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Two coaxial solenoids are made by winding thin insulated wire over pipe of cross-sectional area $A = 10 c m^{2}$ and length $= 20 c m$ . If one of the solenoids has 500 turns and the other 400 turns their mutual inductance is $n \times 10^{- 4} H$ . The value of $n$ is_______(take $π = 3.14$ )

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answer is 12.56.

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Detailed Solution

mutual inductance $M = \frac{μ o N_{1} N_{2} A}{l}$
$M = \frac{4 π \times 10^{- 7} \times 500 \times 400 \times 10 \times 10^{- 4}}{20 \times 10^{- 2}}$
$= 12.56 \times 10^{- 4}$

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