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JEE Main 2008 (AIEEE) · Mutual Inductance

Mutual Inductance question from JEE Main 2008 (AIEEE)

Q61JEE Main 2008 · AIEEEMedium

Mutual Inductance

Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A=10 cm2A = 10\text{ cm}^2 and length =20 cm= 20\text{ cm}. If one of the solenoid has 300300 turns and the other 400400 turns, their mutual inductance is

(μ0=4π×10−7 T m A−1)\left(\mu_0 = 4\pi \times 10^{-7}\text{ T m A}^{-1}\right)

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Answer: (d)

  1. For coaxial solenoids on the same pipe: M=μ0N1N2AlM = \frac{\mu_0 N_1 N_2 A}{l}
  2. M=4π×10−7×300×400×10×10−40.20M = \frac{4\pi\times10^{-7}\times 300 \times 400 \times 10\times10^{-4}}{0.20}
  3. M=4π×10−7×600=2.4π×10−4 HM = 4\pi\times10^{-7}\times 600 = 2.4\pi\times10^{-4}\text{ H}
  4. So the answer is (d).
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