Irodov 3.308: Eddy Field Beyond the Winding – Amit & Deepali Sinha Foundation

Problem Irodov 3.308

A long straight solenoid has circular cross-section of radius $a$ and $n$ turns per unit length. The current in its winding changes at a constant rate $\dot{I}$ (in A/s). Find the magnitude of the induced (eddy) electric field as a function of the distance $r$ from the solenoid’s axis, both inside and outside the winding, and sketch a graph of this dependence.

Cross-section of the solenoid, radius a: the field B inside points out of the page and grows, since the current grows. — Cross-section of the solenoid, radius $a$ : the field $\vect B$ inside points out of the page and grows, since the current grows.

Given

$a$
$n$
$\dot{I} = \text{const}$

Find

E(r)

for

r < a

and

r > a

, and its graph

Solution

Step 1The changing field inside a long solenoid

Inside a long solenoid the field is uniform and axial, and outside it is zero. Differentiating in time with the current growing at the rate $\dot I$ :

B = \mu_0 n I \quad\Longrightarrow\quad \dot B = \mu_0 n \dot I \tag{1}

This rate is the same at every point of the cross-section and does not change in time.

Step 2Faraday’s law on a coaxial circle

Take a circle of radius $r$ centred on the axis, in a cross-sectional plane. With $E$ tangent and uniform on it, the circulation is $2\pi r E$ , and Faraday’s law equates it to minus the rate of change of the magnetic flux through the circle:

\oint \vect E\cdot d\vect l = 2\pi r\,E = -\frac{d\Phi}{dt} \tag{2}

The minus sign fixes the sense of circulation (Lenz: clockwise as seen with the growing $\vect B$ pointing at the viewer); for the magnitude only $|d\Phi/dt|$ matters.

Eddy field lines are closed circles about the axis, inside and outside the winding. E is drawn tangent at one point of each, its length set by E(r) there. The glyphs mark the growing B, nonzero inside the winding only. — Eddy field lines are closed circles about the axis, inside and outside the winding. $\vect E$ is drawn tangent at one point of each, its length set by $E(r)$ there. The glyphs mark the growing $\vect B$ , nonzero inside the winding only.

Step 3Inside the winding,

r < a

The circle encloses an area $\pi r^2$ of the uniform changing field, so using (1) the flux changes at the rate

\left|\frac{d\Phi}{dt}\right| = \pi r^2 \dot B = \pi r^2 \mu_0 n\dot I

Dividing by the circumference $2\pi r$ , as (2) requires:

E = \frac{\pi r^2 \mu_0 n\dot I}{2\pi r} \tag{3}

Step 4Outside the winding,

r > a

Now the circle encloses the whole cross-section $\pi a^2$ , and beyond $r = a$ there is no field to add, so the rate of change of the flux stops growing with $r$ :

\left|\frac{d\Phi}{dt}\right| = \pi a^2 \dot B = \pi a^2 \mu_0 n\dot I

Dividing again by $2\pi r$ through (2):

E = \frac{\pi a^2 \mu_0 n\dot I}{2\pi r} \tag{4}

Step 5The field at every distance

Cancelling $\pi$ and one power of $r$ in (3), and $\pi$ in (4), gives the two branches together:

E = \begin{cases} \dfrac{\mu_0 n\dot I\, r}{2}, & r < a\\[8pt] \dfrac{\mu_0 n\dot I\, a^2}{2r}, & r > a \end{cases}

(5)

Step 6The graph

The two branches meet at $r = a$ , where both equal the largest value $E_{\mathrm{m}} = \mu_0 n\dot I a/2$ . The field rises linearly from zero on the axis to this peak at the winding, then falls off as $1/r$ outside, with a corner (not a jump) at $r = a$ .

The eddy field magnitude against distance from the axis: a straight line up to the winding at r = a, where it peaks at E_m = μ_0 n I a/2, then a 1/r decay outside. — The eddy field magnitude against distance from the axis: a straight line up to the winding at $r = a$ , where it peaks at $E_{\mathrm{m}} = \mu_0 n\dot I a/2$ , then a $1/r$ decay outside.

Linear rise to a peak at

r = a

, then a

1/r

fall.

Check

At $r = a$ both branches of (5) give $\mu_0 n\dot I a/2$ , so $E$ is continuous, as it must be, since the flux through a circle changes smoothly as the circle crosses the winding. Units: $\mu_0 n\dot I$ is $\dot B$ in T/s, which is $\text{V}/\text{m}^2$ , and multiplying by a length gives V/m, a field strength.

Remark

Outside the winding $B$ is zero at every point, yet the eddy field there is not: it is fixed by the flux enclosed by the circle, not by the field where the circle runs. A ring of wire threaded over a long solenoid therefore has the same induced emf, $\pi a^2\mu_0 n\dot I$ , whatever its radius, which is how a transformer couples its windings through the core alone.