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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Physics
Chapter
Electromagnetic Induction

BITSAT 2013 · 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 cm2\mathrm{A}=10 \mathrm{~cm}^{2} and length =20 cm.=20 \mathrm{~cm} . If one of the solenoid 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{TmA}^{-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=4π×10−7×300×400×100×10−40.2=2.4π×10−4H\begin{aligned} M &=\frac{\mu_{0} \mathrm{ N}_{1} \mathrm{ N}_{2} \mathrm{ A}}{\ell} \\ &=\frac{4 \pi \times 10^{-7} \times 300 \times 400 \times 100 \times 10^{-4}}{0.2} \\ &=2.4 \pi \times 10^{-4} \mathrm{H} \end{aligned} #
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