Q: A solenoid is of length 50 cm and has a radius of 2 cm. It has 500 turns. Around its central section a coil of 50 turns is wound. Calculate the mutual inductance of the system.

Sol: N_{P} = 500, N_{S} = 50 ; A = π × 0.02 × 0.02 m^{2}

μ_{0} = 4π× 10^{-7} Hm^{-1}, l = 50 cm = 0.5 m

$\large M = \frac{\mu_0 N_p N_s A}{l} $

$\large M = \frac{4\pi \times 10^{-7} \times 500 \times 50 \times \pi (0.02)^2}{0.5} $

=789.8 × 10^{-7} H = 78.98 μH.