Question

A flat, circular disk of radius R is uniformly charged with total charge Q. The disk spins at angular velocity $\omega$about an axis through its center. What is the magnetic field strength at the center of the disk?

Short answer

$B=\begin{array}{rcl}& & \frac{{\mu }_0\omega Q}{2\pi R}\end{array}$

Step-by-step solution

Step1. Given information

The radius of the disk is$R$, total charge on the disk is$Q$and the angular velocity f the disk is$\omega$.

Step 2. Calculation

The relation between the linear velocity $\left(v\right)$and the angular velocity $\left(\omega \right)$is given by

$v=\omega R......................\left(1\right)$

The surface charge density $\left(\sigma \right)$of the disk is given by

role="math" localid="1649565802454" $\sigma =\frac{Q}{\pi R^2}.....................\left(2\right)$

Let's consider an infinitesimal ring of thickness $dr$at a distance $r$from the center of the disk.

The amount of charge $\left(dq\right)$contained within this infinitesimal ring is given by

$dq=2\pi r\sigma dr......................\left(3\right)$

According to Biot-Savart's law, the magnetic field$\left(dB\right)$ due to this infinitesimal charge is given by

role="math" localid="1649565540422" $dB=\frac{{\mu }_0}{4\pi }\frac{dqv}{r^2}...................\left(5\right)$

Here, ${\mu }_0$is the permeability of free space.

Substitute the expression for $dq$from equation (3) and the expression for $v$from equation (1) and simplify to obtain the infinitesimal magnetic field.

role="math" localid="1649565738953" $\begin{array}{rcl}dB& =& \frac{{\mu }_0}{4\pi }\frac{\left(2\pi r\sigma dr\right)\left(\omega r\right)}{r^2}\\ & =& \frac{{\mu }_0\sigma \omega }{2}dr............................\left(6\right)\end{array}$

The formula to calculate the net magnetic field $\left(B\right)$is given by

$B={\int }_0^{R}dB..........................\left(7\right)$

Substitute the expression for $dB$from equation (6) into equation (7) and simplify to obtain the magnetic field.

role="math" localid="1649565846430" $\begin{array}{rcl}B& =& {\int }_0^R\frac{{\mu }_0\sigma \omega }{2}dr\\ & =& \frac{{\mu }_0\sigma \omega }{2}R...........................\left(8\right)\end{array}$

Substitute role="math" localid="1649565878042" $\left(\frac{Q}{\pi R^2}\right)$for $\sigma$into equation (8) to obtain the required magnetic field.

role="math" localid="1649565929821" $\begin{array}{rcl}B& =& \frac{{\mu }_0\left(\frac{Q}{\pi R^2}\right)\omega }{2}R\\ & =& \frac{{\mu }_0\omega Q}{2\pi R}\end{array}$

Step 3. Final Answer

The required magnetic field is given by$B=\begin{array}{rcl}& & \frac{{\mu }_0\omega Q}{2\pi R}\end{array}$.