A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of $3,$ keeping the number of turns of the coil per unit length of the frame the same, then the self inductance of the coil:

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Aicharya · Accepted Answer

As total length $L$ of the wire will remain constant $ \mathrm{L}=(3 \mathrm{a}) \mathrm{N} \quad(\mathrm{N}=\text { total turns }) $ and length of winding $=(\mathrm{d}) \mathrm{N}$

(d = diameter of wire) $ \begin{array}{l} \text { self inductance }=\mu_{0} \mathrm{n}^{2} \mathrm{~A} \ell \\ \quad=\mu_{0} \mathrm{n}^{2}\left(\frac{\sqrt{3} \mathrm{a}^{2}}{4}\right) \mathrm{d} \mathrm{N} \\ \quad \propto \mathrm{a}^{2} \mathrm{~N} \propto \mathrm{a}\left[\text { as } \mathrm{N}=\mathrm{L} / 3 \mathrm{a} \Rightarrow \mathrm{N} \propto \frac{1}{\mathrm{a}}\right] \end{array} $ Now 'a' increased to ' $3 \mathrm{a}$ ' So self inductance will become 3 times

A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear…

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