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Question:

A coil of wire of a certain radius has 100 turns and a self-inductance of 15 mH. The self-inductance of a second similar coil of 500 turns will be:

75mH

375mH

15mH

none of these

Solution:

The correct option is B 375mH
Self inductance of a coil of wire =L=μN²A/l
As the number of turns become 5 times, the self inductance must become 25 times the initial value.
Thus new self inductance = 25 × 15 mH = 375 mH