Answer:
The hall voltage is [tex]\epsilon =1.45 *10^{-6} \ V[/tex]
Explanation:
From the question we are told that
The magnetic field is [tex]B = 2.00 \ T[/tex]
The diameter is [tex]d = 2.588 \ mm = 2.588 *10^{-3} \ m[/tex]
The current is [tex]I = 20 \ A[/tex]
The radius can be evaluated as
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{2.588 * 10^{-3}}{2}[/tex]
[tex]r = 1.294 *10^{-3} \ m[/tex]
The hall voltage is mathematically represented as
[tex]\episilon = B * d * v_d[/tex]
where[tex]v_d[/tex] is the drift velocity of the electrons on the current carrying conductor which is mathematically evaluated as
[tex]v_d = \frac{I}{n * A * q }[/tex]
Where n is the number of electron per cubic meter which for copper is
[tex]n = 8.5*10^{28} \ electrons[/tex]
A is the cross - area of the wire which is mathematically represented as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 * [ 1.294 *10^{-3}]^2[/tex]
[tex]A = 5.2611 *10^{-6} \ m^2[/tex]
so the drift velocity is
[tex]v_d = \frac{20 }{ 8.5*10^{28} * 5.26 *10^{-6} * 1.60 *10^{-19} }[/tex]
[tex]v_d = 2.7 *10^{-4 } \ m/s[/tex]
Thus the hall voltage is
[tex]\epsilon = 2.0 * 2.588*10^{-3} * 2.8 *10^{-4}[/tex]
[tex]\epsilon =1.45 *10^{-6} \ V[/tex]
