Magnetic field of a moving charge and coulombs force

AI Thread Summary
The discussion centers on the calculation of the magnetic field produced by a moving charge, specifically addressing the inclusion of the gamma factor (ϒ) from relativity in the equations. One participant questions whether the absence of the gamma factor in some sources indicates an approximation for low speeds, while others argue that it should be included for accurate results at relativistic speeds. The conversation highlights the complexity of the Liénard-Wiechert potential, which accounts for both acceleration and angle dependence in calculating electric and magnetic fields. Ultimately, there is a consensus that the magnetic field does not approach infinity as speed approaches the speed of light, and the field strength scales with the gamma factor. The participants continue to explore the implications of these findings on the behavior of electric and magnetic fields in relativistic contexts.
Hiero
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By relativity+coulombs force, I worked out that the (magnitude of the) magnetic field of a charge Q moving at speed V should be (ϒVV/c2)Q/(4πε0r2) or in terms of mu, ϒVμ0VQ/(4πr2) (where the distance r is measured normal to V).

When I google the answer I only find the same thing but without the gamma factor. Did I make a mistake or are they just using the non-relativistic approximation ϒ≈1?
 
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DuckAmuck said:
ϒV is the gamma factor of relativity associated with speed V

I saw that page when I googled the question; the result is almost the same as mine, but they are lacking the gamma factor.

So my question again: is their result exact and mine wrong, or is their result approximate (for V<< c) and mine correct?

@kuruman Thanks for the reply but it's a bit much; can't you just say yes or no as to if ϒ should appear in the answer?
 
The gamma factor should not be there. As speed of charge approaches c, current does not approach infinity, and magnetic field does not approach infinity.

I changed my mind:
The gamma factor can be there. If a charge passes a physicist at ultra-relativistic speed, the physicist measures: 1: A huge electric field 2: A huge magnetic field.

@Hiero does your Coulomb force approach infinity when speed approaches c?
 
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Hiero said:
@kuruman Thanks for the reply but it's a bit much; can't you just say yes or no as to if ϒ should appear in the answer?
If you look at the equation for ##\mathbf B(\mathbf r,t)## in section "Corresponding values of electric and magnetic fields" in the reference that I gave you, you will see that ##\gamma## appears together with a whole lot other stuff for a relativistically correct expression of what you seek. It may seem a bit much, but it is what it is.
 
The Lienard-Wierchert potential allows for an acceleration of the charge, and gives the angle dependence, which is why it is complicated.
If you remove the acceleration (##\dot{\beta} = 0##) and look at the charge perpendicular to the motion, it is dramatically simplified. Then ##\mathbf{\beta} \times \mathbf{n} = \beta, \mathbf{\beta} \cdot \mathbf{n} = 0##
##B=\frac{\mu_0}{4\pi} \frac{qv(1-\beta^2)}{r^2}=\frac{\mu_0}{4\pi} \frac{qv}{\gamma^2 r^2}##
 
Khashishi said:
The Lienard-Wierchert potential allows for an acceleration of the charge, and gives the angle dependence, which is why it is complicated.
If you remove the acceleration (##\dot{\beta} = 0##) and look at the charge perpendicular to the motion, it is dramatically simplified. Then ##\mathbf{\beta} \times \mathbf{n} = \beta, \mathbf{\beta} \cdot \mathbf{n} = 0##
##B=\frac{\mu_0}{4\pi} \frac{qv(1-\beta^2)}{r^2}=\frac{\mu_0}{4\pi} \frac{qv}{\gamma^2 r^2}##
Well, to this I would say that the gamma factor should not be there, because the magnetic field is not supposed to approach zero when the speed approaches c.
 
Hmm, the fields given in https://en.wikipedia.org/wiki/Liénard–Wiechert_potential don't seem to agree with the fields given here
http://farside.ph.utexas.edu/teaching/em/lectures/node125.html
Note that the latter source assumes a constant velocity.

I think Hiero's answer agrees with Fitzpatrick's for the case where ##u_r=0##, which is the case when viewing a charge where the ray from the apparent position to the viewer is perpendicular to the motion.

I'll have to mull this over. I believe the field strength scales as ##\gamma## as the charge passes by, but the time of the pulse scales as ##1/\gamma##.
The field lines are squashed into a pancake shape.
 
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