Solving Spinning Iron Wire Angular Velocity Change

[monnomestalex]

Homework Statement

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.

Homework Equations

$$\oint \vec{E} \cdot \vec{dl} = - \frac{d}{dt} \int \vec{B} \cdot \vec{da}$$

Poynting vector: $\vec{S} = \mu_0 \vec{E} \times \vec{B}$.

Bohr's magneton might come in handy: $\mu_b = \frac{e \hbar}{2 m_e}$

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 $\vec{r} \times \vec{S}$. Assuming the wire is in the z direction, we can write that the applied magnetic field is $\vec{B} = B \hat{z}$. This would create a magnetic flux through the x-y plane, and hence create an electric field in the $\hat{\phi}$ direction from the Faraday law. But that means that $\vec{S} \approx \hat{\phi} \times \hat{z} = \hat{r}$ 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.

 

[monnomestalex]

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.