Magnetic field of a moving charge and coulombs force

[Hiero]

By relativity+coulombs force, I worked out that the (magnitude of the) magnetic field of a charge Q moving at speed V should be (ϒ_VV/c²)Q/(4πε₀r²) or in terms of mu, ϒ_Vμ₀VQ/(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?

 

[DuckAmuck]

What is ϒ?

Check out this page: http://academic.mu.edu/phys/matthysd/web004/l0220.htm

 

[kuruman]

Perhaps this might be of help.
https://en.wikipedia.org/wiki/Liénard–Wiechert_potential

 

[Hiero]

What is ϒ?

Check out this page: http://academic.mu.edu/phys/matthysd/web004/l0220.htm

ϒ_V 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?

 

[jartsa]

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?

 

[kuruman]

@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.

 

[Khashishi]

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 $\mathbf{\beta} \times \mathbf{n} = \beta, \mathbf{\beta} \cdot \mathbf{n} = 0$
$B=\frac{\mu_0}{4\pi} \frac{qv(1-\beta^2)}{r^2}=\frac{\mu_0}{4\pi} \frac{qv}{\gamma^2 r^2}$

 

[jartsa]

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.

 

[Khashishi]

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.