Ampère's Law — Question 10

Question 10

A cylindrical surface current produces a field jump that a volume current cannot

A long cylinder of radius a carries uniform axial volume current density J inside and an additional infinitesimally thin axial surface current $K\hat z$ on its boundary. K may be negative. There are no other currents. Use vacuum permeability $\mu_0\simeq4\pi\times10^{-7}\,\mathrm{T\,m/A}$ where numerical values are needed.

1. Find the fields immediately inside and outside the boundary and their difference.

2. Choose K so the exterior field vanishes everywhere. Find the remaining interior field.

Question 10 - Solution

A cylindrical surface current produces a field jump that a volume current cannot

(a) Method and calculation

$$B_\phi(a^-)=\frac{\mu_0Ja}{2},\quad I_{\mathrm{total}}=J\pi a^2+K(2\pi a).$$$$B_\phi(a^+)=\mu_0(Ja/2+K),\quad\boxed{B_\phi(a^+)-B_\phi(a^-)=\mu_0K}.$$

(b) Interpretation and further calculation

$$I_{\mathrm{total}}=0\Rightarrow\boxed{K=-Ja/2}.$$$$\boxed{B_\phi(r)=\mu_0Jr/2\ (r<a),\qquad B_\phi(r)=0\ (r>a)}.$$

Physical check

A surface-current sheet can sustain a discontinuity of tangential magnetic field; zero exterior field does not imply zero interior field.