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JEE Main 202430 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
30 January
Shift
Morning Shift
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2024 · 30 January · Morning Shift · Physics · Electromagnetic Waves

The electric field of an electromagnetic wave in free space is...

The electric field of an electromagnetic wave in free space is represented as \overset{\to}{E} = {E}_{0}\cos ( \omegat - kz ) \overset{^}{i}. The corresponding magnetic induction vector will be :

Options

  1. A\overset{\to}{B} = {E}_{0}Ccos ( \omegat - kz ) \overset{^}{j}
  2. B\overset{\to}{B} = \dfrac{{E}_{0}}{C}\cos ( \omegat - kz ) \overset{^}{j}
  3. C\overset{\to}{B} = {E}_{0}Ccos ( \omegat + kz ) \overset{^}{j}
  4. D\overset{\to}{B} = \dfrac{{E}_{0}}{C}\cos ( \omegat + kz ) \overset{^}{j}

Correct answer

B. \overset{\to}{B} = \dfrac{{E}_{0}}{C}\cos ( \omegat - kz ) \overset{^}{j}

Step-by-step solution

Given \overset{\to}{E} = {E}_{0}\cos ( \omegat - kz ) \overset{^}{i}:

Clearly, (\omegat−kz)( \omegat - kz ) term represents, electromagnetic wave is travelling along +z+ z direction or along \overset{^}{k}.

Now, in vector form, direction of electromagnetive wave is given by \overset{^}{C} = \overset{^}{E} \times \overset{^}{B}. Therefore, magnetic field is along \overset{^}{j}. 

Also, we know that B0=E0C{B}_{0} = \dfrac{{E}_{0}}{C}, where CC is speed of light.

Therefore, \overset{\to}{B} = \dfrac{{E}_{0}}{C}\cos ( \omegat - kz ) \overset{^}{j}.

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