Verified Solution • 2025–26 Session moderate

QJEEM19Ph7ba0: JEE Main 2019 Physics - Class 12 JEEM19Ph7ba0 Solution: The electric field of a plane electromagnetic wave is given by…

The electric field of a plane electromagnetic wave is given by
E→=E0i^cos⁡(kz)cos(ωt)\overrightarrow E = {E_0}\widehat i\cos (kz)cos(\omega t)
The corresponding magnetic field B→\overrightarrow B is then given by

Multiple Choice Options

  • A B→=E0Cj^sin⁡(kz)sin⁡(ωt)\overrightarrow B = {{{E_0}} \over C}\widehat j\sin (kz)\sin (\omega t)
  • B B→=E0Cj^sin⁡(kz)cos⁡(ωt)\overrightarrow B = {{{E_0}} \over C}\widehat j\sin (kz)\cos (\omega t)
  • C B→=E0Cj^cos⁡(kz)sin⁡(ωt)\overrightarrow B = {{{E_0}} \over C}\widehat j\cos (kz)\sin (\omega t)
  • D B→=E0Ck^sin⁡(kz)cos⁡(ωt)\overrightarrow B = {{{E_0}} \over C}\widehat k\sin (kz)\cos (\omega t)

Step-by-Step Verified Solution

∴E→×B→∥v→\therefore \overrightarrow E \times \overrightarrow B \parallel \overrightarrow v

Given that wave is propagating along positive z-axis and E→\overrightarrow E along positive x-axis. Hence B→\overrightarrow B along y-axis.

From Maxwell equation
V→×E→=−∂B∂t\overrightarrow V \times \overrightarrow E = - {{\partial B} \over {\partial t}}

i.e. ∂E∂Z=−∂Bdt{{\partial E} \over {\partial Z}} = - {{\partial B} \over {dt}} and B0=E0C{B_0} = {{{E_0}} \over C}

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