Ampère's Law — Question 9

Question 9

A measured azimuthal field reconstructs the enclosed current and its density

Inside an axially symmetric long conductor, measurements give $B_\phi(r)=kr^2$ for $0\le r\le a$, with k positive. The field outside comes only from this conductor. Use vacuum permeability $\mu_0\simeq4\pi\times10^{-7}\,\mathrm{T\,m/A}$ where numerical values are needed.

1. Infer the axial current density as a function of r and the total current.

2. For k=.500 T/m squared and a=.0200 m, find I and the exterior field at r=.0600 m.

Question 9 - Solution

A measured azimuthal field reconstructs the enclosed current and its density

(a) Method and calculation

$$I_{\mathrm{enc}}(r)=\frac{2\pi rB_\phi}{\mu_0}=\frac{2\pi kr^3}{\mu_0},\quad dI=J_z(r)2\pi r\,dr.$$$$\boxed{J_z(r)=\frac{3kr}{\mu_0},\quad I=\frac{2\pi ka^3}{\mu_0}}.$$

(b) Interpretation and further calculation

$$I=\frac{2\pi(.500)(.0200)^3}{4\pi\times10^{-7}}=\boxed{20.0\,\mathrm A}.$$$$B(.0600)=\frac{\mu_0(20.0)}{2\pi(.0600)}=\boxed{66.6667\,\mu\mathrm T}.$$

Physical check

Differentiating enclosed current, rather than simply dividing total current by area, recovers the nonuniform density.