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Q. A coil is wound on a transformer of rectangular cross-section. If all the linear dimensions of the transformer are increased by a factor 2 and the number of turns per unit length of the coil remains the same, the self inductance increases by a factor of

Jharkhand CECEJharkhand CECE 2013

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}}.\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}} $