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Consider a rectangular sample, a metal or an ntype semiconductor, with a length L, width W, and thickness D. A current I is passed along L, perpendicular to the crosssectional area WD. The face W X L is exposed to a magnetic field intensity B. A voltmeter is connected across the width to read the Hall voltage V_{H}.


Solution

Consider a slab of length L, width W, and thickness D as shown above. Let us take a small element "dX" in the length; its volume is given by dV = WDdX
If n = number of charge carriers per unit volume, then the total number of charge carriers in this volumetric element of thickness dX will be nWDdX
Therefore, total charge (dQ) = enWDdX
Thus, current (I) = dQ/dt = enWD(dX/dt)
Or, I = enWDv (where v is the charge velocity)
Or, ne = I/WDv
Magnetic force (F_{m}) = q(v x B)
Or, F_{m} = nevB = (I/WDv) x (vB) = IB/WD
Electrostatic force (F_{e}) = qE = neE
A potential develops across the width W, and is given by V_{H} = EW. V_{H} is called as the Hall potential.
It gives the value of the electric field E = V_{H}/W
Therefore, F_{e} = neV_{H}/W
In equilibrium, F_{e} = F_{m}
That is, neV_{H}/W = IB/WD
Or, V_{H} = IB/neD
This equation shows that the Hall potential is independent of the length and the width of the conductor slab. 
The atomic weight of gold is 196.97 grams.
i.e., 6.023 x 10^{23} atoms weigh 196.97 grams
or, 1 gram gold has 6.023 x 10^{23}/196.97 atoms
The density of gold is 19.32 g/cm^{3}.
i.e., 1 cm^{3} of gold weighs 19.32 grams.
Therefore, 1 cm^{3} has [6.023 x 10^{23}/196.97] x 19.32 atoms
i.e., n = 5.9077 x 10^{22} atoms/cm^{3} or 5.9077 x 10^{28} atoms/m^{3}
Gold is monovalent (valency = 1), therefore, total number of conduction electrons = total number of atoms
Thus, total number of conduction electrons per m^{3} = 5.9077 x 10^{28}.
Therefore, total charge = ne = 5.9077 x 10^{28} x 1.6 x 10^{19}
V_{H} = IB/neD
Or, B/V_{H} = neD/I = 10^{6} x 5.9077^{28} x 1.6 x 10^{19} / 100 x 10^{3} (Given thickness = 1 micron and current = 100 mA)
B/V_{H} = 94523.2 Tesla/Volt
Or, B/V_{H} = 94523.2 x 10^{6} / 10^{6} Tesla/Volt
Or, B/V_{H}= 0.0945232 Tesla/micro Volt