I Why -e(v/c)H is the magnetic force?

  • Thread starter Haorong Wu
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Why ##-e \left ( v/c \right ) H_0## is the magnetic force?
In a paper I am reading, it reads, "For these orbits the electric force ##-e E_r## almost balances the magnetic force ##-e \left ( v/c \right ) H_0##." where ##-e## is the charge of the electrons, ##v## is the speed of the electrons, ##H_0## is a homogeneous magnetic field, and ##c## is not clearly indicated, but I guess it is the speed of light.

However, according to Lorentz force, ##F=-e v B= -e v \mu _0 H##. I do not see any ##c## or ##\epsilon_0##, so why the paper says ##-e \left ( v/c \right ) H_0## is the magnetic force?

Thanks.

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From the following content, it seems that ##B=\frac {H_0} c##, which I am not familiar with, so what is ##H_0##?
 
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gleem

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The presence of c in the definition of magnetic force is probably due to the use of cgs instead of the the more familiar MKS system. That is the charge is measured in statcoulombs instead of coulombs
 

vanhees71

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It's ##\vec{B}##, not ##\vec{H}## which has to be used in the Lorentz force,
$$\vec{F}=q \left (\vec{E} + \frac{\vec{v}}{c} \times \vec{B} \right).$$
This is valid in Gaussian as well as Heaviside-Lorentz units.

In the SI the eq. reads
$$\vec{F}=Q (\vec{E} + \vec{v} \times \vec{B}).$$
There it's the more important to write correctly ##\vec{B}=\mu_0 \vec{H}## (in vacuo, with ##\mu_0## the permeability of the vacuum, which occurs in the SI as a conversion factor of units).
 

Vanadium 50

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It's B not H which has to be used in the Lorentz force
Are you sure? Does not a charged particle in a region with B = 0 but M <> 0 deflect?
 
The presence of c in the definition of magnetic force is probably due to the use of cgs instead of the the more familiar MKS system. That is the charge is measured in statcoulombs instead of coulombs
Thanks, gleem. You are right. The author used cgs system which I am not familiar with and it is a paper in 1960s so H may mean B in that time.

And thanks to other friends.
 

Vanadium 50

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Good point.
 

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