The current passing through a conducting loop in the form of equilateral triangle of side 4√(3) cm is 2 A. The magnetic field at its centroid is α × 10⁻⁵ T. The value of α is ______. (Given: μ₀ = 4π × 10⁻⁷ SI units) (a) 3√(3) (b) 2√(3) (c) √(3) (d) √(3)/2

Answer: (a). Distance of centroid from each side: d = (a)/(2√(3)) = (4√(3))/(2√(3)) = 2 cm. Each side: B₁ = (μ₀ I)/(4π d)(2sin 60°) = (μ₀ I√(3))/(4π d). B = 3B₁ = 3 × 10⁻⁷ × 2 × √(3)0.02 = 3√(3) × 10⁻⁵ T So α = 3√(3), answer (a).

The current passing through a conducting loop in the form of… | JEE Main 2026 (23 Jan, Shift 2)

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Magnetic Field of Straight Wires

The current passing through a conducting loop in the form of equilateral triangle of side $4\sqrt{3}\text{ cm}$ is 2 A. The magnetic field at its centroid is $\alpha \times 10^{-5}\text{ T}$ . The value of $\alpha$ is ______.

(Given: $\mu_0 = 4\pi \times 10^{-7}$ SI units)

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Answer: (a)

Distance of centroid from each side: $d = \frac{a}{2\sqrt{3}} = \frac{4\sqrt{3}}{2\sqrt{3}} = 2$ cm.
Each side: $B_1 = \frac{\mu_0 I}{4\pi d}(2\sin 60^\circ) = \frac{\mu_0 I\sqrt{3}}{4\pi d}$ .
$B = 3B_1 = 3 \times \frac{10^{-7} \times 2 \times \sqrt{3}}{0.02} = 3\sqrt{3} \times 10^{-5}$ T
So $\alpha = 3\sqrt{3}$ , answer (a).

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