Electromagnetic Waves

JEE Main level, single correct, medium

61 of 165

Question 61 of 165

The magnetic field of a plane electromagnetic wave is given by B⃗=2×10−8sin⁡(0.5×103x+1.5×1011t)j^T\vec{B}=2\times {10}^{-8}\sin \left(0.5\times {10}^{3}x+1.5\times {10}^{11}t\right)\hat{j}T. The amplitude of the electric field would be

Show answer and solution

Answer: (C) 6Vm−16{Vm}^{-1} along zz-axis

The ratio of electric field amplitude to the magnetic field amplitude of an electromagnetic wave is equal to the speed of light in the medium. Therefore,
c=E0B0⇒E0=cB0c=\dfrac{{E}_{0}}{{B}_{0}} \Rightarrow {E}_{0}=c{B}_{0}
⇒E0=3×108×2×10−8\Rightarrow {E}_{0}=3\times {10}^{8}\times 2\times {10}^{-8}
⇒E0=6Vm−1\Rightarrow {E}_{0}=6V{m}^{-1}
According to the given equation B⃗=2×10−8sin⁡(0.5×103x+1.5×1011t)j^T\vec{B}=2\times {10}^{-8}\sin \left(0.5\times {10}^{3}x+1.5\times {10}^{11}t\right)\hat{j}T the wave travels in xx direction. As electric field and magnetic field vibrate perpendicular to the velocity of light, electric field can not be along xx axis.

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Electromagnetic Waves (Electromagnetic Waves and Maxwell's equations), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.