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Question
A coil of wire of a certain radius has 600 turns and a self-inductance is 108 mH. The self-inductance of a second similar coil of 500 turns will be:
Options
74 mH
75 mH
76 H
77 mH
MCQ
Solution
75 mH
Explanation:
As Ln2
∴ L = `108 xx (500/600)^2` mH = 75 mH.
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