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Physics · Electromagnetic Induction

JEE Main 2026 — 22 January, Morning Shift — Question 28

XPQY is a vertical smooth long loop having a total resistance R where PX is parallel to QY and separation between them is ll. A constant magnetic field B perpendicular to the plane of the loop exists in the entire space. A rod CD of length L ( L>l\mathrm{L}>l ) and mass mm is made to slide down from rest under the gravity as shown in figure. The terminal speed acquired by the rod is ____\_\_\_\_ m/s\mathrm{m} / \mathrm{s}. ( g=\mathrm{g}= acceleration due to gravity)

Question figure
  1. Option A:

    2mgRB2l2\frac{2 \mathrm{mgR}}{\mathrm{B}^{2} l^{2}}

  2. Option B:

    8mgRB2l2\frac{8 \mathrm{mgR}}{\mathrm{B}^{2} l^{2}}

  3. Option C:

    2mgRB2 L2\frac{2 \mathrm{mgR}}{\mathrm{B}^{2} \mathrm{~L}^{2}}

  4. Option D:

    mgRB2l2\frac{\mathrm{mgR}}{\mathrm{B}^{2} l^{2}}

    Correct

Answer: D

Step-by-step solution

at equilibrium (Or for terminal velocity) mg=iBℓ⇒mg=(BvℓR)Bℓ\mathrm{mg}=\mathrm{iB} \ell \Rightarrow \mathrm{mg}=\left(\frac{\mathrm{Bv} \ell}{\mathrm{R}}\right) \mathrm{B} \ell V=mgRB2ℓ2\mathrm{V}=\frac{\mathrm{mgR}}{\mathrm{B}^{2} \ell^{2}}

figure

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
22 January, Morning Shift
Subject
Physics
Chapter
Electromagnetic Induction
Topic
Motional EMF

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