A hollow copper wire with an inner diameter of 0.60 mm and an outer diameter of 2.1 mm carries a current of 10 A .
What is the current density in the wire?
A= π(r2-r1)
r2= 2.1/2 = 1.15
r1= 0.6/2 = 0.3
A= π((1.05 * 10-3)2-(0.3*10-3)2
A= 3.18 * 10-6 m2
J = I/A
J = 10/(3.18 * 10-6)
J = 3.14 * 106 A/m2
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