Basic Electrodynamics: Solution to Hall Effect Problem

We now return to the problem to do with the Hall Effect, posted at the end of the third instalment on Basic Electrodynamics. For convenience, I show again the Hall Effect slab referenced for the problem.
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The problem, again, reads:

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.

Solutions:

a) By definition of current: I = nevA

where v is the drift velocity: v = I/neA

Therefore:

v = (200A)/ [(7.4 x 10^28/m^3)(1.6 x 10^-19C)(2 x 10^-5 m^2)]

v = 8.4 x 10^-4 (-x^) m/s


b) E = V_H/ z1 where z1= 0.02m (= 2 cm)

Then: E = (2.5 x 10^-5 V)/ 0.02 m = 1.25 x 10^-3 V/m(-z^)


c) V_H = B v z1 (by def. of Hall voltage using diagram parameters)

Then:

V_H = (1.5T) (8.4 x 10^-4 m/s) (0.02m)

Therefore: V_H = 2.5 x 10^-5 V

Note that this quantity must be obtained before part (b) is attempted, since the quantity V_H figures into the computation of the E-field, E.


Further notes on vector directions:

B = B(y^), I = I(x^), v=(-x^)