(a) What is the current carried by the wire? A (b) What is the potential differe

Question

(a) What is the current carried by the wire?
A

(b) What is the potential difference between two points in the wire 6.2 m apart?
V

(c) What is the resistance of a 6.2 m length of the same wire?
?

A current-carrying gold wire has diameter 0.76 mm. The electric field in the wire is 0.51 V/m. (Assume the resistivity of gold is 2.44 x 10^-8 ? m.) (a) What is the current carried by the wire? A (b) What is the potential difference between two points in the wire 6.2 m apart? V (c) What is the resistance of a 6.2 m length of the same wire? ?

Explanation / Answer

part A:

apply current density J = I/A = E /rho

where i is current

A is area = pi r^2

E is eelctric field

and

rho is resisitivity

so

Current i = EA/rho

i = 0.51 * 3.14 * 0.38e-3*0.38e-3 /(2.44e-8)

i = 9.477 Amps

--------------------------------------------

apply ohms law as V = iR

so

here Resistance R = rho L/A

R = 2.44 e-8 * 6.2/(3.14 * 0.38e-3*0.38e-3)

R = 0.3336 ohms

so

PD V = iR = 9.477 * 0.3336

V = 3.161 Volts

-------------------------------

Resistance R = rho L/A

R = 2.44 e-8 * 6.2/(3.14 * 0.38e-3*0.38e-3)

R = 0.3336 ohms

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