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JEE Main 202623 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Morning Shift
Subject
Physics
Chapter
Electromagnetic Induction

JEE Main 2026 · 23 January · Morning Shift · Physics · Electromagnetic Induction

A 20 m long uniform copper wire held horizontally is allowed to...

A 20 m long uniform copper wire held horizontally is allowed to fall under the gravity (g=10 m/s2g=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 when it travells a vertical distance of 200 m is ____\_\_\_\_ mV.

Options

  1. A201020 \sqrt{10}
  2. B0.2100.2 \sqrt{10}
  3. C20010200 \sqrt{10}
  4. D2102 \sqrt{10}

Correct answer

A. 201020 \sqrt{10}

Step-by-step solution

For a conducting wire falling in a magnetic field, induced EMF is: E=BLv\mathcal{E} = BLv
After falling through height h=200h = 200 m, velocity: v=2gh=2×10×200=4000=2010v = \sqrt{2gh} = \sqrt{2 \times 10 \times 200} = \sqrt{4000} = 20\sqrt{10} m/s
Converting magnetic field: B=0.5B = 0.5 Gauss =0.5×10−4= 0.5 \times 10^{-4} T
Induced EMF: E=(0.5×10−4)×20×2010=200×10−410=0.0210\mathcal{E} = (0.5 \times 10^{-4}) \times 20 \times 20\sqrt{10} = 200 \times 10^{-4}\sqrt{10} = 0.02\sqrt{10} V =2010= 20\sqrt{10} mV
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