Answer :
The Hall voltage when a 2.00-T magnetic field is applied across a 10-gauge copper wire carrying a 20.0-A current can be calculated using the Hall effect equation. This requires the charge carrier density (n) which is generally around 8.47 × 10²⁸ for copper, along with other known values such as the current (I), magnetic field strength (B), charge of an electron (e), and the diameter of the wire (t).
To calculate the Hall voltage when a 2.00-T magnetic field is applied across a 10-gauge copper wire carrying a 20.0-A current, we use the Hall effect equation:
VH = (IB)/(net)
where VH is the Hall voltage, I is the current, B is the magnetic field strength, n is the charge carrier density, e is the charge of an electron, and t is the thickness of the wire.
The diameter of the wire is given as 2.588 mm, so its radius (r) is half of that, which is 1.294 mm or 0.001294 meters. The thickness (t) of the wire for the purpose of the Hall voltage calculation will be its diameter, as the magnetic field is applied perpendicular to the current flow and spans across the entire diameter of the wire.
The charge carrier density for copper (n) is typically around 8.47 × 10²⁸ electrons/m3. Since we are not given a specific value for this problem, we'd use this typical value of n for copper. However, to complete the calculation, we would need the exact value of n from Example 20.6 or other provided resources.
Once the value of n is determined, you can compute the Hall voltage using the above formula by plugging in the values for I, B, n, e, and t.
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]
