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 × 10 - 7 T, outside the plane of triangle (b) 2√(3) × 10 - 7 T, outside the plane of triangle (c) 2√(3) × 10 - 5 T, inside the plane of triangle (d) 3 × 10 - 5 T, inside the plane of triangle

Answer: (d). Equilateral triangle: B = (9μ₀ I)/(2π a) (from 3 × (μ₀ I)/(4π d)(2sin 60°) with d = (a)/(2√(3))). B = 9 × 2 × 10⁻⁷ × 1.50.09 = 3 × 10⁻⁵ T Clockwise current gives a field into the plane. So the answer is (d) 3 × 10⁻⁵ T, inside the plane.

A current of 1.5 A is flowing through a triangle, of side 9 cm each… | JEE Main 2021 (31 Aug, Shift 2)

Q53JEE Main 2021 · 31 AUG · SHIFT 2Medium

Magnetic Field of Straight Wires

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.)

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Answer: (d)

Equilateral triangle: $B = \frac{9\mu_0 I}{2\pi a}$ (from $3 \times \frac{\mu_0 I}{4\pi d}(2\sin 60^\circ)$ with $d = \frac{a}{2\sqrt{3}}$ ).
$B = \frac{9 \times 2 \times 10^{-7} \times 1.5}{0.09} = 3 \times 10^{-5}$ T
Clockwise current gives a field into the plane.
So the answer is (d) $3 \times 10^{-5}$ T, inside the plane.

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Magnetic Field of Straight Wires question from JEE Main 2021 (31 Aug, Shift 2)

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