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Homework Help: Magnetic moments

  1. Aug 7, 2008 #1
    1. The problem statement, all variables and given/known data

    A system of spin 1/2 magnet moments with μ = 1*10^-23 Am² are in a magnetic field of 0.2T. At what temperature will 90% of the magnets be aligned parallel to B?

    3. The attempt at a solution
    I think I have found the relevant equation:

    U=-μB(N↑ - N↓)

    where U is the magnetic potential energy, μ is the magnetic moment, B is the magnetic field, N↑ is the number of magnets oritentated up and N↓ is the number of magnets orientated down. My questions now are, how do I know which direction is parallel to B and if this equation is relevant, how do I then convert U to temperature?
     
  2. jcsd
  3. Aug 8, 2008 #2
    From the way the question is phrased, I would believe the B-field is oriented in the "up" direction; this is because in general the energy of a magnetic dipole in a magnetic field is:

    [tex]-\mu \cdot B[/tex]

    so the dipole will want to align with the field to get to the lowest energy. In this case, lower energy is achieved if B is pointing "up" rather than "down".

    This much I know, the rest I'm not 100% sure, but I think the following would work:

    Suppose we have N such spins, then we have [tex] N = N\uparrow - N\downarrow [/tex]
    Let [tex] 2 S = N\uparrow - N\downarrow [/tex] be the spin difference. With this we can solve for
    [tex] N\uparrow = S + \frac{N}{2} [/tex]

    and get

    [tex] \frac{N\uparrow}{N} = \frac{S}{N} + \frac{1}{2} [/tex]

    Taking the average:


    [tex] \frac{\left\langle N\uparrow \right\rangle}{N} = \frac{\left\langle S \right\rangle}{N} + \frac{1}{2} [/tex]

    We would like this ratio to be 90%.


    For one single magnet, the partition function is given by

    [tex] Z_{1} = 2 cosh (\beta\mu B) [/tex].

    where [tex] \beta = \frac{1}{k_b T} [/tex]

    For N such magnets, our partition function will be

    [tex] Z = \left(Z_{1}\right)^{N} [/tex]

    The average energy is given by:

    [tex] \left\langle E \right\rangle = -\frac{1}{Z}\frac{\partial Z}{\partial\beta} [/tex]

    But we also know that [tex] \left\langle E \right\rangle = - \mu B \left\langle S \right\rangle [/tex]

    so now we can solve for the temperature "beta" at which our desired ratio is reached.
     
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