The current passing through a conducting loop in the form of equilateral triangle of side $4 \sqrt{3} \mathrm{~cm}$ is 2A. The magnetic field at its centroid is $\alpha \times 10^{-5} \mathrm{~T}$. The value of $\alpha$ is \_\_\_\_. (Given : $\mu_{\mathrm{o}}=4 \pi \times 10^{-7}$ SI units)

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Magnetic Effects of Current

The current passing through a conducting loop in the form of equilateral triangle of side

43 cm

is 2A. The magnetic field at its centroid is

α \times 1 0^{- 5} T

. The value of

α

is ____.
(Given :

μ_{o} = 4 π \times 1 0^{- 7}

SI units)

Answer: (C)

For an equilateral triangle of side

a = 43 cm

, the perpendicular distance from the centroid to each side is

d = \frac{a}{2 3} = \frac{4 3}{2 3} = 2 cm = 2 \times 1 0^{- 2} m

. Each side subtends

6 0^{\circ}

on either side of the foot of this perpendicular, so the field due to one side is

B_{1} = \frac{μ _{0} I}{4 π d} (sin 6 0^{\circ} + sin 6 0^{\circ}) = \frac{1 0 ^{- 7} \times 2}{2 \times 1 0 ^{- 2}} \times 3 = 3 \times 1 0^{- 5} T

. The three sides contribute equally and add, so

B = 3 B_{1} = 33 \times 1 0^{- 5} T

, giving

α = 33

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Single correct option