 3 \cdot \frac{(10^{-7} \, \text{T m/A}) \cdot 2 \cdot \sqrt{3}}{4\pi \cdot (0.03\sqrt{3})} \]

### Step 7: Simplify the Expression
Calculating the constants:
\[ B_{\text{net}} = \frac{6 \cdot 10^{-7} \cdot \sqrt{3}}{4\pi \cdot 0.03\sqrt{3}} = \frac{6 \cdot 10^{-7}}{4\pi \cdot 0.03} \]

### Step 8: Final Calculation
Now, calculate the numerical value:
\[ B_{\text{net}} = \frac{6 \cdot 10^{-7}}{0.12\pi} = \frac{50 \cdot 10^{-7}}{\pi} \approx \frac{5 \cdot 10^{-6}}{3.14} \approx 1.59 \cdot 10^{-6} \, \text{T} \]

Thus, the magnetic field at the centroid of the triangle is approximately:
\[ B_{\text{net}} \approx 1.59 \, \mu \text{T} \]