A circular coil of radius ' R ' has ' N ' turns of a wire. The coefficient of self induction of the coil will be ( μ₀ = permeability of free space) (a) (μ₀ N π R²)/(2) (b) (μ₀ N π R)/(4) (c) (μ₀ N² π R)/(2) (d) (μ₀ N π R)/(2)

Answer: (c). Field at centre of the coil: B = (μ₀ N I)/(2R). Flux linkage Nφ = N· B·π R² = (μ₀ N² I π R)/(2). L = (Nφ)/(I) = (μ₀ N² π R)/(2) So the answer is (c) (μ₀ N² π R)/(2).

A circular coil of radius ' R ' has ' N ' turns of a wire. The… | MHT CET 2021 (22 Sep, Shift 1)

Q68MHT CET 2021 · 22 SEP · SHIFT 1Easy

Self-Inductance & Energy

A circular coil of radius ' $R$ ' has ' $N$ ' turns of a wire. The coefficient of self induction of the coil will be ( $\mu_0 =$ permeability of free space)

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Answer: (c)

Field at centre of the coil: $B = \dfrac{\mu_0 N I}{2R}$ .
Flux linkage $N\phi = N\cdot B\cdot\pi R^2 = \dfrac{\mu_0 N^2 I \pi R}{2}$ .
$L = \dfrac{N\phi}{I} = \dfrac{\mu_0 N^2 \pi R}{2}$
So the answer is (c) $\dfrac{\mu_0 N^2 \pi R}{2}$ .

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Self-Inductance & Energy question from MHT CET 2021 (22 Sep, Shift 1)

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