Exercise 4.1.15: Another given field

With $H_{\varphi }(\rho )$ only, $J=\nabla \times H=\frac{1}{\rho }\frac{d(\rho H_{\varphi })}{d\rho }{a}_z=\frac{1}{\rho }\frac{d}{d\rho }\left(\frac{H_0{\rho }^3}{b^2}\right)a_z$:

This is the current density of Home exercise 4.1.5 with $J_0=3H_0/b$.

From the source: $I={\int }_0^{2\pi }{\int }_0^b\frac{3H_0\rho }{b^2}\rho d\rho d\varphi =\frac{2\pi \cdot 3H_0}{b^2}\frac{b^3}{3}=2\pi b{H}_0$. From the field on the circle $\rho =b$: $I=\oint H\cdot dl=H_0\cdot 2\pi b$, the same.

For $\rho >b$, $I_{\mathrm{enc}}=2\pi b{H}_0$, so $H_{\varphi }=\frac{2\pi b{H}_0}{2\pi \rho }$.

Numbers: $I=12.6$ A, $J=1.5\times 10^6$ A/m${}^2$ at the surface and half of that at $\rho =b/2$.

Units: $H_0/b$ is A/m${}^2$. Any $H_{\varphi }(\rho )$ that is finite at $\rho =0$ and continuous at $\rho =b$ is a possible magnetostatic field, and the source is read off from the curl. A field like $H_{\varphi }\propto 1/\rho$ down to the axis, on the other hand, would require a line current on the axis ($J$ singular there).