# Torque per unit length on infinite rotating charged cylinder

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):

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

I had exactly the same problem as you and just solved it. We are interested in the force the electric field exerts on the surface charge which - as you correctly pointed out - is given by $$\vec{F} = q\vec{E}_{surface} = \sigma 2\pi Rl(-\frac{1}{2}\mu_0\sigma R^2 \frac{d\omega}{dt}\hat{\phi} = -\mu_0\pi l\sigma^2 R^3\frac{d\omega}{dt}\hat{\phi}$$ where the electric field is evaluated at the surface since this is where the charge is. The torque on the cylinder is, by definition: $$\vec{N} = \vec{r}\times\vec{F} = -\mu_0\pi l\sigma^2 R^3\frac{d\omega}{dt}\sqrt{R^2+z^2}(-\hat{\theta})$$ where, on the cylindrical surface $\vec{r} = \sqrt{R^2+z^2}\hat{r}$ and $\hat{r}\times\hat{\phi} = -\hat{\theta} = -\cos\theta(\cos\phi\hat{x} + \sin\phi\hat{y})+\sin\theta\hat{z}$. Now, we are going to integrate the above expression over $\phi$ from $0$ to $2\pi$ to get the work, only the $\hat{z}$ component of $\hat{\theta}$, namely $\sin\theta = \frac{R}{\sqrt{R^2+z^2}}$, will remain nonzero. Thus, $$N_z = -\mu_0\pi l\sigma^2 R^3\frac{d\omega}{dt}\sqrt{R^2+z^2}\frac{R}{\sqrt{R^2+z^2}} = -\mu_0\pi l\sigma^2 R^4\frac{d\omega}{dt}$$ Finally, we are ready to get the torque. Noting that $\frac{d\phi}{dt} = \omega$ by definition, and that we must exert a torque opposite to the field, $$W = \int_0^{2\pi} -N d\phi = \int_0^{2\pi} -N_z d\phi = \int_0^{2\pi}\mu_0\pi l\sigma^2 R^4\frac{d\omega}{dt} d\phi = \mu_0\pi l\sigma^2 R^4\int_0^{2\pi} d\omega\frac{d\phi}{dt} = \mu_0\pi l\sigma^2 R^4\int_0^{\omega_f} \omega d\omega = \frac{1}{2} \mu_0\pi l\sigma^2 R^4 \omega_f^2$$ just as we recieved from the more straight-forward manner in (b). I just want to note that since $N = I\frac{d\omega}{dt}$ by Newton's second law applied to rotation, this cylinder's motion can be viewed as that of a massive body with moment of inertia $I = \mu_0\pi l\sigma^2 R^4$. The kinetic energy of such a body is $KE = \frac{1}{2}I\omega_f^2 = \frac{1}{2}\mu_0\pi l\sigma^2 R^4\omega_f^2$ in agreement with the result from electrodynamics.