Question

The single most important use of the element gold is for wires that join integrated circuits to the conducting pads that then connect microelectronic chips with external circuitry. A typical application uses gold wires 1.0 milmil (0.001 inch) in diameter and 45 milmil long. Although the current under normal conditions is much less, such a wire will fuse (that is, melt and interrupt the current) when the current exceeds some 660 mAmA.

A) Find the current density in the wire when this maximum current is flowing.

B) Find the electric field in the wire when this maximum current is flowing.

C) Find the voltage between the ends of the wire when this maximum current is flowing.

Answer 1

Part A.

Current density is given by:

J = Current/Cross-section Area = I/A

I = max Current in wire = 660 mA = 0.660 A

A = Cross-sectional area of wire = pi*d^2/4

d = diameter of wire = 0.001 inch = (0.001 inch)*(0.0254 m/1 inch) = 2.54*10^-5 m

So,

J = 0.660*4/(pi*(2.54*10^-5)^2)

J = 1.303*10^9 A/m^2

Part B.

Relation between current density and Electric field in wire is given by:

J = E

= conductivity of material = 1/resistivity of material = 1/(2.44*10^-8)

So,

E = J/

E = 1.303*10^9*2.44*10^-8

E = 31.79 N/C

Part C.

Using ohm's law:

V = I*R

R = resistance of wire = rho*L/A

rho = resistivity of gold = 2.44*10^-8 ohm-m

L = length of wire = 45 mil = 0.045 inch = 0.045*0.0254 = 0.001143 = 1.143*10^-3 m

So,

V = I*rho*L/A = 4*I*rho*L/(pi*d^2)

V = 4*0.660*2.44*10^-8*1.143*10^-3/(pi*(2.54*10^-5)^2)

V = Voltage between ends of the wire = 3.63*10^-2 V