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JEE Main 2024 (31 Jan, Shift 1) · Mutual Inductance

Mutual Inductance question from JEE Main 2024 (31 Jan, Shift 1)

Q35JEE Main 2024 · 31 JAN · SHIFT 1Medium

Mutual Inductance

A small square loop of wire of side ll is placed inside a large square loop of wire of side L(L=l2)L \left(L = l^2\right). The loops are coplanar and their centers coincide. The value of the mutual inductance of the system is x×10−7 H\sqrt{x} \times 10^{-7}\text{ H}, where x=x = _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 128

  1. For a small square inside a large square (centres coinciding): M=22 μ0l2πLM = \frac{2\sqrt{2}\,\mu_0 l^2}{\pi L}
  2. With L=l2L = l^2: M=22 μ0π=22×4π×10−7π=82×10−7 HM = \frac{2\sqrt{2}\,\mu_0}{\pi} = \frac{2\sqrt{2}\times 4\pi\times10^{-7}}{\pi} = 8\sqrt{2}\times10^{-7}\text{ H}
  3. 82=1288\sqrt{2} = \sqrt{128}, so M=128×10−7 HM = \sqrt{128}\times10^{-7}\text{ H}
  4. So x=128x = 128.
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