Calculating the Magnetic Field Due to a Current — Question 9

Question 9

A rotating charged annulus is a continuum of circular current loops

A thin insulating annulus occupies $a\le r\le b$ and carries uniform positive surface charge density sigma. It rotates rigidly at angular speed omega about +z. Assume nonrelativistic motion and neglect fields caused by mechanical supports. Use vacuum permeability $\mu_0\simeq4\pi\times10^{-7}\,\mathrm{T\,m/A}$ where numerical values are needed.

1. Treat a narrow ring as a current loop and derive the central field.

2. Express the result using total charge Q, and compare it with placing the same Q on a ring of radius b at the same omega.

Question 9 - Solution

A rotating charged annulus is a continuum of circular current loops

(a) Method and calculation

$$dq=2\pi\sigma r\,dr,\quad dI=\frac{\omega}{2\pi}dq=\sigma\omega r\,dr.$$$$\vec B=\int_a^b\frac{\mu_0dI}{2r}\hat z=\boxed{\frac{\mu_0\sigma\omega}{2}(b-a)\hat z}.$$

(b) Interpretation and further calculation

$$Q=\pi\sigma(b^2-a^2)\Rightarrow B=\frac{\mu_0Q\omega}{2\pi(a+b)}.$$$$B_{\mathrm{ring}}=\frac{\mu_0Q\omega}{4\pi b},\quad\boxed{B/B_{\mathrm{ring}}=2b/(a+b)}.$$

Physical check

At fixed Q and angular speed, charges nearer the center provide a larger central field per unit charge; the thin-ring limit gives a ratio of one.