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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 d7 10^{-3} Wb. The self-inductance of the solenoid is

1. 2.5 H
2. 2.0 H
3. 1.0 H
4. 4.0 H

Correct answer: 1.0 H

Solution

Self-inductance is defined by L = N rac{ ext{flux linkage}}{I} . Here flux linkage = 500 d7 4 d7 10^{-3} = 2 Wb-turn, and dividing by current 2 A gives L = 1 H .

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