Question

A potential difference of 1.47 V will be applied to a 22.6 m length of 18-gauge copper wire (diameter = 0.0400 in.). Calculate (a) the current, (b) the magnitude of the current density, (c) the magnitude of the electric field within the wire, and (d) the rate at which thermal energy will appear in the wire.

Answer 1

$Voltage = 1.47\\L= 22.6m$

$r=\frac{d}{2} = \frac{0.04in}{2} = 0.02in = 0.508mm = 0.508*10^{-3}m$

So we can now calculate the area of the wire

$A= \pi *r^2 = \pi (0.508*10^{-3})^2 = 8.107*10^{-7}m^2$

We need also the resistivity specific of cupper, that is:

$p_{cupper}=1,72*10^{-8}$

a) We have to calculate the current that is given by,

$R=\frac{pL}{A}\\R=\frac{(1.72*10^{8})(22.6)}{(8.107*10^{-7})}\\R=0.4\Omega$

b)$I=\frac{V}{R}$

$I=\frac{1.47}{0.4} = 3.675A$

c) current density

$j= \frac{I}{A}\\j= \frac{3.675}{8.107*10^{-7}}\\j= 4.533*10^5 A/m^2$

d) Rate of thermal energy

$\alpha = I^2*R\\\alpha = 3.675^2*0.4 = 5.4W$