Why does magnetic field strength decrease faster than gravity at a distance?

[Proof.Beh]

why does magnetic field strength decrease much faster than the inverse square of the distance like gravity? Is it because the force lines have to curve back to the source? Is magnetism also carried by an "imaginary" particle?

Thanks.
Mr Beh

 

[Dick]

Your first guess is basically correct. In the absence of monopoles the force law of magnetism is (at best) that of a dipole. I have no idea what an "imaginary" particle would mean.

 

[Proof.Beh]

If we test that around of Earth or a larger mass, space-time can help to force lines that curve to other sides according to GR?

 

[adriansd]

I think he means 'virtual partice', like in a Feynman diagram?

 

[Dick]

If we test that around of Earth or a larger mass, space-time can help to force lines that curve to other sides according to GR?

What does this have to do with the original question? There is no fundamental difference between electrical and magnetic forces except for the lack of monopoles in the latter case. GR does distort space and certainly affects the solution to EM equations. But has little to do with the original post.

 

[Hurkyl]

Wait a minute; the magnetic field does obey an inverse square law...

 

[Dick]

Wait a minute; the magnetic field does obey an inverse square law...

Yes, it does. From a monopole. And from a dipole but you have to sum the inverse squares. Which subtract to give an inverse cube. I hope that is what the poster is referring to.

 

[Proof.Beh]

Hey guys, you didn't answer to elementary question. I am saying, if forces lines curve in a magnetic field then can we say that it is stronger than the strength space-time curvature at curving light around a large mass (for example sun) or not?

Of course I agree with adriansd, but what is it really? Does it identical with 'virtual partice' In a Feynman diagram?

Thanks.
Mr Beh

 

[Hurkyl]

Yes, it does. From a monopole. And from a dipole but you have to sum the inverse squares. Which subtract to give an inverse cube. I hope that is what the poster is referring to.

I was referring to the fact that the magnetic field generated by a current is proportional to the inverse square of the distance from the source.

 

[Dick]

I was referring to the fact that the magnetic field generated by a current is proportional to the inverse square of the distance from the source.

Oh yeah, that. But that's a infinite linear source. The E field from an infinite line charge isn't inverse square either (even though the underlying point force law is). It doesn't seem to really matter though since the poster doesn't seem to be able to make a clear statement of what he does mean.

 

[Hurkyl]

Oh yeah, that. But that's a infinite linear source. The E field from an infinite line charge isn't inverse square either (even though the underlying point force law is). It doesn't seem to really matter though since the poster doesn't seem to be able to make a clear statement of what he does mean.

The magnetostatic field from an infinite wire is inverse-linear, just like the electrostatic field; they're essentially the same calculation! The only difference is that one has q\hat{r} in the numerator, and the other has \vec{I} \times \hat{r}. (And there's probably another constant factor I'm forgetting)