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For homework I have the following question, but I am stuck on finding the torque on the cylinder.

An infinite cylinder of radius $R$ carries a uniform surface charge $\sigma$. We start rotating the cylinder to a final rotating speed of $\omega_{f}$. How much work will this take per unit length? Do it in the following two ways:

a) Find the magnetic field and the induced electric field (quasistatic approximation). Then calculate the torque you must exert and from that obtain the work done per unit length. ($W=\int N d\theta$)

b) Use the energy of the final magnetic field to find the energy per unit length stored in the field.

So far I have been able to calculate the magnetic and induced electric field:

$$\oint\vec{B}\cdot d\vec{l}=\mu_{0}I_{encl}\Longrightarrow \vec{B}(t)=\mu_{0}\sigma R\omega(t)\hat{z}$$

Which gives rise to flux $\Phi=\mu_{0}\pi r^{2}\sigma R \omega(t)$ for $r<R$ and $\Phi=\mu_{0}\pi R^{3}\sigma \omega(t)$ for $r\geq R$. Then I can calculate the induced electric field with:

$$\oint\vec{E}\cdot d\vec{l}=-\frac{d\Phi}{dt}\Longrightarrow \vec{E}= \left\{ \begin{array}{ll} -\frac{1}{2}\mu_{0}r\sigma R\frac{d\omega}{dt}\hat{\theta} & \mbox{if } r<R \\ -\frac{\mu_{0}\sigma R^{3}}{2r}\frac{d\omega}{dt}\hat{\theta} & \mbox{if } r > R \end{array} \right.$$

But I am not sure how to find the torque on the cylinder from this. My guess is that $dF=dqE$ with $dq=\sigma Rd\theta dz$ and just forget about $dz$ because we are interested in the answer per unit length. But everything I try I always have the factor $\frac{d\omega}{dt}$ which should not be in the final answer because one can calculate the energy stored in the field per unit length easily by (which is also the answer to b):

$$W=\frac{1}{2\mu_{0}}\int^{2\pi}_{0}\int^{R}_{0}(\mu_{0}\sigma R \omega_{f})^{2}rdrd\theta=\frac{1}{2}\pi \mu_{0} \sigma^{2} R^{4} \omega_{f}^{2}$$

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1 Answer 1

You'll need to first calculate the Poynting vector or the momentum density of the fields. The angular momentum density is L= r x p. Once you have that, you can find torque easily.

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This would be a fine comment. It's not an answer. –  Brandon Enright Dec 15 at 7:18
    
The question asks the OP to do the problem a specific way. This may be an alternative that may be relevant, if you flesh it out a bit more, but I think one pretty much needs to work things out as asked in the question before one finds the momentum density and that which follows from it. –  WetSavannaAnimal aka Rod Vance Dec 15 at 10:41

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