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In magnetostatic the continuity equation is:

$$\frac {\partial \rho}{\partial t} + \nabla \vec j(\vec r)=0$$.

In my script the following is said:

Since we are in magnetostatic $\frac {\partial \rho}{\partial t} =0$ and therefore $\nabla \vec j(\vec r)=0$. Now I understand that this $\nabla \vec j(\vec r)=0$ means that in the medium we are observing there are no sources of the electric current density. But at the same time, since a current density is present, there should be a region in which a charge is changing over time. If the current density is going towards a region that means we are having a sink there, and if it is going out of a place we are having a source. How would we have current without a change in the value of some charge over time? What am I understanding incorrectly ?

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    $\begingroup$ The current flows in a closed loop. $\endgroup$
    – mike stone
    Nov 26 '21 at 22:28
  • $\begingroup$ So the charged particles leave and return to the same location? $\endgroup$
    – imbAF
    Nov 26 '21 at 22:32
  • $\begingroup$ Yes. usually a battery or generator, but in a superconductor you can have a closed loop without any power supply. $\endgroup$
    – mike stone
    Nov 26 '21 at 23:15
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The presence of a current does mean charge is moving around, but not necessarily that the charge density is changing anywhere. If $\nabla \cdot \vec{j} \neq \vec{0}$, then we'd have sinks and/or sources. Since it is zero, all we have is charge flowing around with more charge coming in to replace it.

Imagine, for example, there is an infinite line of people walking at a constant speed. Of course, everyone is changing places all the time. However, the density of people is always the same, for whenever someone walks forward, the person behind fills in the space that would be left open. It is the very same idea. While charges are moving around, there is always some more charge to fill in the places that would be left open, and charges are always leaving a spot where another charge is entering. Hence, charge does move around with the current, but no sources or sinks ever occur.

A similar analogy is to observe the flow of water on a bathtub, for example. While water can move around, it has no sinks nor sources.

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  • $\begingroup$ Can you give me an example of a physical object where we see this whirlpool-like movement of charges? $\endgroup$
    – imbAF
    Nov 26 '21 at 22:50
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    $\begingroup$ @imbAF Pick an uniformly charged, rotating sphere, for example. You have a current, since the charges are rotating with the sphere, but the charge density is always constant. $\endgroup$ Nov 26 '21 at 22:55
  • $\begingroup$ Another possibility, though a less physical one, is a constant current, such as $\vec{j} = a \hat{x}$. All of the charges are going to the right (let's assume $a > 0$ for simplicity), but the charge density is constant, because there is always another charge coming in from the left to fill in the gap. In both examples, we get $\nabla \cdot \vec{j} = 0$ $\endgroup$ Nov 26 '21 at 22:56
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If you let $\rho$ be the density of water instead of charge then an example would be a whirlpool. The density of water is constant throughout the whirlpool but the water itself is still moving. The same holds true for your case.

Note that we made a continuum approximation here. Similar to how fluid density is also an approximation; the fluid is actually made up of small particles. If you were to look at individual charges moving then you would see the density changing rapidly but if you have enough charges and if you squint a bit you can approximate this as a density field.

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  • $\begingroup$ I don't understand the 2nd part of your comment. Why would I see the density changing if I am observing the individual charges? Is there a difference between the charge density in a macroscopic sense and in microscopic sense? I don't know what you mean with density field $\endgroup$
    – imbAF
    Nov 26 '21 at 22:48
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    $\begingroup$ Imagine you have a large number of equally spaced charges moving along a wire in a straight line. Each charge approximately produces a coulomb potential and because they are moving you would see the electric field (and magnetic field) change rapidly near the wire. The continuum approximation is then to not consider the individual charges but to consider a charge density $\rho(x)$ along the wire. See also the link from my last edit. Then the electric field would no longer be changing if the density is constant. Remember that a field is just any quantity that depends on position. $\endgroup$ Nov 26 '21 at 22:58
  • $\begingroup$ I see. That was a nice short explanation. Even though it helps me understand more, at the same time I have more questions. But this is a nice start. One more question though. Should the charged objects accelerate in order to have an EM-wave or it's not necessary ? Even by simply moving with constant velocity, we can have an EM-wave $\endgroup$
    – imbAF
    Nov 26 '21 at 23:03
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    $\begingroup$ Glad it helped :) . And charges should definitely accelerate to produce EM waves (that carry energy at least). A charge that is moving at constant velocity can also be viewed in a frame of reference where it is stationary. If it doesn't radiate energy in its stationary frame it also shouldn't radiate in any other frame of reference. Remember that only non-accelerating frames are proper frames of reference. Electromagnetic fields transform a bit complicated when you change your frame of reference and to understand it fully you would need some special relativity. $\endgroup$ Nov 26 '21 at 23:10

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