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Homework Help: Spinning iron wire

  1. Aug 20, 2013 #1
    1. The problem statement, all variables and given/known data

    Iron atoms (atomic mass 56) contain two free electron spins that can align with an external magnetic field. An iron wire 3 cm long and 1 mm in diameter is suspended vertically and is free to rotate about its axis. A strong magnetic field parallel to the wire's axis is applied. How large is the resulting change in its angular velocity.

    2. Relevant equations

    [tex] \oint \vec{E} \cdot \vec{dl} = - \frac{d}{dt} \int \vec{B} \cdot \vec{da} [/tex]

    Poynting vector: [itex] \vec{S} = \mu_0 \vec{E} \times \vec{B} [/itex].

    Bohr's magneton might come in handy: [itex] \mu_b = \frac{e \hbar}{2 m_e} [/itex]

    3. The attempt at a solution

    My first thought was to find the Poynting vector because the angular momentum contained in the fields is proportional to [itex] \vec{r} \times \vec{S} [/itex]. Assuming the wire is in the z direction, we can write that the applied magnetic field is [itex] \vec{B} = B \hat{z} [/itex]. This would create a magnetic flux through the x-y plane, and hence create an electric field in the [itex] \hat{\phi} [/itex] direction from the Faraday law. But that means that [itex] \vec{S} \approx \hat{\phi} \times \hat{z} = \hat{r} [/itex] and therefore there would be no angular momentum change in the fields.

    I feel this is too simplistic and possibly wrong, especially since we didn't use any of iron's properties.
  2. jcsd
  3. Aug 20, 2013 #2
    So as always I overcomplicate stuff. It is easy to just calculate L = N hbar/2, where N is the number of spin, given the density of iron. L = I w, where I is the moment of inertia. Super simple. Ugh.
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