Monday, April 17, 2023

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 = VH/ 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) VH = B v z1 (by def. of Hall voltage using diagram parameters)

Then:

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

Therefore: VH  = 2.5 x 10 -5 V


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