A long solenoid has 500 turns. When a current of 2 A is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4×10⁻³ Wb. The self-inductance of the solenoid is

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(1) 1.0 henry. The flux quoted is per turn, so the total flux linkage is Nφ=500×4×10⁻³=2 Wb. Self-inductance is defined by Nφ=LI: L=2/2=1.0 H. Forgetting the factor N is the usual slip; the flux per turn alone would give only 2 mH.

A long solenoid has 500 turns. When a current of 2 A is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4×10⁻³ Wb. The self-inductance of the solenoid is

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