Asked Oct 7, 2019

A 0.510-mm-diameter silver wire carries a 30.0 mA current. What is the electric field in the wire?


Expert Answer

Step 1

Given values:

Radius of silver wire, r = diameter/2 = 0.51/2 =0.255 mm =0.255 x 10-3m = 2.55 x 10-4 m

Current in the wire, I = 30.0 mA = 30 x 10-3 A

Resistivity of silver = 1.59 × 10−8 Ω m

Step 2

The relation between current density and electric field:


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J =GE Where J=Current density = Current Area А 1 1 G=ConductivityResistivity = ! E Electricfield I -E A P pI Е- A pI

Step 3

 Plugging in the...


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1.59 x 10 x30 x 10 E = 3.14x(2.55x10) 47.7x 1011 20.417x 10 = 2.33 x 103 N/c


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