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The length of a wire required to manufacture a solenoid of length l and self-induction L is (cross-sectional area is negligible)

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$\sqrt{\frac{μ_{0} L l}{4 π}}$

$\sqrt{\frac{μ_{0} L l}{2 π}}$

$\sqrt{\frac{2 π L l}{μ_{0}}}$

$\sqrt{\frac{4 π L l}{μ_{0}}}$

answer is C.

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Detailed Solution

$L = \frac{μ_{0} N^{2} A}{l}$

If x is the length of the solenoid with r as radius, then

$x = 2 π r N, A = π r^{2}$

$∴ L = μ_{0} (\frac{x^{2}}{4 π^{2} r^{2}}) \frac{π r^{2}}{l} [∴ N = \frac{x}{2 π r}]$

$∴ x = \sqrt{\frac{4 π L l}{μ_{0}}}$

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