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Particle movement in inhomogeneous magnetic field

  • Thread starter Dishsoap
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Homework Statement

Show that for the case of a general inhomogeneous magnetic field, $$\dot{\vec{v}}=\frac{e}{2mc} (\vec{v} \times \vec{B} - \vec{B} \times {v})$$

The attempt at a solution

I think I am oversimplifying things. I used that, for an electron in a magnetic field, [itex]m \frac{d \vec{v}}{dt}=e \vec{v} \times \vec{B}[/itex], and that [itex]\vec{v} \times \vec{B} = - \vec{B} \times \vec{v} [/itex]

Doing this, I find that [itex]RHS = \frac{1}{c} \dot{\vec{v}}[/itex]
 

TSny

Homework Helper
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The presence of ##c## in the equation is just due to the choice of units.

May I ask where this problem came from?
 
1,001
299
The presence of ##c## in the equation is just due to the choice of units.

May I ask where this problem came from?
I figured as such, but units for what? B? v? e?

This problem was not out of a book but was just on a homework sheet, and I was unable to find it elsewhere.
 

TSny

Homework Helper
Gold Member
12,046
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I figured as such, but units for what? B? v? e?
Compare the Gaussian system of units with the SI units here: https://en.wikipedia.org/wiki/Gaussian_units#Maxwell.27s_equations
The units for B, v, and e are all different in the two systems: https://en.wikipedia.org/wiki/Gaussian_units#Electromagnetic_unit_names

This problem was not out of a book but was just on a homework sheet, and I was unable to find it elsewhere.
OK. It seems odd since, as you say, you can always just rewrite ##\bf{B} \times \bf{v}## as ##-\bf{v} \times \bf{B}## .
 

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