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Physics · Electromagnetic Induction

JEE Main 2026 — 23 January, Morning Shift — Question 39

A 20 m long uniform copper wire held horizontally is allowed to fall under the gravity ( g=10 m/s2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2} ) through a uniform horizontal magnetic field of 0.5 Gauss perpendicular to the length of the wire. The induced EMF across the wire it travells a vertical distance of 200 m is ____\_\_\_\_ mV .

  1. Option A:

    0.2100.2 \sqrt{10}

  2. Option B:

    201020 \sqrt{10}

    Correct
  3. Option C:

    2102 \sqrt{10}

  4. Option D:

    20010200 \sqrt{10}

Answer: B

Step-by-step solution

ε=vBℓ\varepsilon=\mathrm{vB} \ell v=2gh=2×10×200=2010\mathrm{v}=\sqrt{2 \mathrm{gh}}=\sqrt{2 \times 10 \times 200}=20 \sqrt{10} ε=(2010)(0.5×10−4)20\varepsilon=(20 \sqrt{10})\left(0.5 \times 10^{-4}\right) 20 =2010×10−3=2010mV=20 \sqrt{10} \times 10^{-3}=20 \sqrt{10} \mathrm{mV}

Answer key and solution verified before publishing.

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

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