A coil of wire of a certain radius has 600 turns and a self-inductance of 108 mH. The self inductance of another similar coil with same radius and length but 500 turns for the same current flow would be-
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 75 mH
For a long coil (solenoid) the self-inductance is \( L = \mu_0 N^{2} A/l \). With the same radius and length, \( L \propto N^{2} \); the current does not affect L.
\( \displaystyle \frac{L_2}{L_1} = \left(\frac{N_2}{N_1}\right)^{2} \)
\( \displaystyle L_2 = 108\,\mathrm{mH} \times \left(\frac{500}{600}\right)^{2} = 108\,\mathrm{mH} \times \frac{25}{36} \)
\( \displaystyle L_2 = 75\,\mathrm{mH} \)