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Suppose a 1.95 T field is applied across a 10-gauge copper wire (2.588 mm in diameter) carrying a 20.5 A current. What would the Hall voltage be, in microvolts, if the number density of free electrons in copper is 8.34 × 1028m-3?

Respuesta :

boffeemadrid

Answer:

1.47 microvolts

Explanation:

B = Magnetic field = 1.95 T

I = Current in wire = 20.5 A

n = Number density of free electrons in copper = [tex]8.34\times 10^{28}\ /m^3[/tex]

D = Diameter of wire = 2.588 mm

A = Area = [tex]\pi \left(\frac{D}{2}\right)^2[/tex]

e = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]

Hall voltage is given by

[tex]V_h=\frac{BID}{neA}\\\Rightarrow V_h=\frac{1.95\times 20.5\times 2.588\times 10^{-3}}{8.34\times 10^{28}\times 1.6\times 10^{-19}\pi \left(\frac{2.588\times 10^{-3}}{2}\right)^2}\\\Rightarrow V_h=1.47\times 10^{-6}\ V[/tex]

The Hall voltage is 1.47 microvolts

okpalawalter8

The hall voltage is  is mathematically given as

Vh=1.47* 10^{-6} V

Hall voltage

Question Parameters:

Suppose a 1.95 T field is applied across a 10-gauge copper wire (2.588 mm in diameter) carrying a 20.5 A current

the number density of free electrons in copper is 8.34 × 1028m-3?

Generally the equation for the area   is mathematically given as

A=\pi(d/2^2

Where The hall volatge is

[tex]V_h=\frac{BID}{neA}[/tex]

[tex]V_h=\frac{1.95*20.5*2.588*10^{-3}}{8.34* 10^{28}*1.6* 10^{-19}\pi *({2.588* 10^{-3}/2^2})}[/tex]

Vh=1.47* 10^{-6} V

For more information on Voltage

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