To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.
The drift velocity is given by the equation:
[tex]V_d = \frac{I}{nAq}[/tex]
Where
I = current
n = Number of free electrons
A = Cross-Section Area
q = charge of proton
Our values are given by,
[tex]I = 25 A[/tex]
[tex]A= 1.2*20 *10^{-6} m^2[/tex]
[tex]q= 1.6*10^{-19}C[/tex]
[tex]N = 8.47*10^{19} mm^{-3}[/tex]
[tex]V_d =\frac{25}{(1.2*20 *10^{-6})(1.6*10^{-19})(8.47*10^{19} )}[/tex]
[tex]V_d = 7.68*10^{-5}m/s[/tex]
The hall voltage is given by
[tex]V=\frac{IB}{ned}[/tex]
Where
B= Magnetic field
n = number of free electrons
d = distance
e = charge of electron
Then using the formula and replacing,
[tex]V=\frac{(2.5)(25)}{(8.47*10^{28})(1.6*10^{-19})(1.2*10^{-3})}[/tex]
[tex]V = 3.84*10^{-6}V[/tex]
