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 mapart; (c) the resistance of a 6.4 mlength of this wire?

Short answer

(a) The current carried by the wire, l = 11.125 A

(b) The potential difference between two points in the wire, V = 3.136 V

(c) The resistance of the wire,$R=0.282\Omega$

Step-by-step solution

Determine the area of the wire

$$\begin{gathered} A=\frac{\mathrm{\pi }}{4}{d}^2 \\ A=\frac{\mathrm{\pi }}{4}\times {\left(0.84\times {10}^{-3}\right)}^2 \\ A=5.54\times {10}^{-7}{m}^2 \end{gathered}$$

Determine the formula to find current carried by the wire

We know,

$E=pJ$

And

$J=\frac{l}{A}$

Therefore,

$l=\frac{EA}{p}$

Determine the current carried by the wire

(a) Use the formula found above,

$$\begin{gathered} l=\frac{Ea}{p} \\ l=\frac{\left(0.49V/m\right)\left(5.54\times {10}^{-7}{m}^2\right)}{2.44\times {10}^{-8}\Omega m} \\ l=11.125A \end{gathered}$$

Therefore, the current carried by the wire is 11.125 A

Determine the potential difference between two points in the wire

(b) Use the formula

$V=EL$

Substitute the given values

$$\begin{gathered} V=\left(0.49V/m\right)\left(6.4m\right) \\ V=3.136V \end{gathered}$$

Thus, the potential difference between two points in the wire 6.4m apart is 3.136 V

Determine the resistance of the wire

Use the formula of resistance

$R=\frac{\rho L}{A}$

Substitute the values

$$\begin{gathered} R=\frac{\left(2.44\times {10}^{-8}\Omega m\right)\left(6.4m\right)}{5.54\times {10}^{-7}{m}^2} \\ R=0.282\Omega \end{gathered}$$

Therefore, the resistance of a 6.4 M length of this wire is $0.282\Omega$