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JEE Main 2008
Exam
JEE Main
Year
2008
Subject
Physics
Chapter
Electromagnetic Induction

JEE Main 2008 · Physics · Electromagnetic Induction

Two coaxial solenoids are made by winding thin insulated wire over...

Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross sectional area A=10 cm2A= 10 \mathrm{~cm}^2 and length =20 cm=20 \mathrm{~cm}. If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is (μ0=4π×10−7TmA−1)\left(\mu_0=4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}^{-1}\right)

Options

  1. A2.4π×10−5H2.4 \pi \times 10^{-5} \mathrm{H}
  2. B4.8π×10−4H4.8 \pi \times 10^{-4} \mathrm{H}
  3. C4.8π×10−5H4.8 \pi \times 10^{-5} \mathrm{H}
  4. D2.4π×10−4H2.4 \pi \times 10^{-4} \mathrm{H}

Correct answer

D. 2.4π×10−4H2.4 \pi \times 10^{-4} \mathrm{H}

Step-by-step solution

M=μ0N1N2Aℓ=\frac{\mu_0 N_1 N_2 A}{\ell}=2.4 π×\pi \times10^{-4} H\mathrm{H}
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