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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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The correct answer is:
increases by a factor of 3
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
$$
\mathrm{L}=(3 \mathrm{a}) \mathrm{N} \quad(\mathrm{N}=\text { total turns })
$$
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
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