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Grade 12Electric Current

The magnitude of J of the current density in a certain lab wire with a circular wire cross section of radius 2.00mm is given by (J=3*10^8)*r^2 with J in amperes per square metre. what is the current through outer section bounded by r=0.900R and r=R??

Profile image of sameeha
8 Years agoGrade 12
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1 Answer

Profile image of Arun
8 Years ago
Current = ∫ J dA = (2π)(3 x 10^8) ∫ r^3 dr from 1.8 x 10^-3 to 2 x 10^-3 

= [(6π x 10^8)/4][16 - 1.8^4][10^-12] = 2.59 x 10^-3 amps.