Question

A current of 1.5 A is flowing through a triangle, of side 9 cm each. The magnetic field at the centroid of the triangle is : (Assume that the current is flowing in the clockwise direction.)

• A. \(3 \times 10^{-5}\) T, inside the plane of triangle
• B. \(3 \times 10^{-7}\) T, outside the plane of triangle
• C. \(2\sqrt{3} \times 10^{-5}\) T, inside the plane of triangle
• D. \(2\sqrt{3} \times 10^{-7}\) T, outside the plane of triangle

Hint:

The field at the center of an equilateral triangle is always \( \frac{9 \mu_0 I}{2 \pi a} \). Using this direct formula can save time in exams.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The magnetic field at the centroid of a current-carrying polygon is the sum of the fields produced by each individual side. For an equilateral triangle, all three sides contribute equally and in the same direction.
Step 2: Key Formula or Approach: 1. Field due to a finite straight wire: \(B = \frac{\mu_0 I}{4\pi r} (\sin \theta_1 + \sin \theta_2)\).
2. For equilateral triangle side \(a\), distance to centroid \(r = \frac{a}{2\sqrt{3}}\) and \(\theta_1 = \theta_2 = 60^{\circ}\).
Step 3: Detailed Explanation: Given \(I = 1.5\,\text{A}\), \(a = 0.09\,\text{m}\). Field due to one side: \[ B_1 = \frac{10^{-7} \times 1.5}{0.09 / (2\sqrt{3})} (\sin 60^{\circ} + \sin 60^{\circ}) \] \[ B_1 = \frac{10^{-7} \times 1.5 \times 2\sqrt{3}}{0.09} (\sqrt{3}) = \frac{10^{-7} \times 1.5 \times 6}{0.09} = 10^{-5}\,\text{T} \] Total field at centroid \(B = 3 \times B_1 = 3 \times 10^{-5}\,\text{T}\). Direction: Since the current is clockwise, by the right-hand thumb rule, the magnetic field at the center points into (inside) the plane of the triangle.
Step 4: Final Answer:
The magnetic field is \(3 \times 10^{-5}\) T, inside the plane.