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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Electromagnetic Induction

Two coaxial solenoids are made by winding their insulated wire over a pipe of cross sectional area $A= 10\;cm^2$ length $20\;cm$ .One of the solenoids has 200 turns and another 400 turns. the mutual inductance is .

$(a)\;2.4 \times 10^{-4} \pi H \\(b)\;2 \times 10^{-4} \; \pi H \\ (c)\;1.6 \times 10^{-4} \;\pi \;H \\ (d)\;8 \times 10^{-4} \; \pi H $

1 Answer

$M= \large\frac{\mu_0 N_1N_2 A}{l}$
$\quad= \large\frac{4 \pi \times 10^{-7} \times 200 \times 400 \times 10 \times 10^{-4}}{0.2}$
$\quad = 1.6 \pi \times 10^{-4} \;H$
Hence c is the correct answer.
answered Mar 27, 2014 by meena.p
 

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