A certain volume of copper is drawn into a wire of radius a and is wrapped in shape of a helix having radius $r (> > a)$ . The winding are as close as possible without overlapping. Self-inductance of the inductor so obtained is $L_{1}$ . Another wire of radius $2 a$ is drawn using same volume of copper and wound in the fashion as described above. This time the inductance is $L_{2}$ . Find $\frac{L_{1}}{L_{2}}$

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Detailed Solution

Length of wire $l = \frac{v o l u m e (V)}{π a^{2}}$
Number of turns $N = \frac{l}{2 π r}$
Length of the helix $b = 2 a . N = \frac{a l}{π r}$
Number of turns per meter length $n = \frac{1}{2 a}$
$∴$ Self-inductance
$\begin{array}{l} L = π μ_{0} n^{2} r^{2} b \\ = π μ_{0} . (\frac{1}{2 a}) . r^{2} . \frac{a l}{π r} \\ = \frac{1}{4} μ_{0} r \frac{l}{a} = \frac{1}{4} μ_{0} r \frac{V}{a (π a^{2})} \\ = \frac{μ_{0} r}{4 π} \frac{V}{a^{3}} \\ \frac{L_{1}}{L_{2}} = {(\frac{2 a}{a})}^{3} \end{array}$