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QJEEAdv24Ph3d74: JEE Advanced 2024 Physics - Class 12 JEEAdv24Ph3d74 Solution: A region in the form of an equilateral triangle (in plane) of…

A region in the form of an equilateral triangle (in x−yx-y plane) of height LL has a uniform magnetic field B⃗\vec{B} pointing in the +z+z -direction. A conducting loop PQR\mathrm{PQR} , in the form of an equilateral triangle of the same height LL , is placed in the x−yx-y plane with its vertex P\mathrm{P} at x=0x=0 in the orientation shown in the figure. At t=0t=0 , the loop starts entering the region of the magnetic field with a uniform velocity v⃗\vec{v} along the +x+x -direction. The plane of the loop and its orientation remain unchanged throughout its motion.

JEE Advanced 2024 Paper 2 Online Physics - Electromagnetic Induction Question 1 English

Which of the following graph best depicts the variation of the induced emf (E)(E) in the loop as a function of the distance (x)(x) starting from x=0x=0 ?

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/13999f4550a3f89aa0591f6105bbbe55857fb61154d2aa6ba7fe363ab4e94438.webp
  • B https://ik.imagekit.io/novatra/photos/1c02f565818ee63ea103505c897f43bc0dc755e6d8457e0e0a601f2af23abb23.webp
  • C https://ik.imagekit.io/novatra/photos/d8d917a12a395a7cce8f99842ca6c365efa80ad5e2e6ff91289a17fff52f7994.webp
  • D https://ik.imagekit.io/novatra/photos/a75cd338b968a6ac50bf89a8205462b03a3053feedda59558d324efea1ad2604.webp

Step-by-Step Verified Solution

0 to L

JEE Advanced 2024 Paper 2 Online Physics - Electromagnetic Induction Question 1 English Explanation 1

ε=Bℓentv=B×x3v\varepsilon=\mathrm{B} \ell_{\mathrm{ent}} \mathrm{v}=\mathrm{B} \times \frac{\mathrm{x}}{\sqrt{3}} \mathrm{v}

L to 2 L

JEE Advanced 2024 Paper 2 Online Physics - Electromagnetic Induction Question 1 English Explanation 2

amp;∣emf⁡∣=B(L3−x03)v−B2x03vamp;=BvL3−3Bvx0amp;=Bv[L3−3(x−L)]amp;=Bv3[L−3x+3L]amp;=Bv3[4L−3x]amp; at x=4L3amp; emf =0\begin{aligned}& |\operatorname{emf}|=B\left(\frac{L}{\sqrt{3}}-\frac{x_0}{\sqrt{3}}\right) v-B \frac{2 x_0}{\sqrt{3}} v \\& =\frac{B v L}{\sqrt{3}}-\sqrt{3} B v x_0 \\& =B v\left[\frac{L}{\sqrt{3}}-\sqrt{3}(x-L)\right] \\& =\frac{B v}{\sqrt{3}}[L-3 x+3 L] \\& =\frac{B v}{\sqrt{3}}[4 L-3 x] \\& \text { at } x=\frac{4 L}{3} \\& \text { emf }=0\end{aligned}

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