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 \[L=\pi {{\mu }_{0}}{{n}^{2}}{{r}^{2}}b\] \[=\pi {{\mu }_{0}}{{\left( \frac{1}{2a} \right)}^{2}}.{{r}^{2}}.\frac{al}{\pi r}\] \[=\frac{1}{4}{{\mu }_{0}}r\frac{l}{a}=\frac{1}{4}{{\mu }_{0}}r\frac{V}{a(\pi {{a}^{2}})}\] \[=\frac{{{\mu }_{0}}r}{4}\frac{V}{{{a}^{3}}}\] \[\therefore \,\,\,\,\,\,L\propto \frac{1}{{{a}^{3}}}\] \[\therefore \,\,\,\,\,\,\frac{{{L}_{1}}}{{{L}_{2}}}={{\left( \frac{2a}{a} \right)}^{3}}=\frac{8}{1}\]You need to login to perform this action.
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