Calculating the magnetic moment of an electron

  • Thread starter PsychoDash
  • Start date
  • #1
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Homework Statement



Assume that an electron is a sphere of uniform mass density
[itex]\rho_m=\frac{m_e}{\frac{4}{3} \pi r_e^3}[/itex], uniform charge
density [itex]\rho_e=\frac{-e}{\frac{4}{3} \pi r_e^3}[/itex], and
radius [itex]r_e[/itex] rotating at a frequency [itex]\omega[/itex]
about the z-axis. [tex]m_e=9.109*10^{-31}[/tex] kg and
[tex]e=1.602*10^{-19}[/tex] C

Using the formula [tex]\vec{m}=\frac{1}{2} \int \vec{r} \times
\vec{J(\vec{r})} d\tau[/tex], compute the magnetic moment of this
electron. Your answer should depend on e, [tex]\omega[/tex] and
[tex]r_e[/tex]

Homework Equations



Given above.

The Attempt at a Solution



Ok, so I know that in general, [tex]\vec {J}=\rho_e \vec {v}[/tex]. I'm not sure how to proceed from here since writing J in terms of omega yields [tex]\vec {J}=\frac {\rho_e \omega}{\vec {r}}[/tex]. I've always heard that dividing by a vector is not strictly defined in a math sense. Either I'm not approaching this in the right way, or putting that funkiness into the cross product above yields some magnificence that I am, as of now, incapable of seeing.

Help will be greatly appreciated.
 

Answers and Replies

  • #2
322
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yeah integrating vectors gets really really hairy, when i get home i will help you out. There is a way to define some new quantity "C" per-say and it makes it easier

Unless someone helps you first
 
  • #3
16
1
The assignment has already been turned in. I ended up just not using the given formula. Got the correct answer, even if not really in the correct way. Thanks though.
 

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