Solutions To Basic Electrodynamics and E&M Problems (3)

 

(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 m²)]

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