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Why magnetic field from a current carrying conductor obey inverse-square law?

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NANDHU001
#1
Nov14-12, 11:46 AM
P: 22
I have read that the electric field from a point charge fall off as 1/(r*r) since it is analogous to
variation of intensity of radation from source (whose geometric proof depends on solid-angle), similarily is there any geometric explanation why magnetic field in the stated case fall off as 1/(r*r).
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Enthalpy
#2
Nov14-12, 02:05 PM
P: 661
Bizarre "proof" that the static electric field is analoguous to a radiation... What about the gluon force? It increases over distance. What tells the previous reason in this case?

As for the static magnetic field... It cannot decrease as 1/R^2 because this would need a permanent current in an open wire. Either it's static, and then you need to close the circuit, and this loop creates a field as 1/R^3, or you have an antenna which accepts only AC current, and radiates an electromagnetic field, not a static magnetic one.

So 1/R^2 exists only as a computation intermediate of static magnetic fields.
nasu
#3
Nov14-12, 02:07 PM
P: 1,970
Quote Quote by NANDHU001 View Post
similarily is there any geometric explanation why magnetic field in the stated case fall off as 1/(r*r).
It doesn't, does it?
The magnetic field of an infinite, linear conductor goes like 1/r where r is the distance from the wire (along the radius of a cylinder coaxial with the wire).
Maybe you mean a different geometry of the current carrying conductor?


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