Respuesta :
Answer:
[tex]V_H=7.3544\times 10^{-6} \,V[/tex]
Explanation:
Given that:
thickness of the metal strip, [tex]d=150\times 10^{-6} m[/tex]
width of the metal strip, [tex]w=4.5\times 10^{-3}m[/tex]
magnitude of the perpendicular magnetic field, [tex]B=0.65 \,T[/tex]
current through the strip, [tex]I=23\,A[/tex]
charge density, [tex]n=8.47\times 10^{28} \,electrons\,per\,m^3[/tex]
Hall voltage is a transverse voltage given by:
[tex]V_H=\frac{I.B}{n.e.d}[/tex]
putting the respective values
[tex]V_H=\frac{23\times 0.65}{(8.47\times 10^{28})\times 1.6\times 10^{-19}\times (150\times 10^{-6} )}[/tex]
[tex]V_H=7.3544\times 10^{-6} \,V[/tex]
Note:
V = E d is used when a charge is moved in a uniform electric field between the two oppositely charged plates, it can't be used here because it is the case of Hall effect where the small voltage develops transverse to the direction of current flow.
