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Electromagnetic field strength 
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#1
Mar1912, 07:32 AM

P: 34

hello world. it is know that electrostatic (coulomb's law) and magnetostatic (biotsavart 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.



#2
Mar1912, 07:42 AM

Mentor
P: 17,329




#3
Mar1912, 07:48 AM

P: 34




#4
Mar1912, 08:03 AM

Mentor
P: 17,329

Electromagnetic field strength
Both Coulomb's law and the BiotSavart 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 BiotSavart 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. 


#5
Mar2012, 04:46 AM

PF Gold
P: 956

One way to shed light on this is to note that the 1/r fields (unlike the 1/r^{2} 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/r^{2} for the 1/r propagating field. But this 1/r^{2} 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 r^{2}.



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