A very long coaxial cable consists of a solid cylindrical inner conductor of rad
ID: 1874467 • Letter: A
Question
A very long coaxial cable consists of a solid cylindrical inner conductor of radius R1 surrounded by an outer cylindrical conductor with inner radius R2 and outer radius R3. The region between the two conductors is filled with a waxlike insulating material to keep the conductors from touching each other.
Part A If the inner and outer conductors carry equal currents I in opposite directions, use Ampère's law to derive an expression for the magnetic field as a function of r (the distance from the central axis) at points between the two conductors (R1R3).
Part B
If the inner and outer conductors carry equal currents I in opposite directions, use Ampère's law to derive an expression for the magnetic field as a function of r (the distance from the central axis) at points outside the cable (r>R3).
Explanation / Answer
Using Ampere's Law, this problem becomes very simple. If you draw your Amperian loop as a circle around the
center cable, the field is perfectly symmetrical. The enclosed current will just be I, so Ampere's law is just:
B*2pi*r = uo*I
Solving this for B,
B = uo*I / 2pi*r
Using the right hand rule, B is in the theta direction (curls around the axis of the cable). This equation is only valid for
R1 < r < R2.
Outside of the outer conductor, total enclosed current is zero. That is because the two enclosed currents are equal in
magnitude but opposite in direction. Therefore, adding them gives you a result of zero. So for r > R3:
B*2pi*r = 0
B = 0
For each of the cases above, the 2pi*r is just the circumference of the amperian loop. You can use this value for the
integral in Ampere's Law because the wire has cylindrical symmetry.