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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.

Giorgi Lagidze
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2 Answers2

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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.

Dale
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  • 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 – Giorgi Lagidze May 30 '23 at 12:04
  • @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” – Dale May 30 '23 at 12:12
  • Probably better asked question about this can be here: https://physics.stackexchange.com/questions/766068/why-are-not-kinks-produced-for-constant-speed-charge would you mind answering ? – Giorgi Lagidze May 30 '23 at 12:16
  • I can see you commented there and I updated what I meant by kink – Giorgi Lagidze May 30 '23 at 12:16
  • @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 – Dale May 30 '23 at 12:56
  • The thing is i could not understand then what you meant by field pointing in the wrong direction then – Giorgi Lagidze May 30 '23 at 13:30
  • @GiorgiLagidze are you familiar with Coulomb’s law, the Biot Savart law, and Poynting’s theorem? Alternatively Coulomb’s law, the Lorentz transform, and Poynting’s theorem. Or even better would be the Lienard Wiechert fields, and Poynting’s theorem – Dale May 30 '23 at 15:00
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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.

hyportnex
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