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Consider a silver wire with a cross-sectional area of 1mm2carrying0.3Aof current. The conductivity of silver is6.3×107(A/m2)(V/m). Calculate the magnitude of the electric field required to drive this current through the wire.

Short Answer

Expert verified

4.76×10-3V/m

Step by step solution

01

Given Data

AreaA=1mm2CurrentI=0.3AConductivityσ=6.3×107A/m2V/m

02

Concept

The force as per the test charge is known as the magnitude of electric field.

03

Calculate the magnitude of the electric field 

First we have to find the resistivity,

σ=1ÒÏwhere,ÒÏisresistivityÒÏ=1σ=16.3×107=1.58×10-8Ω³¾

Electric Field,

localid="1662180919419" E=ÒÏ1A(whereEiselectricfield)=1.58×10-8×0.3A1×10-6m2=4.76×10-3V/m

Hence, the magnitude of the electric field required to drive this current through the wire is 4.76×10-3V/m

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