Question

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 \(4 \times 10^{-3} \mathrm{~Wb}\). The selfinductance of the solenoid is
(a) \(1.0\) henry
(b) \(4.0\) henry
(c) \(2.5\) henry
(d) \(2.0\) henry

Answer 1

Correct option (a) \(1.0\) henry
Explanation:
For a long solenoid, \(B=\mu_{0} n i=\mu_{0} \frac{N}{l} \cdot i\)
Flux \(=\mu_{0} \frac{N}{l} \cdot i \cdot A\)
given flux per turn \(=4 \times 10^{-3}, i=2 \mathrm{~A}\)
\(\therefore\) Total flux \(=4 \times 10^{-3}\)
\(L=\left(\mu_{0} \frac{N}{l} \cdot N A\right)=\frac{4 \times 10^{-3} \times 500}{2}=1\) henry

Answer 2

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Answer 3

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