please dont briefly answer, explain every step of the answer. A slab of semicond
ID: 1560117 • Letter: P
Question
please dont briefly answer, explain every step of the answer.
A slab of semiconductor has a rectangular cross-section with a 10 cm width and 1 mm thickness. The slab is placed under a uniform magnetic field of 0.8 T and carries a current of I = 0.1 mA. When the potential difference between points a and b are measured, it is found that V_ab = +1mV (Hall voltage). The geometry and the directions are shown on the figure. (a) Find the electric field E due to the Hall field. (Note that there is an additional electric field along the x-direction which drives the current. Ignore this and find the Hall field only.) (b) Find the drift velocity, v_d. (c) What is the sign of the charge of the charge carriers? (d) Find the current density J. (e) Find the density of the charge carriers, n.Explanation / Answer
a)
electric field E = Vab/ab = 1*10^-3/(0.01) = 0.1 V/m
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b)
In the slab
magnetic force = electric force
q*vd*B = E*q
vd = E/B
vd = 0.1/0.8 = 0.125 m/s
======================
(c)
negative charge
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(d)
current density J = I/A
J = 0.1*10^-3/(0.01*10^-3) = 10 A/m^2
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(e)
vd = I/(n*e*A)
Vd = J/(n*e)
e = charge of electron
0.125 = 10/(n*1.6*10^-19)
n = 5*10^20
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