Question

11. The resistivity of gold is 2.H 10-50 mat room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries a current of 940 mA. What is the electric field in the wire? hwina n of 4.00.600 and 100 are connected in parallel. If the combination is

Answer 1

For a wire of length l and radius r, resistance of the wire is

$R=\rho\frac{l}{\pi r^2}$

rho is the resisitivity of the wire

Radius of the wire is r=0.9/2 mm=0.45X10^-3 m

Putting the values,

$R=2.44\times10^{-8}\times\frac{14\times10^{-2}}{\pi\times(0.45\times10^{-3})^2}\ \Omega=5.37\times10^{-3}\ \Omega$

So, potential difference across the wire is

$V=Current\times Resistance=940\times10^{-3}\times5.37\times10^{-3}\ volts$

$=5.05\times10^{-3}\ V$

We know, electric field over a distance l is given by

$E=\frac{V}{l}=\frac{5.05\times10^{-3}}{14\times10^{-2}}\ V/m=0.036\ V/m$