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JEE Main 2023 (6 Apr, Shift 2) · Mutual Inductance

Mutual Inductance question from JEE Main 2023 (6 Apr, Shift 2)

Q46JEE Main 2023 · 6 APR · SHIFT 2Medium

Mutual Inductance

Two concentric circular coils with radii 1 cm1\text{ cm} and 1000 cm1000\text{ cm}, and number of turns 10 and 200 respectively are placed coaxially with centers coinciding. The mutual inductance of this arrangement will be ___________ ×10−8 H\times 10^{-8}\text{ H}. (Take, π2=10\pi^2 = 10)

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

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Answer: 4

  1. Field of the big coil at its centre: B=μ0N2I2r2B = \frac{\mu_0 N_2 I}{2 r_2}; flux through the small coil: N1Bπr12N_1 B \pi r_1^2.
  2. M=μ0N1N2πr122r2=4π×10−7×10×200×π×10−42×10M = \frac{\mu_0 N_1 N_2 \pi r_1^2}{2 r_2} = \frac{4\pi\times10^{-7}\times 10 \times 200 \times \pi \times 10^{-4}}{2\times 10}
  3. M=4π2×10−7×0.01=40×10−9=4×10−8 HM = 4\pi^2 \times 10^{-7} \times 0.01 = 40\times10^{-9} = 4\times10^{-8}\text{ H}
  4. So the answer is 4.
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