Electromagnetic Induction

JEE Main level, single correct, medium

116 of 153

Question 116 of 153

A circular current loop of radius RR is placed inside square loop of side length LL (L>>R)(L \gt \gt R) such that they are co-planar and their centers coincide. The permeability of free space is μ0\mu_0. The mutual inductance between circular loop and square loop is _______.

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Answer: (D) 22 μ0R2L2\sqrt{2} \, \dfrac{\mu_0 R^2}{L}

Using the reciprocity theorem, assume a current II flows through the larger square loop and compute the flux through the smaller circular loop. The mutual inductance is given by M=ΦIM = \dfrac{\Phi}{I}.

Magnetic field at the centre of the square loop:
For a square of side LL, the perpendicular distance from the centre to any side is d=L2d = \dfrac{L}{2}, and each side subtends angles θ1=θ2=45∘\theta_1 = \theta_2 = 45^\circ at the centre.

The field due to one finite straight segment is:
Bsegment=μ0I4πd(sin⁡θ1+sin⁡θ2)B_{\text{segment}} = \dfrac{\mu_0 I}{4\pi d}(\sin\theta_1 + \sin\theta_2)

Bsegment=μ0I4π⋅L2(12+12)=μ0I2πL⋅2=2 μ0I2πLB_{\text{segment}} = \dfrac{\mu_0 I}{4\pi \cdot \frac{L}{2}}\left(\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}}\right) = \dfrac{\mu_0 I}{2\pi L} \cdot \sqrt{2} = \dfrac{\sqrt{2}\,\mu_0 I}{2\pi L}

All four sides contribute equally in the same direction at the centre, so:
Btotal=4⋅2 μ0I2πL=22 μ0IπLB_{\text{total}} = 4 \cdot \dfrac{\sqrt{2}\,\mu_0 I}{2\pi L} = \dfrac{2\sqrt{2}\,\mu_0 I}{\pi L}

Flux through the circular loop:
Since L≫RL \gg R, the field BtotalB_{\text{total}} can be treated as uniform over the area A=πR2A = \pi R^2 of the circular loop.

Φ=Btotal⋅A=22 μ0IπL⋅πR2=22 μ0R2IL\Phi = B_{\text{total}} \cdot A = \dfrac{2\sqrt{2}\,\mu_0 I}{\pi L} \cdot \pi R^2 = \dfrac{2\sqrt{2}\,\mu_0 R^2 I}{L}

Mutual inductance:
M=ΦI=22 μ0R2LM = \dfrac{\Phi}{I} = \dfrac{2\sqrt{2}\,\mu_0 R^2}{L}

Hence, the correct option is (4) 22 μ0R2L(4)\ \dfrac{2\sqrt{2}\,\mu_0 R^2}{L}.

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