A) halved
B) doubled
C) 1/4 times the original value
D) unaffected
Correct Answer: B
Solution :
(b) doubled We know that, \[L=\frac{{{\mu }_{0}}{{N}^{2}}A}{l}\] Now, \[N'=2N\] and \[l'=2l\] \[\therefore l'=\frac{{{\mu }_{0}}{{(2N)}^{2}}A}{2l}={{\mu }_{0}}\frac{4{{N}^{2}}A}{2l}\] \[=\frac{2{{\mu }_{0}}{{N}^{2}}A}{l}=2L\]
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