A hollow copper wire with an inner diameter of 1.2 mm and an outer diameter of 2.0 mm carries a current of 11
a.What is the current density in the wire?
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Current density = Current/(Area of cross section of wire)
Area = πr22 - πr12
= π[r22 - r12]
= π [(0.0021)2 - (0.0009)2]
= 11.304*10-6 m2
Area = 11.304*10-6 m2
Current density = 8/11.304*10-6
= 7.08A/m2
Current density = 7.08A/m2
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