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Really hard Physics problem! Help ASAP

Learning Goal: To learn to apply the concept ofcurrent density and microscopic Ohm's law.

A "gauge 8" jumper cable has a diameter d of 0.326 centimeters. Thecable carries a currentIof 30.0 amperes. The electric fieldEin the cable is 0.062 newtons per coulomb.What is the material of the cable?
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Answer #1
WE KNOW THAT CURRENT DENSITY j = σE--1σ = SURFACE CHARGE DENSITYE = electric field = 0.062 newtons per coulombcurrent density j = I/AREAI = 30 AAREA = π(radius)2=π(0.163)2m2=3.14*(0.163)2m2radius = 0.326m/2 = 0.163 mj = 30/3.14*(0.163)2m2= 360A/m2from 1 360 A/m2 = σ*0.062 N/Cσ = 360/0.062 C/m2 = 5806C/m2by know ing charge density we can determine nature ofm,aterial(refer standard values of charge densities and see forσ = 5806C/m2 confirmwhich material it is)
answered by: miil
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Answer #2

muklo

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Answer #3

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answered by: tothemoon
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