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Say we are dealing with a wire that has a current $I$ flowing through it, i.e. $I$ is the free current. Why must it then hold that the net bound current, that is, the bound volume current, $J_b$, and the bound surface current, $K_b$, adds up to zero?

I cannot get an intuitive grasp on why this is true, especially considering the fact that if you, on the other hand, have a uniformly magnetized cylinder, the net bound currents do not add up to zero, since there's no bound volume current, but there is a bound surface current. Similarly, in a solenoid you also get a bound surface current, but there's no bound volume current, so the two again do not add up to zero.

Any help with this would be greatly appreciated, since I'm just stuck in trying to wrap my head around this. I can calculate the damn things, but I have no idea why this would be the case on a qualitative level.

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" Why must it then hold that the net bound current, that is, the bound volume current, Jb, and the bound surface current, Kb, adds up to zero?" Where did you read this? Please specify author, name of the book and the section/page. –  Ján Lalinský Jul 20 at 11:55

1 Answer 1

According to Ampere's Circuital Law, ∫B.dl=μi. [i.e. circular integral of product of magnetic field along imaginary circulating path and path length is equal to permeability of medium multiplied with current flowing through the imaginary cross section made by circulating path]

So, ∑i must be zero, hence net current comes zero, i.e.Jb+Kb=0. This implies that both are mutually opposite in direction. enter image description here

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