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When charge moves at a constant speed, it is said that it produces static magnetic and electric field.

Why static ? If it moves, at every points, electric field and magnetic field change as they are dependent on a distance to the point. So even if charge moves at constant speed, we still should have electric and magnetic fields changing in every point in space.

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Indeed, neither the electric nor the magnetic fields are static in the case of a charge moving with constant velocity. However, the fields, although they are not static, they also do not radiate.

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  • $\begingroup$ So i am correct that they are not static as at every point, they change due to distance increasing or decreasing. If so we got a time varying electric and magnetic field even if charge moves at constant velocity. If so, why not radiate ? I am asking because they explain em radiation is as changing electric and magnetic field which we also got for constant speed charge $\endgroup$
    – Giorgi Lagidze
    Commented May 30, 2023 at 12:04
  • $\begingroup$ @GiorgiLagidze it doesn’t radiate because the fields point in the wrong direction for radiation. Energy flows perpendicular to both the E and B fields. For a uniformly moving charge that direction is “forward” not “outward” $\endgroup$
    – Dale
    Commented May 30, 2023 at 12:12
  • $\begingroup$ Probably better asked question about this can be here: physics.stackexchange.com/questions/766068/… would you mind answering ? $\endgroup$
    – Giorgi Lagidze
    Commented May 30, 2023 at 12:16
  • $\begingroup$ I can see you commented there and I updated what I meant by kink $\endgroup$
    – Giorgi Lagidze
    Commented May 30, 2023 at 12:16
  • $\begingroup$ @GiorgiLagidze sorry, I don’t know much about these kinks. I think this question here is much better than those. I don’t think I can help much regarding the other question $\endgroup$
    – Dale
    Commented May 30, 2023 at 12:56
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Change does not mean radiation, to get radiation a necessary condition is that the field components vary as $\mathcal O(r^{-1})$, Then $S=E\times B$ will be $\mathcal O(r^{-2})$ which will give a nonzero value when multiplied by $4\pi r^2$, a surface area far away. This cannot happen if there is no work done on the source, see, energy conservation. A charge moving with a constant velocity all by itself is neither doing work nor having work done on.

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