A long, cylindrical conductor is solid throughout and has a radius R . Electric

Question

A long, cylindrical conductor is solid throughout and has a radiusR. Electric charges flow parallel to the axis of thecylinder and pass uniformly through the entire cross section. Thearrangement is, in effect, a solid tube of currentI0. The current per unit cross-sectional area(i.e., the current density) isI0(R2). UseAmpère's law to show that the magnetic field inside theconductor at a distance r from the axis is0I0r/(2R2).(Hint: For a closed path, use a circle of radius rperpendicular to and centered on the axis.

Explanation / Answer

consider a circle of radius r perpendicular to and centered on theaxis,use the Ampere's law for closed path that is the circle: o I= B.2.r compute I : I=Io /(R^2) ..r^2=Io .(r/R)^2 => B=o .Io .(r/R)^2 /(2r)=o .Io .r/(2R^2)

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