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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 $4 \times 10^{- 3} Wb .$ The self – inductance of the solenoid is:

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1.0 henry

4.0 henry

2.5 henry

2.0 henry

answer is A.

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Detailed Solution

$L = \frac{Nϕ}{i} = \frac{4 \times 10^{- 3} \times 500}{2} = 1 H$

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