From what I know, In electromagnetic waves, the components electric field and magnetic field generate each other i.e. magnetic field generated from moving electric charge generates another electric field and which also generates another magnetic field and so on thus, creating an electromagnetic field. So each of them gets weaker with distance which results in weaker EMF. Am I correct here?


2 Answers 2


The weakening or otherwise of EM radiation is purely a matter of geometry.

If you have an isolated point charge the field lines travel radially outwards from the charge. At some distance $r$ from the charge the field lines are spread out over a sphere of area $4 \pi r^2$ so the field line density, i.e the field strength, is proportional to 1/$r^2$. Hence the inverse square law.

On the other hand, if you had a perfectly collimated laser its brightness would be unchanged no matter how far the beam travelled. A perfectly collimated laser is impossible for various reasons, but if you've been to any rock festival you'll know that lasers stay bright for a large distance (though they may be attenuated by the herbal smoke they travel through).

  • 5
    $\begingroup$ The herbal smoke remark deserves a bonus reputation score :) $\endgroup$
    – ktorn
    Apr 13, 2015 at 12:13

The EMW weakening is due to decreasing the photon flux, just like flux of particles decrease in vacuum after explosion. Kinetic energy of a given particle is the same, but the number of particles per square meter per second decreases with distance. Photons are still capable of exiting atoms at large distances - photon energy (frequency) does not change with distance.

  • $\begingroup$ I like the reminder of the particle nature of light. Really helps to visualize in the context of the question. $\endgroup$
    – John
    Jan 17, 2012 at 22:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.