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The magnetic field vector of an electromagnetic wave is given by B=B_0 i+ j √2 cos(kz-ω t) where i, j represents unit vector along x and y-axis…

JEE Main 2021 — Physics Electromagnetism

2021mcqmedium

The magnetic field vector of an electromagnetic wave is given by B=B0i^+j^2cos(kz−ωt)B={B}_{0}\frac{\hat{i}+\hat{j}}{\sqrt{2}}\mathrm{cos}(kz-\omega t)B=B0​2​i^+j^​​cos(kz−ωt) where i^,j^\hat{i},\hat{j}i^,j^​ represents unit vector along xxx and yyy-axis respectively. At t=0s,t=0s,t=0s, two electric charges q1{q}_{1}q1​ of 4π4\pi4π coulomb and q2{q}_{2}q2​ of 2π2\pi2π coulomb located at (0,0,πk)(0,0,\frac{\pi }{k})(0,0,kπ​) and (0,0,3πk),(0,0,\frac{3\pi }{k}),(0,0,k3π​), respectively, have the same velocity of 0.5ci^,0.5c\hat{i},0.5ci^, (where ccc is the velocity of light ). The ratio of the force acting on charge q1{q}_{1}q1​ to q2{q}_{2}q2​ is :

Official previous-year question

Held on 31 Aug 2021 · Verified 6 Jul 2026.

Options

  1. A

    1:21:\sqrt{2}1:2​

  2. B

    22:12\sqrt{2}:122​:1

  3. C

    2:1\sqrt{2}:12​:1

  4. D

    2:12:12:1

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