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Chapter 26: Problem 76

A disk of radius \(a\) carries uniform surface charge density \(\sigma\) and rotates with angular speed \(\omega\) about the disk axis. Show that the magnetic field at the disk's center is \(\frac{1}{2} \mu_{0} \sigma \omega a\)

Short Answer

Expert verified
The magnetic field at the disk's center is \(\frac{1}{2} \mu_{0} \sigma \omega a\)

Step by step solution

01

Define the effective current

As the disk is rotating, the charges on it are effectively circulating, equivalent to a current flow. The surface charge density is \(\sigma\), its area is \(\pi a^2\) and it completes one cycle in \( \frac{2\pi}{\omega} \) seconds. Using \(I=\frac{Q}{t}\), the effective current \(I\) is \(\frac{\sigma\pi a^{2}\omega}{2\pi}\) which simplifies to \(I=\frac{1}{2}\sigma \omega a^{2}\).
02

Apply the Biot-Savart Law

The magnetic field \(B\) at the center due to a small current element can be described by the Biot-Savart Law as \(d B=\frac{\mu_{0}}{4 \pi} \frac{I d l \sin \theta}{r^{2}}\). Since we're looking for the magnetic field at the center of the disk, the angle \(\theta\) between \(d l\) and \(r\) is 90 degrees and \( \sin \theta = 1\). Also, as \(d l\) is the length of the current element, for a disk of radius \(a\), we consider \(d l = 2\pi rdr\). Substituting \(\sin \theta = 1\), \(I = \frac{1}{2}\sigma \omega a^{2}\) and \(d l = 2\pi rdr\), the expression for dB becomes \(d B=\frac{\mu_{0} I}{2 a^{2}} r d r\).
03

Integrate to find the total magnetic field

Integrate this expression from 0 to \(a\) to get the total magnetic field at the center of the disk. The result is \( B=\frac{\mu_{0} I}{2} \int_{0}^{a} \frac{r}{a^{2}} d r \), which simplifies to \( B=\frac{1}{2} \mu_{0} \sigma \omega a\).

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