The electric field in a 1.6 mm x 1.6 mm square aluminum wire is \(1.8 x 10^{-2} V/m\)
What is the current in the wire?
R_of aluminum = 2.8*10^-8
Current density = 0.018 *10^8 / 2.8 = 6.428*10^5 A/m^2
Current = 1.6*1.6*10^-6 * 6.428*10^5 = 1.6455 A
current density J = i/A = E/rho
so i/A = E/rho
i = EA/rho
rho of Al 2.82
*10^-8
i = 0.018 * 1.6*1.6*10^-6/2.82*10^-8
i = 1.634 A
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