Ampère's Law — Question 3

Question 3

A coaxial return current confines the field despite a nonzero inner current

An ideal infinite coaxial cable has a solid inner conductor of radius a carrying +I uniformly along z. A concentric outer conductor occupies b less than r less than c and carries -I uniformly, where b>a. All media have permeability mu0. Use vacuum permeability $\mu_0\simeq4\pi\times10^{-7}\,\mathrm{T\,m/A}$ where numerical values are needed.

1. Derive B(r) in all four radial regions.

2. Find the radius within the outer conductor at which half the return current is enclosed. Is the field there half of B(b)?

Question 3 - Solution

A coaxial return current confines the field despite a nonzero inner current

(a) Method and calculation

$$I_{\mathrm{enc}}=\begin{cases}Ir^2/a^2,&r<a,\\I,&a<r<b,\\I(c^2-r^2)/(c^2-b^2),&b<r<c,\\0,&r>c.\end{cases}$$$$\boxed{\vec B(r)=\frac{\mu_0I_{\mathrm{enc}}(r)}{2\pi r}\hat\phi};\quad\vec B=0\text{ outside by cylindrical symmetry}.$$

(b) Interpretation and further calculation

$$\frac{r_h^2-b^2}{c^2-b^2}=\frac12\Rightarrow\boxed{r_h=\sqrt{(b^2+c^2)/2}}.$$$$B(r_h)=\frac{\mu_0I}{4\pi r_h},\quad\boxed{B(r_h)/B(b)=b/(2r_h)<1/2}.$$

Physical check

The return-current fraction sets enclosed current, but the circumference factor also changes with radius; both enter the field.