A solenoid is of length 50 cm and has a radius of 2cm. It has 500 turns. Around its central section a coil of 50 turns is wound. Calculate the mutual inductance of the system. (in $μ H$ ) (Round off to the Nearest Integer)

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Detailed Solution

$N_{P} = 500, N_{S} = 50; A = π \times 0.02 \times 0.02 m^{2}$

$μ_{0} = 4 π \times 10^{- 7} H m^{- 1}, 1 = 50 c m = 0.5 m$

Now, $M = \frac{μ_{0} N_{P} N_{S} A}{l} = \frac{4 π \times 10^{- 7} \times 500 \times 50 \times π \times {(0.02)}^{2}}{0.5} H$

$= 789.8 \times 10^{- 7} H = 78.98 μ H$

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