Irodov 3.247: Every Latitude Adds to the Center – Amit & Deepali Sinha Foundation

Problem Irodov 3.247

An insulating sphere of radius $R = 50$ mm carries a uniform surface charge of density $\sigma = 10.0$ $\mu$ C/m $^2$ and spins with angular velocity $\omega = 70$ rad/s about an axis through its center. Find the magnetic induction at the center of the sphere.

The insulating sphere of radius R and center O, its surface charged with density σ, spinning with angular velocity ω about the vertical axis through O (drawn for σ > 0). — The insulating sphere of radius $R$ and center $O$ , its surface charged with density $\sigma$ , spinning with angular velocity $\omega$ about the vertical axis through $O$ (drawn for $\sigma > 0$ ).

Given

$R = 50$ mm
$\sigma = 10.0$ $\mu$ C/m $^2$
$\omega = 70$ rad/s

Find

B

at the center

Solution

Step 1The rotating surface as ring currents

Measure the polar angle $\theta$ from the rotation axis. The band of the surface between $\theta$ and $\theta + d\theta$ is a ring of radius $R\sin\theta$ and width $R\,d\theta$ , so it carries the charge

dq = \sigma\,(2\pi R\sin\theta)(R\,d\theta) = 2\pi\sigma R^2\sin\theta\,d\theta

Turning once every $2\pi/\omega$ , the ring is a circular current

dI = \frac{\omega\,dq}{2\pi} = \sigma\omega R^2\sin\theta\,d\theta \tag{1}

The band of the sphere at the polar angle θ is a current loop dI; its field dB at the center O points along the axis, in the sense of ω for σ > 0. — The band of the sphere at the polar angle $\theta$ is a current loop $dI$ ; its field $d\vect{B}$ at the center $O$ points along the axis, in the sense of $\omega$ for $\sigma > 0$ .

Step 2The field of one ring at the center

The center lies on the ring’s axis, at the distance $R\cos\theta$ from its plane and $R$ from every point of the ring. The on-axis field of a loop of radius $a$ , seen from a distance $\rho$ to its points, is $\mu_0\,dI\,a^2/(2\rho^3)$ along the axis; here $a = R\sin\theta$ and $\rho = R$ , so with (1)

dB = \frac{\mu_0\,dI\,(R\sin\theta)^2}{2R^3} = \frac{\mu_0\sigma\omega R^2\sin\theta\,d\theta\;R^2\sin^2\theta}{2R^3}

dB = \frac{\mu_0\sigma\omega R}{2}\,\sin^3\theta\,d\theta

(2)

Step 3Adding the rings

Integrating (2) from pole to pole, with $\sin^3\theta = \sin\theta\,(1 - \cos^2\theta)$ :

\int_0^{\pi}\sin^3\theta\,d\theta = \left[-\cos\theta + \frac{\cos^3\theta}{3}\right]_0^{\pi} = \left(1 - \frac{1}{3}\right) - \left(-1 + \frac{1}{3}\right) = \frac{4}{3}

so that

B = \frac{\mu_0\sigma\omega R}{2}\cdot\frac{4}{3}

Numerically,

B = \frac{2}{3}\,(4\pi\times 10^{-7}\text{ H/m})(10.0\times 10^{-6}\text{ C/m}^2)(70\text{ s}^{-1})(0.050\text{ m})

B = \frac{2}{3}\,\mu_0\sigma\omega R \approx 29\text{ pT}

(3)

For $\sigma > 0$ the field points along $\vect{\omega}$ .

Check

A second route, by charge rather than by angle. A charge $dq$ on a ring of radius $R\sin\theta$ turning with $\omega$ is the current $\omega\,dq/(2\pi)$ , which gives at $O$ the field $\mu_0\omega\,dq\sin^2\theta/(4\pi R)$ : every charge counts with the weight $\sin^2\theta$ . Were the whole charge $q = 4\pi R^2\sigma$ on the equator, the field would be $\mu_0\omega q/(4\pi R)$ , which is $\mu_0\sigma\omega R$ . Spread uniformly over the sphere, the weight $\sin^2\theta$ averages to $2/3$ , so the field is $2/3$ of that, as in (3). Dimensionally, $\sigma\omega R$ is a surface current in A/m and $\mu_0$ times it is a field in tesla.

Remark

The field is tiny, about $3\times 10^{-11}$ T, because $\sigma\omega R$ is only $3.5\times 10^{-5}$ A/m of surface current. Rotating charge is a weak source of magnetism; Rowland’s experiment of 1876, which first showed that moving charge makes a magnetic field, had to spin a charged disc very fast and detect the effect with a sensitive compass needle.