question_answer

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

A)             \[8\]                 

B) \[\frac{1}{8}\]

C) \[\frac{1}{6}\]                          

D) \[6\]

Correct Answer: A

Solution :

[a] Length of wire \[I=\frac{volume\,(V)}{\pi {{a}^{2}}}\] Winding is as shown in the figure. Number of turns \[N=\frac{l}{2\pi r}\] Length of the helix \[b=2a.\,N=\frac{al}{\pi r}\] Number of turns per meter length \[n=\frac{1}{2a}\] \[\therefore \]  Self-inductance \[L=\pi {{\mu }_{0}}{{n}^{2}}{{r}^{2}}b\] \[=\pi {{\mu }_{0}}{{\left( \frac{1}{2a} \right)}^{2}}.{{r}^{2}}.\frac{al}{\pi r}\] \[=\frac{1}{4}{{\mu }_{0}}r\frac{l}{a}=\frac{1}{4}{{\mu }_{0}}r\frac{V}{a(\pi {{a}^{2}})}\] \[=\frac{{{\mu }_{0}}r}{4}\frac{V}{{{a}^{3}}}\] \[\therefore \,\,\,\,\,\,L\propto \frac{1}{{{a}^{3}}}\] \[\therefore \,\,\,\,\,\,\frac{{{L}_{1}}}{{{L}_{2}}}={{\left( \frac{2a}{a} \right)}^{3}}=\frac{8}{1}\]