CBSE Class 12-science Answered
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
Expert Answer
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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