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A semiconductor slab of thickness t=0.5 mm with length l=30 cm carrying current I=2A placed in a magnetic field B=0.25T directed into the page, perpendicular to the flat surface of the slab carries 7*10^24 electrons per m^3. Calculate the Hall voltage across the slab. Indicate which edge (top or bottom) is at higher potential and why?

Asked by arjunsah797 | 26 Jan, 2022, 02:56: PM
Figure shows a semiconductor slab of length l , width w and thickness t .

Let us assume cartesian coordinate system so that length of slab is along y-direction and width is along z-direction as shown in figure.

Let Current I is passing through the semiconductor slab along y-direction .

Let magnetic field B is directed into the page, i.e. along -x-axis direction.

Since current is passing along y-axis direction, drift velocity vd of electron is along -y-axis direction as shown in figure.

Force acting on electron F = -e ( vd × B )

where e is magnitude of charge on electron

Direction of force is along z-direction . Hence electrons accumulate at top of slab and gives -ve potential  ,
while bottom of slab gets +ve potential as shown in figure.

Hall voltage,  VH = ( vd × B × w )  ...............................(1)

Current and drift velocity are related as, I = ne × e  × vd × A  ............................(2)

where ne is electron density  and A = ( w × t ) is area of cross section

Substituting vd from eqn.(2) , we get Hall voltage from eqn.(1) as

Answered by Thiyagarajan K | 26 Jan, 2022, 08:15: PM

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