Displacement Current — Question 9

Question 9

A nonuniform displacement-current profile changes enclosed-current fractions

Between circular capacitor plates, the axial displacement-current density is $J_d(r)=J_0(1-r^2/a^2)$ for r<a. Assume axisymmetry and quasistatic fields.

1. Find total current and the fraction enclosed at radius a/2.

2. Find the magnetic field at a/2 in terms of total current.

Question 9 - Solution

A nonuniform displacement-current profile changes enclosed-current fractions

(a) Method and calculation

$$I=2\pi J_0\int_0^a(1-r^2/a^2)r\,dr=\pi J_0a^2/2.$$$$\boxed{I(r<a/2)/I=2(1/2)^2-(1/2)^4=7/16}.$$

(b) Interpretation and further calculation

$$B(a/2)\,2\pi(a/2)=\mu_0(7I/16).$$$$\boxed{B(a/2)=7\mu_0I/(16\pi a)}.$$

Physical check

A uniform-area estimate would give one-quarter; the center-weighted profile instead encloses seven-sixteenths.