Torque of charged rotating disc in uniform magnetic field

[charlief]

Hi, I'm having an issue with finding a starting point for the following problem. I've looked through my magnetism and electrodynamic notes but was unable to find a starting point. Could someone please help direct me on the correct path? Thank you.

Homework Statement

A thin circular disc of radius $R$ rotates with angular speed $\omega$ about an axis through
the centre of the disc and perpendicular to the plane of the disc. A uniform magnetic field
$\mathbf{B}$ acts parallel to the plane of the disc. If the disc is uniformly charged with a surface
density $\sigma$, derive an expression for the magnitude of the torque on the disc, in terms of
$|\mathbf{B}|$, $R$, $\omega$ and $\sigma$.

 

[Asim Riaz]

Homework Equations The relevant equations for this problem are the equation for the Lorentz Force $$\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B})$$ and the equation for the torque $$\mathbf{\tau} = \mathbf{r} \times \mathbf{F}$$ The Attempt at a Solution The first step in solving this problem is to calculate the electric field due to the uniformly charged disc. Since the charge is distributed uniformly over the surface of the disc, we can use Gauss's Law to find the electric field:$$\oint \mathbf{E}\cdot d\mathbf{A} = \frac{Q}{ \epsilon_0}$$where Q is the total charge on the disc and $\epsilon_0$ is the permittivity of free space. We can rewrite this equation as $$\int_0^{2\pi}\int_0^R E \cdot r \,dr \,d\theta = \frac{Q}{\epsilon_0}$$From here, we can solve for the magnitude of the electric field, which is given by $$E = \frac{\sigma R}{2\epsilon_0}$$ where $\sigma$ is the surface density of the charge.Next, we can calculate the force on the disc by using the Lorentz Force equation. Since the magnetic field is acting parallel to the plane of the disc, the cross product term will be zero. Thus, the force is just $$\mathbf{F} = q(\mathbf{E}) = q\left(\frac{\sigma R}{2\epsilon_0}\right)$$Finally, we can calculate the torque on the disc by using the equation for torque $$\mathbf{\tau} = \mathbf{r} \times \mathbf{F} = \frac{q\sigma R^2}{2\epsilon_0}\,\hat{\mathbf{z}}$$ where ##\hat{\mathbf{z