Question

A long, cylindrical shell has an inner radius a and outer radius b and carries a current i parallel to the central axis. assume that within the material the current density is uniformly distributed. Use Ampere's law to find expressions for the magnetic field for (a) 0<r<a, (b) a<r<b (c) r>b. Draw sketches and carefully explain your work.

Answer 1

surface current density=current/total surface area

==>rho=I/(pi*(b^2-a^2))

part a:

for 0<r<a:

current enclosed=0

hence using ampere's law;

B*2*pi*r=mu*current enclosed=0

==>B=0

part b:

for a<r<b:

current enclosed=current density*surface area

=rho*pi*(r^2-a^2)

if magnetic field is B,

then using ampere's law:

B*2*pi*r=mu*current enclosed=mu*rho*pi*(r^2-a^2)

==>B=mu*rho*pi*(r^2-a^2)/(2*pi*r)

replacing rho with I/(pi*(b^2-a^2))

B=mu*I*(r^2-a^2)/(2*pi*r*(b^2-a^2))

for r>b:

total current enclosed=I

if magnetic field is B,

B*2*pi*r=mu*I

==>B=mu*I/(2*pi*r)

plot: