A coaxial cable is made from a cylindrical inner conductor with radius 1.5 mm th

Question

A coaxial cable is made from a cylindrical inner conductor with radius 1.5 mm that is enclosed by a concentric cylindrical conducting tube of inner radius 3.0 mm and outer radius 4.0 mm. The inner conductor delivers a current of 1.0 A to an electrical load while the outer conductor returns that same current to complete a circuit. Assume that the currents are uniformly distributed across the conducting areas. What is the magnitude of the magnetic field B at a distance of 1.9 mm from the axis of the cable? Express your answer using two significant figures. What is the magnitude of the magnetic field B at a distance of 4.6 mm from the axis of the cable? Express your answer using two significant figures. What is the magnitude of the magnetic field B at a distance of 0.40 mm from the axis of the cable (inside the center conductor)? Express your answer using two significant figures.

Explanation / Answer

from maxwell's equations:

integration of H.dl=total current enclosed

where H=magnetic field strength

and magnetic field density B can be found by B=mu*H

where mu=magnetic permeability of free space=4*pi*10^(-7)

part A:

consider a coaxial cylinder of radius 1.9 mm .

then total current enclosed within that surface=current flowing in the inner conductor=1 A

hence if magnetic field strength is H, then H*2*pi*1.9*0.001=1

==>H=1/(2*pi*1.9*0.001)=83.7657

==>B=mu*H=1.05*10^(-4) T

part B:

as 4.6 mm is outside the entire cable, net current enclosed=1-1=0

hence magnetic field outside the cable=0

hence magnetic field at 4.6 mm is 0.

part C:

at 0.4 mm, current enclosed=(total current/total area)*area of radius 0.4 mm

=(1/(pi*0.0015^2))*pi*0.0004^2=0.0711 A

hence using maxwell's equation,

H*2*pi*0.0004=0.0711

==>H=0.0711/(2*pi*0.0004)=28.2897

==>B=mu*H=4*pi*10^(-7)*28.2897=3.56*10^(-5) T

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