3- The cross-section of an infinitely long circular cylinder- whose ra.docx

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3. The cross-section of an infinitely long circular cylinder, whose radius is a, is shown in the figure. A current, with a density of J (22-a2e [A/m2] flows through the cylinder. Find the magnetic field vector H in whole space. i Solution radial distance be r. current density=(r^2-a^2) along +ve z axs. for r<a: consider a small segment of length dr at radial distance r. cross section area=2*pi*r*dr current enclosed=current density*cross section area =2*pi*r*dr*(r^2-a^2) total current enclosed=integration of 2*pi*(r^3-a^2*r)*dr from 0 to r =2*pi*((r^4/4)-(a^2*r^2/2) =0.5*pi*(r^4-2*a^2*r^2) using ampere\'s law, line integral of magnetic field vector=current enclosed ==>H*2*pi*r=0.5*pi*(r^4-2*a^2*r^2) ==>H=0.25*(r^3-a^2*r) field vector is along counter clockwise direction. for r>a: current enclosed=0.5*pi*(r^4-2*a^2*r^2) with r=a =0.5*pi*(a^4-2*a^4) =-0.5*pi*a^4 using ampere\'s law, H*2*pi*r=-0.5*pi*a^4 hence H=-0.25*a^4/r field vector is along counter clockwise direction .

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for r>a:
current enclosed=0.5*pi*(r^4-2*a^2*r^2) with r=a
=0.5*pi*(a^4-2*a^4)
=-0.5*pi*a^4
using ampere's law,
H*2*pi*r=-0.5*pi*a^4
hence H=-0.25*a^4/r
field vector is along counter clockwise direction

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