Verified Solution • 2025–26 Session moderate

QJEEM20Phe26f: JEE Main 2020 Physics - Class 12 JEEM20Phe26f Solution: The electric field of a plane electromagnetic wave propagating…

The electric field of a plane electromagnetic wave propagating along the x direction in vacuum is
E→=E0j^cos⁡(ωt−kx)\overrightarrow E = {E_0}\widehat j\cos \left( {\omega t - kx} \right) .
The magnetic field B→\overrightarrow B , at the moment t = 0 is :

Multiple Choice Options

  • A B→=E0μ0∈0cos⁡(kx)j^\overrightarrow B = {{{E_0}} \over {\sqrt {{\mu _0}{ \in _0}} }}\cos \left( {kx} \right)\widehat j
  • B B→=E0μ0∈0cos⁡(kx)k^\overrightarrow B = {{{E_0}} \over {\sqrt {{\mu _0}{ \in _0}} }}\cos \left( {kx} \right)\widehat k
  • C B→=E0μ0∈0cos⁡(kx)k^\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos \left( {kx} \right)\widehat k
  • D B→=E0μ0∈0cos⁡(kx)j^\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos \left( {kx} \right)\widehat j

Step-by-Step Verified Solution

E→=E0 cos⁡(ωt−kx)j^\overrightarrow E = {E_0}\,\cos (\omega t - kx)\widehat j

We know, E=BCE = BC

B0=E0C=E01μ0∈0B_{0} = {E_{0} \over C} = {{{E_0}} \over {{1 \over {\sqrt {{\mu _0}{ \in _0}} }}}}

⇒\Rightarrow B0=E0μ0∈0B_{0} = {E_0}\sqrt {{\mu _0}{ \in _0}} JEE Main 2020 (Online) 3rd September Evening Slot Physics - Electromagnetic Waves Question 101 English Explanation

You can see direction of B→\overrightarrow B is along z axis.

∴\therefore B→=E0μ0∈0cos⁡(ωt−kx)k^\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos (\omega t - kx)\widehat k

at t = 0

B→=E0μ0∈0cos⁡(kx)k^\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos (kx)\widehat k

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring