Question

A disk of radius \(a\) lies in the \(x y\) plane, with the \(z\) axis through its center. Surface charge of uniform density \(\rho_{s}\) lies on the disk, which rotates about the \(z\) axis at angular velocity \(\Omega \mathrm{rad} / \mathrm{s}\). Find \(\mathbf{H}\) at any point on the \(z\) axis.

Short answer

Answer: The magnetic field \(\mathbf{H}\) at any point on the \(z\)-axis due to the rotating charged disk is given by the expression: \(\mathbf{H}(z) = \frac{\rho_{s} \Omega z}{2\pi} \left(1 - \frac{a^2}{\sqrt{a^2 + z^2}}\right) \mathbf{\hat{z}}\).

Step-by-step solution

Use the Biot-Savart law to find magnetic field due to rotating disk

First, let's find the magnetic field \(\mathbf{H}_{d}\) produced by the rotating disk. To do this, we will use the Biot-Savart law: \(\mathbf{H}_{d}(\mathbf{r}) = \frac{1}{4 \pi} \int \frac{\rho_{s} \Omega \mathbf{v} \times \mathbf{dl}}{|\mathbf{r} - \mathbf{r'}|^3}\) where \(\mathbf{v}\) is the velocity of the charge at the element \(\mathbf{dl}\), and \(\mathbf{r'}\) is the position of the element, \(\mathbf{r}\) is the position where we want to find the magnetic field, and \(\mathbf{r} - \mathbf{r'}\) is the vector pointing from the element to the observation point. The integral is taken over the whole disk.

Simplify the expression

We are interested in finding the magnetic field on the \(z\) axis. So, let's choose the observation point \(\mathbf{r}=(0,0,z)\). The position of the charged element on the disk \(\mathbf{r'}=(x',y',0)\). We can express the velocity of the charged element \(\mathbf{v}\) in polar coordinates: \(\mathbf{v} = \Omega [-y' \mathbf{\hat{i}} + x' \mathbf{\hat{j}}]\) Now, let's replace this expression for \(\mathbf{v}\) into the Biot-Savart law expression: \(\mathbf{H}_{d}(z) = \frac{1}{4 \pi} \int \frac{\rho_{s} \Omega [-y' \mathbf{\hat{i}} + x' \mathbf{\hat{j}}] \times \mathbf{dl}}{[(x'^2 + y'^2 + z^2)]^{3/2}}\) We now need to convert this integral into polar coordinates: \(x' = r' \cos \theta'\) \(y' = r' \sin \theta'\) \(\mathbf{dl} = r' \mathbf{\hat{r}'} d\theta' + \mathbf{\hat{z}} dz'\) Where \(r'\) and \(\theta'\) are the polar coordinates of the charged element, and \(\mathbf{\hat{r}'}\) and \(\mathbf{\hat{z}}\) are the unit vectors in the polar coordinate system.

Simplify the integral

Now we substitute the polar coordinates expressions, and the Biot-Savart law becomes: \(\mathbf{H}_{d}(z) = \frac{\rho_{s} \Omega}{4 \pi} \int_{0}^{a} \int_{0}^{2 \pi} \frac{[-r' \sin \theta' \mathbf{\hat{i}} + r' \cos \theta' \mathbf{\hat{j}}] \times (r' \mathbf{\hat{r}'} d\theta' + \mathbf{\hat{z}} dz')}{[(r'^2 + z^2)]^{3/2}}\) Notice that due to symmetry, the \(x\)-component and \(y\)-component of \(\mathbf{H}_{d}(z)\) will cancel each other out. Also, \(dz'=0\) (since we're not integrating along the \(z\) direction), and thus only need to consider the \(z\)-component of \(\mathbf{H}_{d}(z)\). \(\mathbf{H}_{d,z}(z) = \frac{\rho_{s} \Omega}{4 \pi} \int_{0}^{a} \int_{0}^{2 \pi} \frac{(-r' \sin \theta' \mathbf{\hat{i}} + r' \cos \theta' \mathbf{\hat{j}}) \times (r' \mathbf{\hat{r}'} d\theta')}{(r'^2 + z^2)^{3/2}}\) Now, perform the cross product and integrate: \(\mathbf{H}_{d,z}(z) = \frac{\rho_{s} \Omega}{4 \pi} \int_{0}^{a} \int_{0}^{2 \pi} \frac{r'^2 \cos \theta' \mathbf{\hat{i}} \times \mathbf{\hat{r}'} + r'^2 \sin \theta' \mathbf{\hat{j}} \times \mathbf{\hat{r}'} }{(r'^2 + z^2)^{3/2}} d\theta' dr'\) After integrating with respect to \(\theta'\), \(\mathbf{H}_{d,z}(z) = \frac{\rho_{s} \Omega z}{2\pi} \int_{0}^{a} \frac{r'^2}{(r'^2 + z^2)^{3/2}} dr'\) Now, perform the remaining integral with respect to \(r'\): \(\mathbf{H}_{d,z}(z) = \frac{\rho_{s} \Omega z}{2\pi} \left(-\frac{a^2 z}{\sqrt{a^2+z^2}}+\frac{z}{\sqrt{z^2}}\right)\) Finally, we get the expression for the magnetic field on the \(z\)-axis: \(\mathbf{H}(z) = \frac{\rho_{s} \Omega z}{2\pi} \left(1 - \frac{a^2}{\sqrt{a^2 + z^2}}\right) \mathbf{\hat{z}}\) This is the magnetic field \(\mathbf{H}\) at any point on the \(z\)-axis due to the rotating charged disk.