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A long solenoid has 500 turns. When a current of 2 ampere 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

1. 2.5 henry
2. 2.0 henry
3. 1.0 henry
4. 40 henry

Correct answer: 2.5 henry

Solution

The self-inductance (L) of the solenoid is given by L = NΦ/I, where N is the number of turns, Φ is the magnetic flux per turn, and I is the current. Substituting N = 500, Φ = 4×10^-3 Wb, and I = 2 A, we get L = (500 × 4×10^-3) / 2 = 2.5 H.

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