Problem 2

 
2) EQUAL but OPPOSITE currents I travel in the inner and outer wires of a coaxial cable. The radius of the inner wire is R₁. The inside radius of the outer wire is R₂ and its outside radius is R₃. As a function of the distance from the central axis, find the magnetic field (a) inside the inner wire (10 pts); (b) in the region between the two wires (7 pts); (c) outside the outer wire (8 pts). Assume that the current is uniformly distributed in the two wires.  

Answer:
The magnetic field in the wire as a function of the current is given by
$\oint \vec{B} \cdot d\vec{s} = \mu_{0} I_{enclosed}$
Doing the line integral since we have circular symmetry, we find that
$B(r) = \frac{\mu_{0}}{2\pi r} I_{enclosed}$
So, the question comes down to what the enclosed current is. For the region outside the wire r>R₂, we find that
I_enclosed = I + -I
I_enclosed = 0
So, similarly,
B(r) = 0
So, outside the outer radius of the wire, the magnetic field is zero.
For the region in between the central wire and the outer shell, we see that
I_enclosed = I
So, that the magnetic field falls as $\frac{1}{r}$
$B(r) = \frac{\mu_{0}I}{2\pi r}$
Inside the central wire we have to find the enclosed current. We do this by setting up a ratio of the amount of cross sectional area bounded by a circle of radius r to the total amount of cross sectional area in the center wire. This is similar to saying that all the current I flows through an area $\pi R_{1}^{2}$, so if we look at a smaller amount of area, a fraction of the current will be flowing throught that area. In equation format, we are saying,
$I_{enclosed} = \frac{\pi r^{2}}{\pi R_{1}^{2}} I$
So if we plug this in to find the magnetic field for r<R₁, we find that
$B(r) = \frac{\mu_{0}}{2\pi r} \frac{\pi r^{2}}{\pi R_{1}^{2}} I$
$B(r) = \frac{\mu_{0}I \pi r^{2}}{2\pi^{2} r R_{1}^{2}}$
$B(r) = \frac{\mu_{0}I}{2\pi R_{1}^{2}} r$
So, the magnetic field is linear in r in the region r<R₁.