electromagnetic field strength

roboticmehdi

hello world. it is know that electrostatic (coulomb's law) and magnetostatic (biot-savart law) fields lose their strength like 1/r^2. why do they say that electromagnetic field falls like 1/r ? is that true ? if yes how, can you explain please ? after all energy radiated from a point source must fall like 1/r^2, because the area of surface of a sphere increases like r^2.

DaleSpam

Can you provide a reference for this? It is hard to say one way or the other without knowing the details.

roboticmehdi

http://en.wikipedia.org/wiki/Larmor_formula there in the part ''Derivation 2: Using Edward M. Purcell approach'' it says stuff related to this.

DaleSpam

Both Coulomb's law and the Biot-Savart law are approximations for 0 velocity and 0 acceleration respectively. The full general field produced by a point charge moving with arbitrary velocity and acceleration is given by the Lienard Wiechert potential:
http://en.wikipedia.org/wiki/Li%C3%A...hert_potential

If you look at the formula for the LW fields you see that for a stationary charge you get a 0 B field and a 1/r² E field, corresponding with Coulomb's law. If you look at the formula for the LW fields for a moving but not accelerating charge you get a 1/r² B field, corresponding with the Biot-Savart law. However, if you look at the formula for an accelerating charge you also get a 1/r E and a 1/r B field.

Philip Wood

One way to shed light on this is to note that the 1/r fields (unlike the 1/r² fields) are propagating away from the source, carrying energy with them. In a wave, the intensity (energy per unit time per unit normal area) is proportional to the square of the amplitude, so to 1/r² for the 1/r propagating field. But this 1/r² intensity law is just what we get by assuming energy not to be lost from the wave as it propagates outwards through larger and larger spherical surfaces – whose areas are proportional to r².