Irodov 3.249: Two Opposite Currents Cancel at Center

Problem Irodov 3.249

A long dielectric cylinder of radius $R$ carries a frozen-in polarization $\vect P = \alpha\vect r$ at every point, where $\alpha$ is a positive constant and $\vect r$ is the radius vector drawn perpendicular to the axis from the axis to the point. The cylinder is made to spin about its axis with angular velocity $\vect\omega$ . Find the magnetic induction $\vect B$ at the center of the cylinder.

The long cylinder of radius R seen from the side, spinning with angular velocity ω about its axis. — The long cylinder of radius $R$ seen from the side, spinning with angular velocity $\omega$ about its axis.

Given

radius $R$ , long
$\vect P = \alpha\vect r$ , $\alpha > 0$
angular velocity $\vect\omega$

Find

\vect B

at the center

Solution

Step 1The bound charges

The polarization is radial and grows with $r$ , so it leaves bound charge inside, with the volume density $-\nabla\cdot\vect P$ , which in cylindrical coordinates reads

\begin{align*} \rho' &= -\frac{1}{r}\frac{d}{dr}\bigl(r P_r\bigr) = -\frac{1}{r}\frac{d}{dr}\bigl(\alpha r^2\bigr) \\ &= -2\alpha \tag{1}\end{align*}

and on the side surface, where the outward normal is $\vect r/R$ , the surface density $P_n$ :

\sigma' = P_r(R) = \alpha R \tag{2}

The bound charges in the cross-section: a uniform negative volume charge and a uniform positive charge on the surface, equal and opposite per unit length.

Step 2Rotation turns the charges into currents

Every charge moves on a circle with speed $\omega r$ . The surface charge (2) becomes a surface current per unit length of the cylinder, running round the axis,

K = \sigma'\omega R = \alpha\omega R^2 \tag{3}

and the volume charge (1) a current density round the axis, in the opposite sense because $\rho' < 0$ :

j_\varphi = \rho'\omega r = -2\alpha\omega r \tag{4}

These are the only currents: $\vect P$ depends only on the distance from the axis, so at every fixed point it stays the same as the cylinder turns, and there is no polarization current $\partial\vect P/\partial t$ . The cylinder is long, so the azimuthal currents act as a nest of coaxial solenoids.

Step 3The field of a nest of solenoids

A long cylindrical sheet carrying the azimuthal current $K$ per unit length makes a uniform axial field inside it and none outside: by symmetry $\vect B$ is axial and depends only on $r$ , and a rectangular Amperian loop of length $l$ with one long side inside and one far outside gives

B\,l = \mu_0 K l

The thin shell of the volume between $r'$ and $r' + dr'$ is such a sheet with $K = j_\varphi\,dr'$ , so at the distance $r$ from the axis only the surface current and the shells outside $r$ count. With (3) and (4):

\begin{aligned} B(r) &= \mu_0\left(K + \int_r^R j_\varphi\,dr'\right) = \mu_0\left(\alpha\omega R^2 - 2\alpha\omega\int_r^R r'\,dr'\right)\\ &= \mu_0\bigl(\alpha\omega R^2 - \alpha\omega\,(R^2 - r^2)\bigr) \end{aligned}

B(r) = \mu_0\alpha\omega r^2

(5)

At the center, $r = 0$ :

\vect B = 0

(6)

Check

Just inside the surface no volume current lies farther out, so the surface sheet alone must set the field there, $\mu_0 K$ ; (5) at $r = R$ gives $\mu_0\alpha\omega R^2$ , which is $\mu_0 K$ by (3). Integrated over the whole cross-section, the volume current per unit length is $\int_0^R (-2\alpha\omega r)\,dr = -\alpha\omega R^2$ , minus the surface current, so the cylinder carries no net azimuthal current, consistent with its being neutral. Dimensionally, $\alpha$ is a charge per volume, so $\alpha\omega r^2$ is a current per length and $\mu_0$ times it a field.

Remark

The profile (5) is the field of a magnetization $\vect M = \vect P\times\vect v$ , of magnitude $\alpha\omega r^2$ along $\vect\omega$ : a moving polarized body looks magnetized. Inside a long body that field is $\mu_0 M$ , zero on the axis because both $\vect P$ and $\vect v$ vanish there, and largest at the surface. For $\alpha > 0$ it points along $\vect\omega$ , the sense of the surface current, which wins everywhere except at the axis.