Electromagnetic field strength

In summary: So we see that the 1/r fields do carry energy away with them, but only to the extent that the energy is not lost in the wave propagation process.
  • #1
roboticmehdi
34
0
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.
 
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  • #2
roboticmehdi said:
why do they say that electromagnetic field falls like 1/r ?
Can you provide a reference for this? It is hard to say one way or the other without knowing the details.
 
  • #3
DaleSpam said:
Can you provide a reference for this? It is hard to say one way or the other without knowing the details.

http://en.wikipedia.org/wiki/Larmor_formula there in the part ''Derivation 2: Using Edward M. Purcell approach'' it says stuff related to this.
 
  • #4
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/Liénard–Wiechert_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.
 
  • #5
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.
 

1. What is electromagnetic field strength?

Electromagnetic field strength refers to the intensity or magnitude of an electromagnetic field, which is a type of physical energy that is created by the movement of electrically charged particles.

2. How is electromagnetic field strength measured?

Electromagnetic field strength is typically measured in units of volts per meter (V/m) or tesla (T), depending on the type of field being measured. Specialized instruments such as gaussmeters or magnetometers are used to measure these fields.

3. What factors affect electromagnetic field strength?

The strength of an electromagnetic field is affected by several factors, including the distance from the source of the field, the type and amount of electrically charged particles involved, and the frequency or wavelength of the field.

4. How does electromagnetic field strength impact health?

There is ongoing research regarding the potential health effects of exposure to electromagnetic fields. Some studies have suggested a possible link between high levels of electromagnetic field exposure and certain health conditions, but more research is needed to fully understand the impact.

5. Can electromagnetic field strength be shielded or reduced?

Yes, electromagnetic field strength can be shielded or reduced by using materials such as lead or copper, which are known to block or absorb electromagnetic energy. Additionally, keeping a safe distance from sources of electromagnetic fields can also reduce exposure.

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