Question

A disk of radius $a$ lies in the $xy$ 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 $\mathrm{\Omega }\mathrm{rad}/\mathrm{s}$. 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: $\mathbf{H}(z)=\frac{{\rho }_s\mathrm{\Omega }z}{2\pi }(1-\frac{a^2}{\sqrt{a^2+z^2}})\hat{\mathbf{z}}$.

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: ${\mathbf{H}}_d(\mathbf{r})=\frac{1}{4\pi }\int \frac{{\rho }_s\mathrm{\Omega }\mathbf{v}\times \mathbf{dl}}{|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3}$ where $\mathbf{v}$ is the velocity of the charge at the element $\mathbf{dl}$, and ${\mathbf{r}}^{\prime }$ is the position of the element, $\mathbf{r}$ is the position where we want to find the magnetic field, and $\mathbf{r}-{\mathbf{r}}^{\prime }$ 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 ${\mathbf{r}}^{\prime }=(x^{\prime },y^{\prime },0)$. We can express the velocity of the charged element $\mathbf{v}$ in polar coordinates: $\mathbf{v}=\mathrm{\Omega }[-y^{\prime }\hat{\mathbf{i}}+x^{\prime }\hat{\mathbf{j}}]$ Now, let's replace this expression for $\mathbf{v}$ into the Biot-Savart law expression: ${\mathbf{H}}_d(z)=\frac{1}{4\pi }\int \frac{{\rho }_s\mathrm{\Omega }[-y^{\prime }\hat{\mathbf{i}}+x^{\prime }\hat{\mathbf{j}}]\times \mathbf{dl}}{[(x^{\prime 2}+y^{\prime 2}+z^2){]}^{3/2}}$ We now need to convert this integral into polar coordinates: $x^{\prime }=r^{\prime }\cos {\theta }^{\prime }$ $y^{\prime }=r^{\prime }\sin {\theta }^{\prime }$ $\mathbf{dl}=r^{\prime }{\hat{\mathbf{r}}}^{\prime }d{\theta }^{\prime }+\hat{\mathbf{z}}d{z}^{\prime }$ Where $r^{\prime }$ and ${\theta }^{\prime }$ are the polar coordinates of the charged element, and ${\hat{\mathbf{r}}}^{\prime }$ and $\hat{\mathbf{z}}$ 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: ${\mathbf{H}}_d(z)=\frac{{\rho }_s\mathrm{\Omega }}{4\pi }{\int }_0^a{\int }_0^{2\pi }\frac{[-r^{\prime }\sin {\theta }^{\prime }\hat{\mathbf{i}}+r^{\prime }\cos {\theta }^{\prime }\hat{\mathbf{j}}]\times (r^{\prime }{\hat{\mathbf{r}}}^{\prime }d{\theta }^{\prime }+\hat{\mathbf{z}}d{z}^{\prime })}{[(r^{\prime 2}+z^2){]}^{3/2}}$ Notice that due to symmetry, the $x$-component and $y$-component of ${\mathbf{H}}_d(z)$ will cancel each other out. Also, $d{z}^{\prime }=0$ (since we're not integrating along the $z$ direction), and thus only need to consider the $z$-component of ${\mathbf{H}}_d(z)$. ${\mathbf{H}}_{d,z}(z)=\frac{{\rho }_s\mathrm{\Omega }}{4\pi }{\int }_0^a{\int }_0^{2\pi }\frac{(-r^{\prime }\sin {\theta }^{\prime }\hat{\mathbf{i}}+r^{\prime }\cos {\theta }^{\prime }\hat{\mathbf{j}})\times (r^{\prime }{\hat{\mathbf{r}}}^{\prime }d{\theta }^{\prime })}{(r^{\prime 2}+z^2{)}^{3/2}}$ Now, perform the cross product and integrate: ${\mathbf{H}}_{d,z}(z)=\frac{{\rho }_s\mathrm{\Omega }}{4\pi }{\int }_0^a{\int }_0^{2\pi }\frac{r^{\prime 2}\cos {\theta }^{\prime }\hat{\mathbf{i}}\times {\hat{\mathbf{r}}}^{\prime }+r^{\prime 2}\sin {\theta }^{\prime }\hat{\mathbf{j}}\times {\hat{\mathbf{r}}}^{\prime }}{(r^{\prime 2}+z^2{)}^{3/2}}d{\theta }^{\prime }d{r}^{\prime }$ After integrating with respect to ${\theta }^{\prime }$, ${\mathbf{H}}_{d,z}(z)=\frac{{\rho }_s\mathrm{\Omega }z}{2\pi }{\int }_0^a\frac{r^{\prime 2}}{(r^{\prime 2}+z^2{)}^{3/2}}d{r}^{\prime }$ Now, perform the remaining integral with respect to $r^{\prime }$: ${\mathbf{H}}_{d,z}(z)=\frac{{\rho }_s\mathrm{\Omega }z}{2\pi }(-\frac{a^{2}z}{\sqrt{a^2+z^2}}+\frac{z}{\sqrt{z^2}})$ Finally, we get the expression for the magnetic field on the $z$-axis: $\mathbf{H}(z)=\frac{{\rho }_s\mathrm{\Omega }z}{2\pi }(1-\frac{a^2}{\sqrt{a^2+z^2}})\hat{\mathbf{z}}$ This is the magnetic field $\mathbf{H}$ at any point on the $z$-axis due to the rotating charged disk.

Key concepts

Magnetic Field

The magnetic field, often denoted as $\mathbf{H}$ or $\mathbf{B}$, is a fundamental concept in physics that describes how magnetic forces are distributed in space. In this problem, we are looking at the magnetic field created by a rotating disk of charge. Utilizing the Biot-Savart Law, which helps us calculate the magnetic field generated by a current, is crucial.
To apply Biot-Savart Law in this scenario, consider that the movement of the charged particles as the disk rotates is equivalent to an electric current. This current is what generates the magnetic field $\mathbf{H}$.
The expression for the magnetic field ${\mathbf{H}}_d(z)$ along the $z$-axis due to the disk is derived from integrating all contributions to the magnetic field (from each small piece of the disk) and considering symmetry properties. It makes it more straightforward since only the $z$-component of the magnetic field remains after accounting for symmetry.

Surface Charge Density

Surface charge density, represented as ${\rho }_s$, is an important quantity that describes how much electric charge is distributed over a surface area. It's measured in coulombs per square meter (C/m²). In this exercise, the disk possesses a uniform surface charge density, which means the charge is spread evenly over its entire surface.
The surface charge density impacts the magnitude of the magnetic field generated by the rotating disk. Higher ${\rho }_s$ means more charge is involved in the rotation, and therefore, the resulting magnetic field will be stronger. In the formula for $\mathbf{H}(z)$, ${\rho }_s$ directly influences the overall expression through multiplication, demonstrating its role in scaling the magnetic field intensity.

Angular Velocity

Angular velocity, denoted by $\mathrm{\Omega }$, describes how fast an object rotates about an axis. It is measured in radians per second (rad/s). For a rotating body like the disk in this problem, the angular velocity determines how quickly the disk is spinning.
The angular velocity is critical in calculating the magnetic field because it dictates the speed at which the charged particles are moving. This speed translates into an equivalent electric current, which is fundamental in generating the magnetic field via the Biot-Savart Law. Thus, $\mathrm{\Omega }$ appears in the solution formula for the magnetic field, highlighting its direct effect on the magnitude of $\mathbf{H}(z)$.

Polar Coordinates

Polar coordinates offer a convenient way to describe positions on a plane using a distance from a reference point (origin) and an angle from a reference direction. In this exercise, polar coordinates are used to represent the positions of infinitesimal charge elements on the disk.
Switching to polar coordinates simplifies integrating the contributions to the magnetic field, especially when considering rotational symmetry. Expressing $x^{\prime }$ and $y^{\prime }$ in polar form as $x^{\prime }=r^{\prime }\cos {\theta }^{\prime }$ and $y^{\prime }=r^{\prime }\sin {\theta }^{\prime }$ allows transforming the integral into a form that's more manageable and reflective of the disk's geometric nature.
Overall, polar coordinates help streamline the solution process by taking advantage of symmetry and making mathematical expressions more tractable.