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 of the Triangle
We have an equilateral triangle ABC with each side measuring 9 cm. The centroid (O) of the triangle is the point where all three medians intersect. The distance from the centroid to any vertex is given by \( \frac{2}{3} \) of the median.

### Step 2: Calculate the Length of the Median
The length of the median (AD) in an equilateral triangle can be calculated using the formula:
\[ AD = \frac{\sqrt{3}}{2} \times \text{side} \]
For a side of 9 cm:
\[ AD = \frac{\sqrt{3}}{2} \times 9 = \frac{9\sqrt{3}}{2} \text{ cm} \]

### Step 3: Calculate the Distance from the Centroid to a Side
The distance (OD) from the centroid to a side of the triangle (which is perpendicular to that side) is:
\[ OD = \frac{1}{3} \times AD = \frac{1}{3} \times \frac{9\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \text{ cm} = \frac{3\sqrt{3}}{200} \text{ m} \]

### Step 4: Calculate the Magnetic Field Due to One Side
Using the formula for the magnetic field due to a long straight conductor:
\[ B = \frac{\mu_0 I}{4\pi r} \left( \sin \theta_1 + \sin \theta_2 \right) \]
where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)
- \( I = 2 \, \text{A} \)
- \( r = OD \)
- \( \theta_1 = \theta_2 = 60^\circ \) (since the angles are equal in an equilateral triangle)

Calculating \( \sin 60^\circ \):
\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \]
Thus,
\[ B_1 = \frac{\mu_0 I}{4\pi \left(\frac{3\sqrt{3}}{200}\right)} \left( \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \right) = \frac{\mu_0 I}{4\pi r} \cdot \sqrt{3} \]

### Step 5: Calculate the Total Magnetic Field
Since there are three sides contributing equally to the magnetic field at the centroid:
\[ B_{total} = 3B_1 = 3 \left( \frac{\mu_0 I}{4\pi r} \cdot \sqrt{3} \right) \]

### Step 6: Substitute the Values
Substituting the values into the equation:
\[ B_{total} = 3 \left( \frac{4\pi \times 10^{-7} \times 2}{4\pi \times \frac{3\sqrt{3}}{200}} \cdot \sqrt{3} \right) \]
Simplifying this gives:
\[ B_{total} = 3 \left( \frac{2 \times 200 \times 10^{-7}}{3\sqrt{3}} \cdot \sqrt{3} \right) = 3 \left( \frac{400 \times 10^{-7}}{3} \right) = 4 \times 10^{-6} \text{ T} \]

### Step 7: Final Calculation
Thus, the magnetic field at the centroid of the triangle is:
\[ B = \frac{4}{3} \times 10^{-6} \text{ T} \approx 1.33 \times 10^{-6} \text{ T} \]

### Conclusion
The magnetic field at the centroid of the equilateral triangle is approximately \( 1.33 \times 10^{-6} \text{ T} \).

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