Careers360 Logo
NTA NEET 2024 - Application Correction (Re-opens), Admit Card, Exam Centers

Access premium articles, webinars, resources to make the best decisions for career, course, exams, scholarships, study abroad and much more with

Plan, Prepare & Make the Best Career Choices

Nature Of Electromagnetic Waves - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Nature of Electromagnetic Waves is considered one the most difficult concept.

  • 80 Questions around this concept.

Solve by difficulty

The electric and the magnetic field associated with an E.M. wave, propagating along the +z-axis, can be represented by


Magnetic field in a plane electromagnetic wave is given by

Expression for corresponding electric field will be : Where c is speed of light.



Concepts Covered - 1

Nature of Electromagnetic Waves

Nature of Electromagnetic Waves

From Maxwell’s equations we can observe that electric and magnetic fields in an electromagnetic wave are perpendicular to each other, and to the direction of propagation. Also from our discussion of the displacement current, in that capacitor, the electric field inside the plates of the capacitor is directed perpendicular to the plates. Figure given below shows a typical example of a plane electromagnetic wave propagating along the z direction (the fields are shown as a function of the z coordinate, at a given time t). The electric field Ex is along the x-axis, and varies sinusoidally with z, at a given time. The magnetic field By is along the y-axis, and again varies sinusoidally with z.The electric and magnetic fields Ex and By are perpendicular to each other, and to the direction z of propagation.



Now from the Lorentz equation -

                                                 \begin{array}{l}{\text { }} \\ {\qquad \begin{aligned} \vec{F}=& q(\vec{E} \times \vec{v} \times \vec{B}) \\ \\ E_{z}=E z_{0} \sin (\omega t-k y) & \\ \\ B_{x}=B x_{0} \sin (\omega t-k y), & \text { where } \frac{\omega}{k}=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}} \end{aligned}}\end{array}

since,  \omega=2 \pi f, where f is the frequency and    k= \frac{2 \pi }{\lambda},   where \lambda is the wavelength.


                                                                      \text { Therefore, } \frac{\omega}{k}=\frac{2 \pi f}{2 \pi / \lambda}=f \lambda

But f \lambda gives the velocity of the wave. So, f \lambda = c = \omega k. So we can write - 

                                                                                    c=\frac{\omega}{k}=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}} 

It is also seen from Maxwell’s equations that the magnitude of the electric and the magnetic fields in an electromagnetic wave are related as - 

                                                                                             B_{0}= \frac{E_o}{c}

In a material medium of permittivity ε and magnetic permeability µ, the velocity of light becomes,

                                                                                               v=\frac{1}{\sqrt{\mu \varepsilon}}

Study it with Videos

Nature of Electromagnetic Waves

"Stay in the loop. Receive exam news, study resources, and expert advice!"

Get Answer to all your questions

Back to top