Irodov 3.351: Displacement Current Around a Solenoid

Problem Irodov 3.351

A long straight solenoid of cross-sectional radius $R$ has $n$ turns per unit length and carries the alternating current $I = I_m\sin\omega t$ . Find the displacement current density as a function of the distance $r$ from the solenoid’s axis.

Cross-section of the solenoid, radius R: the winding carries the current I, drawn counterclockwise at the instant shown (sense chosen). — Cross-section of the solenoid, radius $R$ : the winding carries the current $I$ , drawn counterclockwise at the instant shown (sense chosen).

Given

$n$
$I = I_m\sin\omega t$
radius $R$
slowly varying field, $\omega R \ll c$ (assumed)

Find

j_d(r)

inside and outside the winding

Solution

Step 1The field inside the solenoid

At these frequencies (assumed $\omega R \ll c$ ) the field at each instant is that of a steady current: uniform and axial inside a long solenoid, zero outside it.

B = \mu_0 n I = \mu_0 n I_m\sin\omega t \tag{1}

Step 2The eddy electric field

Take a coaxial circle of radius $r$ , its sense that of the winding current, so that $\vect B$ threads it positively by the right-hand rule. With $E$ the field’s component along that sense, Faraday’s law equates the circulation to minus the rate of change of the flux through the circle:

E\cdot 2\pi r = -\frac{d\Phi}{dt}

Inside the winding the flux is $\pi r^2 B$ ; outside it stops growing at $\pi R^2 B$ . Dividing by $2\pi r$ and using (1):

E = -\frac{\mu_0 n}{2}\,\frac{dI}{dt}\times\begin{cases} r, & r < R\\[2pt] R^2/r, & r > R \end{cases} \tag{2}

Step 3The displacement current density

In vacuum the displacement current density is $\varepsilon_0\,\partial\vect E/\partial t$ , so it too runs along the circles. Differentiating (2) once more in time, with the second derivative of the current

\frac{d^2 I}{dt^2} = -\omega^2 I_m\sin\omega t

the inside branch becomes

j_d = -\frac{\varepsilon_0\mu_0 n r}{2}\,\frac{d^2 I}{dt^2} = \frac{\varepsilon_0\mu_0 n\omega^2 I_m r}{2}\,\sin\omega t

and the outside branch is the same with $r$ replaced by $R^2/r$ :

j_d = \frac{\varepsilon_0\mu_0 n\omega^2 I_m\sin\omega t}{2}\times\begin{cases} r, & r < R\\[2pt] R^2/r, & r > R \end{cases}

(3)

A positive $j_d$ runs along the winding current, so the displacement current circulates round the axis in phase with the current in the turns.

Step 4The graph

The amplitude rises linearly from zero on the axis to its largest value $j_{\mathrm{m}} = \varepsilon_0\mu_0 n\omega^2 I_m R/2$ at the winding, then falls as $1/r$ outside, with a corner, not a jump, at $r = R$ .

The amplitude of the displacement current density against distance from the axis: a straight line up to the winding at r = R, where it peaks at j_m, then a 1/r fall outside. — The amplitude of the displacement current density against distance from the axis: a straight line up to the winding at $r = R$ , where it peaks at $j_{\mathrm{m}}$ , then a $1/r$ fall outside.

Check

Units: $\varepsilon_0\mu_0 = 1/c^2$ , so (3) is $nI_m\cdot\omega^2 r/(2c^2)$ , which is $(\text{A/m})\cdot\text{m}^{-1} = \text{A/m}^2$ , a current density. Continuity: at $r = R$ both branches equal $j_{\mathrm{m}}$ , as they must, since the eddy field (2) is itself continuous there.

Remark

Compared with the winding’s own surface current $nI_m$ , the displacement current through a strip of the cross-section is of order $(\omega R/c)^2$ times smaller, which is why it is neglected in the solenoid’s own field: its own magnetic field is the first correction, making $B$ inside slightly nonuniform, and once $\omega R$ approaches $c$ the quasi-static picture fails altogether.