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(C) The self-inductance $L$ of a solenoid is given by the formula $L = \mu_0 N^2 A / l$.Since the number of turns per unit length $n = N / l$ is const

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$8$

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(C) The self-inductance $L$ of a solenoid is given by the formula $L = \mu_0 N^2 A / l$.Since the number of turns per unit length $n = N / l$ is constant,we can write $N = nl$.Substituting this into the formula,we get $L = \mu_0 (nl)^2 A / l = \mu_0 n^2 l A$.Here,$l$ represents the length of the coil and $A$ represents the cross-sectional area.When all linear dimensions are increased by a factor of $2$,the length $l$ becomes $2l$ and the area $A$ (which is proportional to the square of linear dimensions) becomes $2^2 A = 4A$.Substituting these new values into the expression for $L$,the new inductance $L'$ is $L' = \mu_0 n^2 (2l) (4A) = 8 (\mu_0 n^2 l A) = 8L$.Therefore,the self-inductance increases by a factor of $8$.

$A$ coil is wound as a transformer of rectangular cross-section. If all the linear dimensions of the transformer are increased by a factor of $2$ and the number of turns per unit length of the coil remains the same,the self-inductance increases by a factor of:

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