Introduction to Basic Electrodynamics (3): The Hall Effect

Before getting into Hall current electrodynamics, we begin with some unit vector basics. Along each axes one can define unit vectors: x^, y^ and z^. Then the respective multiplications (by vector directions) yield: x^*y^ = z^, and y^*z^ = x^ and x^*z^ = y^. These rules will always apply for a right handed coordinate system.

Thus, a vector cross product given by: A X B, must always have the directions attached by means of vectors, e.g.:

A(x^) x B(y^) = (A X B) (z^) = C(z^)

Now, as depicted in the accompanying diagram, consider a slab of conducting material through which a current I flows as shown (in direction x^) so that I = I(x^). We consider in turn the effect on positive and negative charge carriers, after having attached the coordinate system x, y, z and thence we specify (in addition to the current I(x^):

The magnetic induction: B = B(y^)

The velocity: v = v(-x^)

(Since the velocity of actual charges is opposite to the direction of conventional current flow)

The magnetic force (F = qvB) acting on a unit (+) charge deflects it toward the upper face, resulting in the accumulation of + charges there, and negative (-) charges on the bottom face.

Expressing the force with appropriate directions:

F(z^) = q v(-x^) x B(y^)

The opposite accumulation of charge (+ to bottom, - to top) gives rise to an electrical force that counteracts the magnetic. Eventually equilibrium occurs when:

Eq = qvB

At this point:

E = V_H/ t

Where V_H is the Hall potential difference.

Then:

(V_H/ t) q = qvB

Or, by directions:

qE(-z^) = q vB(z^)

or V_H = Bvt

The drift velocity can be found from the basic definition of the current:

I = ne v A

Where A is the area A = Lw (length x width of box)

n = number density of charges (per cubic meter)

e = unit of electronic charge = 1.6 x 10^-19 C

Solving for v:

V = I / neLw

Therefore, the Hall potential difference is:

V_H = B{I/neLw} t = BI/ new

Example Problem:

If the magnetic induction B = 1.0 T, and a rectangular slab of material (such as shown) is for copper, with n = 10^29 /m^3, find the Hall current if I = 10A, and the width of the slab is 0.001 m.

V_H = BI/ new

= (1.0T) (10 A)/ {10^29/m^3)(1.6 x 10^-19 C) (0.001m)}

V_H = 0.6 mV


Problem for ambitious and energized readers:

The diagram for this problem (lower graphic) shows a slab of silver with dimensions: z1 = 2 cm, y1 = 1mm, carrying 200 A of current in the +x^ direction. The uniform B-field has a magnitude of 1.5 Tesla. If there are 7.4 x 10^28 free electrons per cubic meter. Find:

a)The electron drift velocity

b)The magnitude and direction of the E-field due to the Hall Effect

c)The magnitude of the Hall EMF.