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 4×10−3Wb. The self-inductance of the solenoid is
A
1.0H
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B
4.0H
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C
2.5H
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D
2.0H
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Solution
The correct option is A1.0H Given flux per turn =4×10−3i=2A
No of turns = 500
Total flux = 500×4×10−3=2A
Flux, ϕ=Li2=L×2L=1H