Two identical current carrying coils with same centre are placed with their planes perpendicular to each other. If current I = √(2) A and radius of the coil is R = 1 m, then magnetic field at centre is equal to ( μ₀ = permeability of free space) (a) μ₀ (b) (μ₀)/(2) (c) 2μ₀ (d) √(2)μ₀

Answer: (a). Each coil: B₁ = (μ₀ I)/(2R) = (μ₀√(2))/(2). Perpendicular planes, so perpendicular fields: B = √(2) B₁ = √(2)·(√(2) μ₀)/(2) = μ₀ So the answer is (a) μ₀.

Two identical current carrying coils with same centre are placed with… | MHT CET 2024 (2 May, Shift 1)

Q39MHT CET 2024 · 2 MAY · SHIFT 1Easy

Biot-Savart Law & Circular Coils

Two identical current carrying coils with same centre are placed with their planes perpendicular to each other. If current $I = \sqrt{2}\text{ A}$ and radius of the coil is $R = 1\text{ m}$ , then magnetic field at centre is equal to ( $\mu_0 =$ permeability of free space)

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Answer: (a)

Each coil: $B_1 = \frac{\mu_0 I}{2R} = \frac{\mu_0\sqrt{2}}{2}$ .
Perpendicular planes, so perpendicular fields: $B = \sqrt{2}\,B_1 = \sqrt{2}\cdot\frac{\sqrt{2}\,\mu_0}{2} = \mu_0$
So the answer is (a) $\mu_0$ .

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Biot-Savart Law & Circular Coils question from MHT CET 2024 (2 May, Shift 1)

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