A silver slab of thickness h=1.1 mm and width w=1.4 cm is connected to the termi

Question

A silver slab of thickness h=1.1 mm and width w=1.4 cm is connected to the terminals of a battery and carries a current of 2.5 A in a region where there is a magnetic field 1.2 T perpendicular to the slab. Due to the magnetic field, charges will accumulate at the sides of the slab, producing a potential difference of 0.334 ?V between A and C.

a)Knowing that the charge carriers inside the silver are electrons, determine the electric field (magnitude and direction) in the x-axis due to the charge separation.

b)What would the electric field be if the charge carriers were positive?

c)Determine the electrons drift velocity in the silver slab knowing that charge separation stops when the magnetic force is counterbalanced by the electric force.

d)Determine the number density of charge carriers inside the silver slab.

Explanation / Answer

h = 1.1 mm , w or d = 1.4 cm , B = 1.2 T , I = 2.5 amp , Vh = .334 ?V

(a) Eh = - Vh / d = - .334 / 1.4 ?V/cm = - 23.85 ?V /m

(b) electron drift velocity , v = Eh /B = - 23.85 /1.2 = 19.88

(c) number density of charge => Vh = -I B / nte

n = -I B / Vhte = 2.5 *1.2 / .334*10-6*1.1*10-3*1.6*10-19

= 51.72*1029

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