Question

A current carrying gold wire has diameter 0.84 mm. The electric field in the wire is 0.49 V/m. What are (a) the current carried by the wire; (b) the potential difference between two points in the wire 6.4 m apart; (c) the resistance of a 6.4-m length of this wire

Answer 1

Explanation:

Given that,

Diameter of the gold wire, d = 0.84 mm

Radius, r = 0.42 mm

Electric field in the wire, E = 0.49 V/m

(a) Electric current density is given by :

$J=\dfrac{I}{A}$

And electric field is :

$E=\rho J$

$\rho$ is resistivity of Gold wire

So,

$E=\dfrac{\rhi I}{A}\\\\I=\dfrac{EA}{\rho}\\\\I=\dfrac{0.49\times \pi (0.42\times 10^{-3})^2}{2.44\times 10^{-8}}\\\\I=11.12\ A$

(b) The potential difference between two points in the wire is given by :

$V=E\times l\\\\V=0.49 \times 6.4\\\\V=3.136\ V$

(c) Resistance of a wire is given by :

$R=\rho \dfrac{l}{A}\\\\R=2.44\times 10^{-8}\times \dfrac{6.4}{\pi (0.42\times 10^{-3})^2}\\\\R=0.281\ \Omega$

Hence, this is the required solution.