Question

As shown in the figure, a current of $2 A$ flowing in an equilateral triangle of side $4 \sqrt{3} cm$ The magnetic field at the centroid $O$ of the triangle is

 (Neglect the effect of earth's magnetic field)

• A. $4 \sqrt{3} \times 10^{-4} T$
• B. $\sqrt{3} \times 10^{-4} T$
• C. $4 \sqrt{3} \times 10^{-5} T$
• D. $3 \sqrt{3} \times 10^{-5} T$

Answer 1

The Correct Option is D

To find the magnetic field at the centroid of an equilateral triangle due to a current flowing along its sides, we can use the Biot-Savart Law. Each side of the triangle contributes to the magnetic field at the centroid.

Given:

• Current, \(I = 2\,A\)
• Side of the triangle, \(a = 4\sqrt{3}\,cm = 0.04\sqrt{3}\,m\)

The formula for the magnetic field at the centroid due to one side of the equilateral triangle is:

\(B = \frac{\mu_0 I}{4\pi R}(\sin \theta_1 + \sin \theta_2)\)

For an equilateral triangle:

• \(\theta_1 = \theta_2 = 30^\circ\)
• \(R = \frac{a}{\sqrt{3}}\)

Thus, the expression becomes:

\(B = \frac{\mu_0 I}{4\pi \frac{a}{\sqrt{3}}}(\sin 30^\circ + \sin 30^\circ)\)

Calculating the sine values:

\(\sin 30^\circ = \frac{1}{2}\)

Therefore, the magnetic field due to one segment is:

\(B = \frac{\mu_0 I}{2\pi \frac{a}{\sqrt{3}}}\)

Substituting the values:

• \(\mu_0 = 4\pi \times 10^{-7} \, T \cdot m / A\)

For one side:

\(B = \frac{4\pi \times 10^{-7} \times 2}{2\pi \times 0.04} = \frac{8 \times 10^{-7} \sqrt{3}}{0.04} = \frac{8 \times 10^{-7} \times \sqrt{3}}{0.04}\)

Simplifying:

\(B = 2 \times 10^{-5} \, \sqrt{3} \, T\)

Since the triangle has three sides, the total magnetic field at the centroid will be:

\(B_{\text{total}} = 3 \times B_{\text{one side}} = 3 \times 2 \times 10^{-5} \, \sqrt{3} \, T = 6 \times 10^{-5} \, \sqrt{3} \, T\)

Therefore, the magnetic field at the centroid is:

\(3\sqrt{3} \times 10^{-5} \, T\), which matches the correct answer.