The coaxial cable shown in the figure ( Figure 1 ) consists of a solid inner con
ID: 1284152 • Letter: T
Question
The coaxial cable shown in the figure (Figure 1) consists of a solid inner conductor of radius a and a hollow outer conductor of inner radius b and thickness c. The two carry equal but opposite currents I, uniformly distributed.
Find an expression for the magnetic field strength as a function of radial position r beyond the outer conductor.
Express your answer in terms of the variables I, a, b, c, r, and constant ?0 (u0).
The coaxial cable shown in the figure (Figure 1) consists of a solid inner conductor of radius a and a hollow outer conductor of inner radius b and thickness c. The two carry equal but opposite currents I, uniformly distributed. Find an expression for the magnetic field strength as a function of radial position r beyond the outer conductor. Express your answer in terms of the variables I, a, b, c, r, and constant ?0 (u0).Explanation / Answer
This is all about Ampere's Law which is difficult to type here but it's
integral of B . dL = mu_0I
For the inner conductor, imagine an Amperian loop inside the wire (just a circle inside the wire), at a distance r from the centre.
Integral of B.dL is just B(2pi r) since an infinitesimal amount of this loop is always in the same direction as the B field and by symmetry the B field is constant.
Now I inside this first wire depends on the amount of wire you have included - if your amperian loop encompasses half the wire you have half the current. The amount of current you have included depends on the area of the loop.
ie. your "mu_0 x I" will be mu_0 x I (pir^2)/(pia^2) = mu_0 I (r/a)^2
So rearranging your Ampere's Law expression for B you get
B = (mu_0 I r^2) / (2pi r a^2)
B = mu_0 I r / (2pi a^2)
Now outside the inner conductor your enclosed current is always I, and the left hand side of Ampere's law is just the same as before so this is easy
B(2pir) = mu_0 I
B = mu_0I / (2pir)
For the outer conductor the left hand side is still the same! But now the current enclosed decreases as r increases because the current is flowing in the opposite direction and it is the net current enclosed in your Amperian loop that matters.
The current enclosed is I at r = b, and 0 at r = b+c
So the right hand side is mu_0 I (1 - fraction of outer wire included)
The fraction of the outer wire included is the area of your amperian loop at a distance r, minus the area of the amperian loop up to the point b (so you're just including an anulus) all divided by the area of the cross section up to the edge of c minus the area of the loop up to b
So that's a pretty horrible expression but it's right, so working through this fraction of the area becomes (r^2 - b^2) / ((b+c)^2 - b^2)
So rearranging your expression you get
B = {mu_0 I (1 - [(r^2 - b^2) / ((b+c)^2 - b^2)] } / 2pir
Yup, hideous, but it's nicer if you write it on a piece of paper. And you can investigate it and see that it does match the required boundary conditions.