To find the magnetic field at the centroid of an equilateral triangle frame of side l carrying a current i, we use the Biot-Savart law.

Step 1: Magnetic Field Due to One Side of Triangle

For a current-carrying finite wire, the magnetic field at a perpendicular distance d from the center of the wire is:

\(B=\frac{{\mu }_{0}i}{4\pi d}[\sin ⁡{\theta }_{1}+\sin ⁡{\theta }_{2}]\)

For an equilateral triangle, the perpendicular distance from the centroid to any side is:

\(d=\frac{l}{2\sqrt{3}}\)

Each side of the equilateral triangle subtends an angle θ = 60° at the centroid.

So,

\(\sin ⁡6{0}^{∘}=\frac{\sqrt{3}}{2}\)

Thus, the field due to one side:

\({B}_{\text{one side}}=\frac{{\mu }_{0}i}{4\pi (l\mathrm{/}2\sqrt{3})}(2\times \frac{\sqrt{3}}{2})\)

\({B}_{\text{one side}}=\frac{{\mu }_{0}i}{2\pi }\times \frac{\sqrt{3}}{l}\times \sqrt{3}\)

\({B}_{\text{one side}}=\frac{{\mu }_{0}i}{2\pi l}\times 3\)

\({B}_{\text{one side}}=\frac{3{\mu }_{0}i}{2\pi l}\)

Step 2: Net Magnetic Field at Centroid

Since all three sides contribute equally to the field at the centroid and add up vectorially:

\({B}_{\text{total}}=3\times {B}_{\text{one side}}\)

\({B}_{\text{total}}=3\times \frac{3{\mu }_{0}i}{2\pi l}\)

\({B}_{\text{total}}=\frac{9{\mu }_{0}i}{2\pi l}\)

Step 3: Comparing with Options

From the given options:

The correct answer is Option A: \(\frac{9{\mu }_{0}i}{2\pi l}\)

Final Answer:

Option C: \(\frac{9{\mu }_{0}i}{2\pi l}\)

An equilateral triangle frame of side \(l\) is carrying current \(i\), find magnetic field at its centroid (Shift - II M…

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