Part A If there are 5.75×10 28 free electrons per cubic meter, find the magnitud

Question

Part A

If there are 5.75×1028 free electrons per cubic meter, find the magnitude of the drift velocity of the electrons in the x-direction.

Part B

Find the magnitude of the electric field in the z-direction due to the Hall effect.

Part C

Find the direction of the electric field in the z-direction due to the Hall effect.

Part D

Find the Hall emf.

z +z The figure (Figure 1) shows a portion of a silver ribbon with z1 = 14.0 mm and y1-0.23 mm , carrying a current of 120 A in the +-direction. The ribbon lies in a uniform magnetic field, in the y-direction, with magnitude 0.92 T. Apply the simplified model of the Hall effect. Figure 1 of 1

Explanation / Answer

(a) The magnitude of the drift velocity
Vd = I/nqA
where A is the cross sectional area
Vd = 120 / (5.75*1028*1.6*10-19*(14*0.23*10-6)) = 4.05*10-3 m/s
(b) Electric field is given by
E = Vd*B = 3.727*10-3 V/m
(c) The direction will be provided by the direction of the cross product of velocity and Magnetic fied which will be +Z direction.
(d) Potential = electric field*distance
VZ = E*DZ = 3.727*10-3*(14*10-3) = 0.052*10-3 V

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