A) \[4\]
B) \[8\]
C) \[12\]
D) \[16\]
Correct Answer: B
Solution :
We have, self-inductance,\[L=\frac{{{\mu }_{0}}{{n}^{2}}A}{l}\] Given, \[n/l\] is constant, doubling the length would double \[n\] also. Besides this, area would become four times on doubling linear dimensions. So, \[\frac{{{L}_{2}}}{{{L}_{1}}}=\frac{n_{2}^{2}}{n_{1}^{2}}\cdot \frac{{{A}_{2}}}{{{A}_{1}}}\times \frac{{{l}_{1}}}{{{l}_{2}}}\] Here, \[\frac{{{L}_{2}}}{{{L}_{1}}}=4\times 4\times \frac{1}{2}=8\] \[\Rightarrow \] \[{{L}_{2}}=8{{L}_{1}}\]
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