A current of 1 A is flowing on the sides of an equilateral triangle of side 4.5 × 10-2 m. The magnetic field at the center of the triangle will be: (a) 2 × 10-5 Wb/m2 (b) Zero (c) 8 × 10-5 Wb/m2 (d) 4 × 10-5 Wb/m2

Answer: (d). Equilateral triangle of side a: B = (9μ₀ I)/(2π a) = 9 × 2 × 10⁻⁷ Ia. B = 18 × 10⁻⁷ × 14.5 × 10⁻² = 4 × 10⁻⁵ Wb/m ² So the answer is (d).

A current of 1 A is flowing on the sides of an equilateral triangle… | JEE Main 2018 (15 Apr, Shift 2)

Q62JEE Main 2018 · 15 APR · SHIFT 2Medium

Magnetic Field of Straight Wires

A current of 1 A is flowing on the sides of an equilateral triangle of side 4.5 $\times$ 10-2 m. The magnetic field at the center of the triangle will be:

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Answer: (d)

Equilateral triangle of side $a$ : $B = \frac{9\mu_0 I}{2\pi a} = \frac{9 \times 2 \times 10^{-7} I}{a}$ .
$B = \frac{18 \times 10^{-7} \times 1}{4.5 \times 10^{-2}} = 4 \times 10^{-5}$ Wb/m $^2$
So the answer is (d).

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