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A coil of 100 turns carries a current of 5 A and links a magnetic flux of 10⁻⁵ T m² per turn. What is its self-inductance L?

1. 0.05 mH
2. 0.10 mH
3. 0.15 mH
4. 0.20 mH

Correct answer: 0.20 mH

Solution

Self-inductance is defined by L = N*phi/I, where phi is the flux per turn. L = (100 * 10⁻⁵)/5 = 10⁻³/5 = 2*10⁻⁴ H = 0.20 mH.

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