Question

A metal strip 2.50 cm wide and 1.80 cm thick carries a current of 29.5 A in a region containing a perpendicular uniform magnetic field of strength 1.51 T. The Hall voltage across the 2.50 cm width of the strip is measured to be 4.03 ?V. The drift speed of the electrons in the strip is 1.07×10-4 m/s.

Find the number density of the charge carriers in the strip (in m-3).

Answer 1

For calculating no. dencity of charge carrier first we have to find drift velocity-

The current in the perpendicular magnetic field experiences a force of magnitude B q vd The electrons therefore migrate across the strip setting up a Hall electric field EH = VH/w (where w is the width) that at equilibrium opposes the magnetic force.

B q vd = q VH / w

vd = VH / B w = 4.03x10^-6/(1.51x0.025) = 1.0675x10^-4=0.10675 mms-1

Now,

The current is given by I = n A q vd

n = I/A q vd = 29.5 /{ (0.025x0.018) x 1.6x10-19 x 0.10675x10^-3} = 3.838x10^28 m^-3 ....... .Ans.

Hope it will help you.

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