Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross sectional area $A=$ $10…

Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross sectional area $A=$ $10 \mathrm{~cm}^2$ and length $=20 \mathrm{~cm}$. If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is $\left(\mu_0=4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}^{-1}\right)$
  1. $2.4 \pi \times 10^{-5} \mathrm{H}$
  2. $4.8 \pi \times 10^{-4} \mathrm{H}$
  3. $4.8 \pi \times 10^{-5} \mathrm{H}$
  4. $2.4 \pi \times 10^{-4} \mathrm{H}$

Solution

$ M=\frac{\mu_0 N_1 N_2 A}{\ell}=2.4 \pi \times 10^{-4} \mathrm{H} $

Asked in: JEE Main 2008

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