Question

A hollow copper wire with an inner diameter of 1.2mmand an outer diameter of 2.3mm carries a current of 12A .

What is the current density of the wire?

Answer 1

apply current density J = Current /area

so

J = i/A

so here Area A = pi*(R^2-r^2)

A = 3.14* (0.0023^2-0.0012^2)/4

A = 3.022 e-6 m^2

so

J = 12/(3.022 e-6)

J = 3.97 *10^6 A/m^3

Answer 2

effective area, A = pi*(d2^2-d1^2)/4

= pi*(2.3^2-1.2^2)*10^-6/4

= 3.02*10^-6 m^2

J = I/A

= 12/3.02*10^-6

= 3.97*10^6 A/m^2

Answer 3

current density= current / (area of cross section of wire)

Area = pi*r?