Irodov 3.361: Charges in a Spinning Metal Cylinder – Amit & Deepali Sinha Foundation

Problem Irodov 3.361

A long solid aluminum cylinder of radius $a = 5.0$ cm spins about its own axis at $\omega = 45$ rad/s in a uniform magnetic field of induction $B = 10$ mT, the angular velocity pointing the same way as $\vect B$ . Ignoring the magnetic field of the charges that appear, find their volume density and their surface density.

The long cylinder of radius a spinning at ω about its axis, in the uniform field B along the axis, pointing the same way as the angular velocity. — The long cylinder of radius $a$ spinning at $\omega$ about its axis, in the uniform field $\vect B$ along the axis, pointing the same way as the angular velocity.

Given

$a = 5.0$ cm
$B = 10$ mT
$\omega = 45$ rad/s
$\vect{\omega}$ along $\vect B$
field of the charges neglected

Find

the volume density

\rho

· the surface density

\sigma

Solution

Step 1The field the electrons need

An electron at the vector distance $\vect r$ from the axis (perpendicular to it) moves with $\vect v = \vect{\omega}\times\vect r$ . The total Lorentz force on it vanishes:

-e\left(\vect E + (\vect{\omega}\times\vect r)\times\vect B\right) = 0

The double product expands as $(\vect a\times\vect b)\times\vect c = \vect b\,(\vect a\cdot\vect c) - \vect a\,(\vect b\cdot\vect c)$ , and $\vect r\cdot\vect B = 0$ because $\vect r$ is perpendicular to the axis while $\vect B$ lies along it:

(\vect{\omega}\times\vect r)\times\vect B = \vect r\,(\vect{\omega}\cdot\vect B) - \vect{\omega}\,(\vect r\cdot\vect B) = \omega B\,\vect r

So the field inside the metal is radial, pointing toward the axis, and grows linearly with $r$ :

\vect E = -\,\omega B\,\vect r

(1)

Step 2The volume charge

Gauss’s law in differential form gives the charge density from the divergence of (1). For a radial field $E_r = -\omega B r$ in cylindrical coordinates:

\rho = \varepsilon_0\nabla\cdot\vect E = \varepsilon_0\,\frac{1}{r}\,\frac{d}{dr}\left(r E_r\right) = \varepsilon_0\,\frac{1}{r}\,\frac{d}{dr}\left(-\omega B r^2\right)

Numerically:

\rho = -2(8.85\times10^{-12})(45)(0.010) \text{ C/m}^3

\rho = -2\varepsilon_0\omega B \approx -8.0 \text{ pC/m}^3

(2)

The volume charge is uniform and negative: the magnetic force has pushed electrons toward the axis.

Step 3The surface charge

Equating the total charge per unit length to zero, with the volume charge over the cross-section $\pi a^2$ and the surface charge over the circumference $2\pi a$ :

\rho\,\pi a^2 + \sigma\cdot 2\pi a = 0 \quad\Longrightarrow\quad \sigma = -\frac{\rho a}{2}

The charges in the spinning cylinder, in a section through its axis: a uniform negative volume charge ρ inside and a positive surface charge σ on the curved surface. — The charges in the spinning cylinder, in a section through its axis: a uniform negative volume charge $\rho$ inside and a positive surface charge $\sigma$ on the curved surface.

Putting in (2), with $a = 0.050$ m:

\sigma = (8.85\times10^{-12})(45)(0.010)(0.050) \text{ C/m}^2

\sigma = \varepsilon_0\omega B a \approx 0.20 \text{ pC/m}^2

(3)

Check

The jump of the radial field across the surface is $0 - E_r(a) = \omega B a$ , and it must equal $\sigma/\varepsilon_0$ , which it does. The neglected centripetal force $m\omega^2 r$ , against the magnetic force $e\omega B r$ , is the fraction

\frac{m\omega}{eB} = \frac{(9.11\times10^{-31})(45)}{(1.60\times10^{-19})(0.010)} \approx 3\times10^{-8},

so dropping it is safe. Reversing $\vect B$ (or $\vect{\omega}$ ) reverses both signs.

Remark

The radial field (1) is the motional emf of a homopolar (Faraday) generator: between the axis and the rim it adds up to $\omega B a^2/2$ , about $0.6$ mV here, and a sliding contact on the rim and another on the axle could draw a steady current from it. The charges found here are what holds that emf in place when no current is drawn.