Electromagnetic Waves

JEE Main level, single correct, medium

97 of 165

Question 97 of 165

A plane electromagnetic wave is propagating along the direction i^+j^2,\dfrac{\hat{i}+\hat{j}}{\sqrt{2}}, with its polarization along the direction k.^\hat{k.} The correct form of the magnetic field of the wave would be (here B0{B}_{0} is an appropriate constant):

Show answer and solution

Answer: (A) B0i^−j^2cos(ωt−ki^+j^2){B}_{0}\dfrac{\hat{i}-\hat{j}}{\sqrt{2}}cos\left(\omega t-k\dfrac{\hat{i}+\hat{j}}{\sqrt{2}}\right)

Polarization of EM Wave means Direction of Its electric Field
Electric field vector is along k^\hat{k}
So, unit electric filed vector, E⃗=k^\vec{E}=\hat{k} (Given)
Direction of wave,c^=i^+j^2\hat{c}=\dfrac{\hat{i}+\hat{j}}{\sqrt{2}} (Given)
B^\hat{B} denotes unit magnetic field vector.
B^=C^×E^\hat{B}=\hat{C}\times \hat{E}
B^=i^+j^2×k^=i^−j^2\hat{B}=\dfrac{\hat{i}+\hat{j}}{\sqrt{2}}\times \hat{k}=\dfrac{\hat{i}-\hat{j}}{\sqrt{2}}
General Form is B=(B0B^)cos⁡(ωt−c^k)B=({B}_{0}\hat{B})\cos \left(\omega t-\hat{c}k\right)
where B0{B}_{0} is Magnitude of Magnetic Field
So,B=B0i^−j^2cos⁡(ωt−ki^+j^2)So,B={B}_{0}\dfrac{\hat{i}-\hat{j}}{\sqrt{2}}\cos \left(\omega t-k\dfrac{\hat{i}+\hat{j}}{\sqrt{2}}\right)

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.