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The resistivity of gold is 2.44×10-8 Ω•m at room temperature. A gold wire that is 1.8 mm in diameter and 40 cm long carries a current of 860 mA. What is the electric field in the wire?

Respuesta :

Answer:

8.25 x 10^-3 V/m

Explanation:

Resistivity, ρ = 2.44 x 10^-8 Ohm m, Diameter = 1.8 mm

radius, r = 0.9 mm = 0.9 x 10^-3 m, l = 40 cm = 0.4 m, i = 860 mA = 0.86 A,

Let V be the potential difference and E be the electric field.

Use the formula for resistance

R = ρ l / A

R = (2.44 x 10^-8 x 0.4) / ( 3.14 x 0.9 x 0.9 x 10^-6)

R =  3.84 x 10^-3 ohm

According to Ohm's law

V = i x R

V = 0.860 x 3.84 x 10^-3 Volt = 3.3 x 10^-3 Volt

E = V / l =  3.3 x 10^-3 / 0.4 = 8.25 x 10^-3 V/m

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