Electromagnetic Induction

JEE Main level, single correct, medium

135 of 153

Question 135 of 153

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 3,3, 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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Answer: (C) increases by a factor of 3

As total length LL of the wire will remain constant

L=(3a)N(N= total turns )\mathrm{L}=(3 \mathrm{a}) \mathrm{N} \quad(\mathrm{N}=\text { total turns })

and length of winding
=(d)N=(\mathrm{d}) \mathrm{N}




(d = diameter of wire)


 self inductance =μ0n2 Aℓ=μ0n2(3a24)dN∝a2 N∝a[ as N=L/3a⇒N∝1a]\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 '
3a3 \mathrm{a} ' So self inductance will become 3 times

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