Irodov 3.276: B Flows On, H Steps Down – Amit & Deepali Sinha Foundation

Problem Irodov 3.276

An infinitely long straight solenoid carries a current. One half of it, the region $x > 0$ along its axis, is filled with a magnetic substance of permeability $\mu$ ; the other half, $x < 0$ , is empty, and $x$ is measured along the axis from the face of the filling at $O$ . Sketch how the magnetic induction $B$ , the field strength $H$ and the magnetization $J$ vary along the axis as functions of $x$ .

The long solenoid in axial section, cut off at both drawn ends, carrying the current I. The magnetic substance (shaded, μ) fills it for x > 0, from the face at O; for x < 0 it is empty. — The long solenoid in axial section, cut off at both drawn ends, carrying the current $I$ . The magnetic substance (shaded, $\mu$ ) fills it for $x > 0$ , from the face at $O$ ; for $x < 0$ it is empty.

Given

infinitely long solenoid
filled for $x > 0$ with permeability $\mu$
empty for $x < 0$
$\mu > 1$ , a paramagnet or iron (assumed)

Find

plots of

B

H

and

J

on the axis against

x

Solution

Step 1Far from the face

Let the winding have $n$ turns per unit length carrying the current $I$ , and radius $a$ . Far from $O$ on either side the solenoid looks infinitely long and uniformly filled (or empty). There the field is uniform and axial inside and zero outside, so the circulation of $\vect{H}$ round a rectangle with one long side inside, length $\ell$ , and one outside equals the free current $nI\ell$ it encloses. On both sides, and in both media,

H = nI \qquad (|x| \gg a)

With $B = \mu\mu_0H$ and $J = (\mu - 1)H$ this gives the far values

\begin{aligned} x \ll -a:&\quad B = \mu_0 nI, \quad H = nI, \quad J = 0\\ x \gg a:&\quad B = \mu\mu_0 nI, \quad H = nI, \quad J = (\mu - 1)\,nI \end{aligned} \tag{1}

Step 2What the face does to each quantity

At $O$ the field is along the axis, normal to the face. The normal component of $\vect{B}$ is continuous across any surface, so $B$ has a single value $B(0)$ at the face. $H$ and $J$ then jump there, because the medium changes:

H(0^-) = \frac{B(0)}{\mu_0}, \qquad H(0^+) = \frac{B(0)}{\mu\mu_0}, \qquad J(0^+) = (\mu - 1)\,H(0^+) \tag{2}

So $H$ drops by the factor $\mu$ crossing the face, and $J$ jumps from zero to $J(0^+)$ .

Step 3Which way the curves bend

The field of the winding alone is $\mu_0nI$ everywhere on the axis. The filling adds the field of its molecular currents, which circle the filling’s surface in the same sense as the winding (for $\mu > 1$ ) and exist only for $x > 0$ . Each ring of them adds a positive axial field everywhere on the axis, more strongly the nearer it is, so, taking the magnetization not to decrease into the filling (which the estimate of Step 4 bears out), their total grows steadily from nothing far to the left to its full value $(\mu - 1)\mu_0nI$ far to the right. $B$ therefore rises monotonically, without a jump, from $\mu_0nI$ to $\mu\mu_0nI$ , and its value at the face lies strictly between:

\mu_0 nI < B(0) < \mu\mu_0 nI

By (2) this puts $H(0^-)$ above $nI$ and $H(0^+)$ below it. So $H$ rises above $nI$ as $x$ approaches the face from the left, drops at the face to below $nI$ , and climbs back to $nI$ deep inside the filling; $J$ , following $H$ on the right, jumps up at the face to a value below $(\mu - 1)nI$ and then rises to it.

Step 4An estimate for the plot

To draw the curves, keep only that pole: the face of the filling carries a uniform effective pole density $-J(0^+)$ over a disc of radius $a$ (the poles spread through the filling where $J$ still grows are neglected; in this estimate their total is $(\mu - 1)/2$ times the face’s, equal to it at $\mu = 3$ and larger beyond, so the estimate fixes the shapes, not the exact values). On the axis such a disc adds to $H$ a field of half the pole density times $1 - |x|/(x^2 + a^2)^{1/2}$ , pointing toward the face, so with $J_f = J(0^+)$

\begin{aligned} H(x) &= nI + \frac{J_f}{2}\left(1 - \frac{|x|}{\sqrt{x^2 + a^2}}\right) \qquad (x < 0)\\[2pt] H(x) &= nI - \frac{J_f}{2}\left(1 - \frac{x}{\sqrt{x^2 + a^2}}\right) \qquad (x > 0) \end{aligned} \tag{3}

The face value must be consistent with the material’s response, $J_f = (\mu - 1)H(0^+)$ . Setting $x = 0^+$ in (3),

J_f = (\mu - 1)\left(nI - \frac{J_f}{2}\right)

and solving for $J_f$ ,

J_f = \frac{2\,(\mu - 1)}{\mu + 1}\,nI

The two one-sided values of (3) at the face are then $H(0^-) = nI + J_f/2$ and $H(0^+) = nI - J_f/2$ , that is

H(0^-) = \frac{2\mu}{\mu + 1}\,nI, \qquad H(0^+) = \frac{2}{\mu + 1}\,nI

and the induction at the face, $\mu_0H(0^-)$ , is in this estimate

B(0) = \frac{2\mu}{\mu + 1}\,\mu_0 nI

(4)

and the three curves, drawn from (3) for $\mu = 3$ , are these:

B

is continuous at

x = 0

(with a kink);

H

and

J

jump there

Check

The estimate keeps $B$ continuous at the face, as it must, because the pole sheet makes $H$ jump by exactly $J_f$ : $\mu\mu_0H(0^+)$ from the inside is also $2\mu\mu_0nI/(\mu + 1)$ , the value (4) found from outside. That value lies between $\mu_0nI$ and $\mu\mu_0nI$ for every $\mu > 1$ , as Step 3 requires. The slope of $B$ on the axis does step up by the factor $\mu$ at the face, the kink in the plot: zero divergence ties it to the radial part of $\vect{B}$ , which is tangential to the face and so $\mu$ times larger inside, where $H_t$ is continuous. With $\mu = 1$ it gives $J_f = 0$ and $B = \mu_0nI$ , $H = nI$ everywhere: no filling, no structure. Far from the face, (3) returns $H = nI$ on both sides, the values of (1).

Remark

For large $\mu$ the estimate’s $B(0) \approx 2\mu_0nI$ is far below the $\mu\mu_0nI$ deep inside, so most of the rise of $B$ happens beyond the face, where the poles the estimate neglects carry it; the numbers are then rough, while the shapes stay those of the sketch. The pole on the face is also why a short iron rod in a coil is magnetized much less than a long one: its two faces carry poles whose field opposes the coil’s inside the rod, the demagnetizing field.