Irodov 3.360: A Moving Plate Charges Its Faces – Amit & Deepali Sinha Foundation

Problem Irodov 3.360

A large plate of non-ferromagnetic material moves at a constant velocity $v = 90$ cm/s through a uniform magnetic field of induction $B = 50$ mT. Both the velocity and the field lie parallel to the plate’s large faces and are perpendicular to each other. Find the surface density of the electric charges that appear on the plate because of its motion.

The plate, its large face in the page and its thickness hatched. The field B and the velocity v both lie parallel to the large face, at right angles to each other. — The plate, its large face in the page and its thickness hatched. The field $\vect B$ and the velocity $\vect v$ both lie parallel to the large face, at right angles to each other.

Given

$v = 90$ cm/s
$B = 50$ mT
$\vect v \perp \vect B$ , both parallel to the large faces
plate conducting, a metal (assumed)

Find

the surface charge density

\sigma

Solution

Step 1The field inside the moving metal

The plate is taken to be a conductor (assumed): its free electrons move with it and can redistribute. The material is non-ferromagnetic, so the induction inside the plate is the applied $\vect B$ . In the steady state the net force on each of them vanishes, so inside the metal the electric field must cancel the magnetic part of the Lorentz force:

q(\vect E + \vect v\times\vect B) = 0

\vect E = -\,\vect v\times\vect B

(1)

Since $\vect v \perp \vect B$ its magnitude is $vB$ , and since both lie in the large faces, $\vect v\times\vect B$ is normal to the plate: the field inside runs straight across the thickness.

Step 2The charges that make that field

The magnetic force $q\,\vect v\times\vect B$ drives positive charge to the face toward which $\vect v\times\vect B$ points and leaves the opposite face negative. Two large sheets with $\pm\sigma$ give the field $\sigma/\varepsilon_0$ between them and none outside:

The plate seen edge on, looking down along B (out of the page), moving with v: the charge σ on the face toward which v· B points, -σ on the other, and E between them. The thickness is drawn schematically. — The plate seen edge on, looking down along $\vect B$ (out of the page), moving with $\vect v$ : the charge $\sigma$ on the face toward which $\vect v\times\vect B$ points, $-\sigma$ on the other, and $\vect E$ between them. The thickness is drawn schematically.

\frac{\sigma}{\varepsilon_0} = vB

Numerically, with $v = 0.90$ m/s and $B = 0.050$ T:

\sigma = (8.85\times10^{-12})(0.90)(0.050) \text{ C/m}^2

\sigma = \varepsilon_0 vB \approx 0.40 \text{ pC/m}^2

(2)

The face toward which $\vect v\times\vect B$ points carries $+\sigma$ , the opposite face $-\sigma$ : in the first figure, the face toward the viewer is positive.

Check

Units: $\varepsilon_0 vB$ is $(\text{F/m})\cdot(\text{m/s})\cdot\text{T} = (\text{F/m})\cdot(\text{V/m}) = \text{C/m}^2$ , since $\text{F}\cdot\text{V} = \text{C}$ . In the plate’s own frame there is no magnetic force on the electrons, which are at rest, but there is the field $\vect E' = \vect v\times\vect B$ of 3.359; the face charges must cancel it inside the metal, which asks for the same $\sigma = \varepsilon_0 vB$ .

Remark

The charge is tiny, about $2.5$ million electrons per square meter, because $vB = 0.045$ V/m is a weak field. The same separation in a long rod is the motional emf: here nothing closes a circuit, so the emf $vB\,h$ across a thickness $h$ is balanced by the face charges and no current flows.