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Gyromagnetic Ratio Question

  1. Nov 3, 2008 #1

    TFM

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    1. The problem statement, all variables and given/known data

    A thin uniform ring carrying a charge Q and mass M, rotates about its axis (see Figure).

    a) Find the ratio of its magnetic dipole moment to its angular momentum. This is called the gyromagnetic ratio.

    b) What is the gyromagnetic ratio for a uniform spinning sphere? (Decompose the sphere in infinitesimal rings and use the result of part a.)

    c) According to quantum mechanics, the angular momentum of a spinning electron is ℏ/2, where ℏ is Planck’s constant (over two pi). What is then the value of the electron’s magnetic dipole moment? (This is the semi-classical value which is actually off by a factor of almost exactly 2)

    2. Relevant equations

    Magnetic Dipole Moment:

    [tex] m = I\int dA = IA[/tex]

    Angular Momentum:

    [tex] L = I \omega [/tex]

    Moment of inertia of a loop:

    [tex] I = mr^2 [/tex]

    3. The attempt at a solution

    Just doing part a) for now:

    Angular Momentum for the ring is:

    [tex] L = mr^2 \omega [/tex]

    Mgnetic dipole moment formula is:

    [tex] m = I\int dA = IA[/tex]

    but what is the area for the loop? does it include the area inside the lop or not?

    If it does include the area inside the loop then it will be:

    [tex] m = I \pi r^2[/tex]

    But I don't think this is right.

    ???

    TFM
     
  2. jcsd
  3. Nov 5, 2008 #2
    Your magnetic dipole is correct. Now you just need to find to I, the current. It's a thin uniform ring spinning and carrying a charge, so what could you treat it as?
     
  4. Nov 5, 2008 #3

    TFM

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    Could you treat it as a ring that is standing still, with a current travelling through it? Possibly using

    [tex] I = \frac{dQ}{dt} [/tex]

    ?

    TFM
     
  5. Nov 5, 2008 #4
    Exactly. The only problem is that in this case dq/dt isn't very helpful. You can also find the current using this equation I [tex]=\lambda[/tex] v where [tex]\lambda[/tex] is the linear charge density and v is the velocity of the charge carriers.
     
  6. Nov 5, 2008 #5

    TFM

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    I am assuming you have to work out the charge density:

    [tex] \lambda = \frac{Q}{2 \pi r} [/tex]

    Also, the Straight speed is related to angular velocity:

    [tex] V = r^2 \omega [/tex]

    So:

    [tex] I = \frac{Q}{2 \pi r}(r^2 \omega) [/tex]

    ?

    TFM
     
  7. Nov 5, 2008 #6
    Thats right, except [tex]v=\omega * r[/tex]. If you go ahead and divide your equations you should get the answer.
     
  8. Nov 5, 2008 #7

    TFM

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    So:

    [tex] I = \frac{Q}{2 \pi r}(r \omega) [/tex]

    The Magnetic Dipole Moment is thus:

    [tex] M = \frac{AQ}{2 \pi r}(r \omega) [/tex]

    [tex] L = mr^2 \omega [/tex]

    so would the ratio of its magnetic dipole moment to its angular momentum be:

    M/L

    Giving:

    [tex] \frac{\frac{AQ}{2 \pi r}(r \omega)}{mr^2 \omega} [/tex]

    ???

    TFM
     
  9. Nov 5, 2008 #8
    Thats be right. Now substitue A for your [tex]\pi*r^{2}[/tex] that you had earlier and then simplify your answer. Hopefully most of the terms will drop out, leaving you with a less complex looking equation.
     
  10. Nov 5, 2008 #9

    TFM

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    So

    [tex] \frac{\frac{\pi*r^{2}Q}{2 \pi r}(r \omega)}{mr^2 \omega} [/tex]

    [tex] \frac{\frac{\pi*r^{2}Qr \omega}{2 \pi r}}{mr^2 \omega} [/tex]

    So cancelling down:

    [tex] \frac{\frac{1}{2}r^2Q \omega}{mr^2 \omega} [/tex]

    [tex] \frac{\frac{1}{2}Q }{m} [/tex]

    So gives

    [tex] \frac{Q}{2M} [/tex]

    Have I done something wring, this seems to have very few units involved

    ???

    TFM
     
  11. Nov 5, 2008 #10
    Yes, your answer is right. It surprised me too.
     
  12. Nov 5, 2008 #11

    TFM

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    Part B:

    b) What is the gyromagnetic ratio for a uniform spinning sphere? (Decompose the sphere in infinitesimal rings and use the result of part a.)


    So for part B, I have to integrate it over a circle

    so would that be:

    [tex] \int_0^{2\pi} \frac{Q}{2M} d\theta [/tex]

    ???

    TFM
     
  13. Nov 5, 2008 #12
    I'm not sure about this part, my 'Physics' calculus is pretty bad. Maybe someone else cold help? I don't think that integral is right though.
     
  14. Nov 6, 2008 #13

    TFM

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    Okay, fair enough.

    what about C) ?

    According to quantum mechanics, the angular momentum of a spinning electron is ℏ/2, where ℏ is Planck’s constant (over two pi). What is then the value of the electron’s magnetic dipole moment? (This is the semi-classical value which is actually off by a factor of almost exactly 2)

    For this, would you use the Gyromagnetic Ratio = M/L

    We know the ratio, and we are given L, to find M?

    ???

    TFM
     
  15. Nov 6, 2008 #14
    Thats right, part C) is just algebra.
     
  16. Nov 6, 2008 #15

    TFM

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    So:

    M = L*Ratio

    We need the answer from B), since an electron is represented by a sphere?

    I have an answer of [tex] \frac{ \pi Q}{m} [/tex]

    I'll use that unless anyone else says its wrong...

    so

    [tex] M = \frac{\hbar}{2} * \frac{ \pi Q}{m} [/tex]

    giving:

    [tex] M = \frac{\hbar \pi Q\piQ}{2m} [/tex]

    That seems easy...

    ???

    TFM
     
  17. Nov 6, 2008 #16
    Your value of L/M for a sphere should be the same as that for a ring. If you slice the sphere into infinitesimal rings of charge dq and mass dm and then sum them all together what do you get?
     
  18. Nov 6, 2008 #17
    I think this is it:

    the magnetic dipole moment is m=Ia, where a is the vector area enclosed by I. When you decompose the spherical shell into infinitesimal rings and then sum, this area will become just the volume of the sphere. The surface current density is just the charge times the velocity divided by the surface area of all the rings-which is just the surface area of a sphere. Put all this in to find m then divide by the angular momentum of a hollow sphere. This leads to q/2m.

    You can do the same for a solid sphere it's just harder.
     
  19. Nov 7, 2008 #18

    TFM

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    Surely the total charge will be q, the total mass m???


    TFM
     
  20. Nov 7, 2008 #19

    TFM

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    That makes sense, although it's strange that you still get the same answer, despite the fact that you have gone from a 2d ring to a 3d Hollow sphere...?

    TFM
     
  21. Nov 8, 2008 #20
    It's because the sphere has uniform charge q and mass m. When you divide it up each ring has caherge dq and mass dm. If you perform the calculation for the ring on each of the sub-rings and then sum all the sub-rings across the whole mass and charge you just get m and q;

    [tex]\int\int dq/(2*dm)=q/2m[/tex]
     
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