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JEE Main 2023 (29 Jan, Shift 1) · Mutual Inductance

Mutual Inductance question from JEE Main 2023 (29 Jan, Shift 1)

Q45JEE Main 2023 · 29 JAN · SHIFT 1Medium

Mutual Inductance

Find the mutual inductance in the arrangement, when a small circular loop of wire of radius 'RR' is placed inside a large square loop of wire of side L (L≫R)L \ (L \gg R). The loops are coplanar and their centres coincide:

Figure

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Answer: (c)

  1. Field at the centre of the square loop: B=4×μ0I4π(L/2)×2sin⁡45∘=22 μ0IπLB = 4 \times \frac{\mu_0 I}{4\pi (L/2)} \times 2\sin 45^\circ = \frac{2\sqrt{2}\,\mu_0 I}{\pi L}
  2. Flux through the small circle (R≪LR \ll L): ϕ=BπR2=22 μ0IR2L\phi = B \pi R^2 = \frac{2\sqrt{2}\,\mu_0 I R^2}{L}
  3. M=ϕ/I=22 μ0R2LM = \phi / I = \frac{2\sqrt{2}\,\mu_0 R^2}{L}
  4. So the answer is (c).
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