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JEE Main 201911 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Evening Shift
Subject
Physics
Chapter
Electromagnetic Induction

JEE Main 2019 · 11 January · Evening Shift · Physics · Electromagnetic Induction

A copper wire is wound on a wooden frame, whose shape is that of an...

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

Options

  1. Adecreases by a factor of 9
  2. Bincreases by a factor of 27
  3. Cincreases by a factor of 3
  4. Ddecreases by a factor of 939 \sqrt{3}

Correct answer

C. increases by a factor of 3

Step-by-step solution

As total length LL of the wire will remain constant L=(\mathrm{L}=(3 a)N(N= total turns )\mathrm{a}) \mathrm{N} \quad(\mathrm{N}=\text{ total turns })and length of winding =(d)N<imgsrc="/api/proxy−image?url=https=(\mathrm{d}) \mathrm{N}<img src="/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fjee_main%252FAwhDsCHiFGmqnF3aQf06y5-65Q23Nslyy8btneTYKEo.original.fullsize.png%3Falt%3Dmedia&clean=1&v=wm2" data-qx-orig-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_main%2FAwhDsCHiFGmqnF3aQf06y5-65Q23Nslyy8btneTYKEo.original.fullsize.png?alt=media" data-qx-storage-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_main%2FAwhDsCHiFGmqnF3aQf06y5-65Q23Nslyy8btneTYKEo.original.fullsize.png?alt=media"> \\ (d = diameter of wire)  self inductance =μ0\mathrmn2 Aℓ=μ0\mathrmn2(3\mathrma24)dN∝\mathrma2 N∝a[ as N=L/3a⇒N∝1a]\begin{array}{l}\text{ self inductance }=\mu_{0} \mathrmn^{2} \mathrm{~A} \ell\\ \quad=\mu_{0} \mathrmn^{2}\left(\frac{\sqrt{3} \mathrma^{2}}{4}\right) \mathrm{d}\mathrm{N}\\ \quad \propto \mathrma^{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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