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Physics · Electromagnetic Waves

JEE Main 2026 — 2 April, Morning Shift — Question 13

An electromagnetic wave travelling in x- direction is described by field equation Ey=300sin⁡ω(t−xc)\mathrm{E}_{\mathrm{y}} = 300 \sin \omega \left(\mathrm{t} - \frac{\mathrm{x}}{\mathrm{c}}\right) . If the electron is restricted to move in y- direction only with speed of 1.5×106 m/s1.5 \times 10^{6} \mathrm{~m} / \mathrm{s} then ratio of maximum electric and magnetic forces acting on the electron is

  1. Option A:

    200

    Correct
  2. Option B:

    150

  3. Option C:

    400

  4. Option D:

    300

Answer: A

Step-by-step solution

FeFm=EvB=cv=3×1081.5×106=200\frac{F_e}{F_m} = \frac{E}{vB} = \frac{c}{v} = \frac{3\times10^8}{1.5\times10^6}=200

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
2 April, Morning Shift
Subject
Physics
Chapter
Electromagnetic Waves
Topic
Displacement Current and Equation of EM Waves
An electromagnetic wave travelling in x- direction is described by… | JEE Main 2026 PYQ with Solution · DhiX AI