A) \[\frac{L}{2}\]
B) L
C) 2L
D) 4L
Correct Answer: C
Solution :
[c] : In case of a solenoid as \[B={{\mu }_{0}}nI,\phi =B(nlS)={{\mu }_{0}}{{n}^{2}}lSI\]and hence \[L=\frac{\phi }{I}={{\mu }_{0}}{{n}^{2}}IS={{\mu }_{0}}\frac{{{N}^{2}}}{l}S\]\[\left( as=\frac{N}{l} \right)\] When N and l are doubled, then \[L'={{\mu }_{0}}\frac{{{(2N)}^{2}}}{2l}S=2{{\mu }_{0}}\frac{{{N}^{2}}}{l}S=2L\] i.e., inductance of the solenoid will be doubled.
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