Irodov 3.262: Half the Jump Pushes the Wall – Amit & Deepali Sinha Foundation

Problem Irodov 3.262

A long cylindrical tube of radius $R$ with a thin wall carries a current $I$ along its length, flowing in the wall. What pressure does the wall of the tube experience?

The long thin-walled tube of radius R, carrying the current I along its wall. — The long thin-walled tube of radius $R$ , carrying the current $I$ along its wall.

Given

current $I$
radius $R$
thin wall, long tube

Find

the pressure on the wall

Solution

Step 1The field on the two sides of the wall

By symmetry $\vect B$ circles the axis and has the same magnitude all round. A circle of radius $r$ about the axis encloses the whole current when $r > R$ and none of it when $r < R$ , so Ampere’s law gives, just outside and just inside the wall,

B_{\mathrm{out}}\cdot 2\pi R = \mu_0 I, \qquad B_{\mathrm{in}} = 0 \tag{1}

The wall carries a current $i$ per unit length of its circumference, the whole current spread round the perimeter:

i = \frac{I}{2\pi R} \tag{2}

Step 2The field that acts on the wall

A small patch of the wall seen edge on, its current out of the page, the outside of the tube above. The patch alone produces the field B_p, the same size on both sides and opposite in sense. — A small patch of the wall seen edge on, its current out of the page, the outside of the tube above. The patch alone produces the field $\vect B_{\mathrm{p}}$ , the same size on both sides and opposite in sense.

The field jumps from $0$ to $B_{\mathrm{out}}$ across the wall, so it is not obvious which value pushes on the current in it. Split the field at a small patch of the wall into the patch’s own part $\vect B_{\mathrm{p}}$ and the part $\vect B'$ of all the other currents. The patch is a flat current sheet on its own scale, whose field has magnitude $\mu_0 i/2$ and reverses across it; $\vect B'$ is smooth across the patch.

\begin{aligned} B_{\mathrm{out}} &= B' + \frac{\mu_0 i}{2}\\ B_{\mathrm{in}} &= B' - \frac{\mu_0 i}{2} \end{aligned}

Adding the two lines eliminates the patch’s own field, and (1) supplies the values:

B' = \frac{B_{\mathrm{in}} + B_{\mathrm{out}}}{2}

B' = \frac{B_{\mathrm{out}}}{2} = \frac{\mu_0 I}{4\pi R}

(3)

Step 3The pressure

The force on the patch per unit area is the surface current times the field acting on it, $i\,B'$ , since the current along the axis is perpendicular to $\vect B'$ . Its direction is that of $\vect i\times\vect B'$ : the current runs along the axis and $\vect B'$ circles it, so by the right-hand rule the force points toward the axis, and the wall is squeezed. Substituting (2) and (3):

p = i\,B' = \frac{I}{2\pi R}\cdot\frac{\mu_0 I}{4\pi R}

p = \frac{\mu_0 I^2}{8\pi^2 R^2}

(4)

The pressure acts inward, compressing the tube.

Check

The pressure should equal the magnetic energy density on the side where the field is, $B_{\mathrm{out}}^2/2\mu_0$ . With (1),

\frac{B_{\mathrm{out}}^2}{2\mu_0} = \frac{1}{2\mu_0}\left(\frac{\mu_0 I}{2\pi R}\right)^2 = \frac{\mu_0 I^2}{8\pi^2R^2},

which is (4). The inward sense agrees with the rule that parallel currents attract: every strip of the wall carries current the same way as all the others. Units: $\mu_0 I^2$ is a force in newtons, so dividing by $R^2$ gives pascals.

Remark

This is the pinch effect: a current flowing along a conducting tube, or through a column of plasma, squeezes it with the pressure $B^2/2\mu_0$ of its own field. Using the full outside field $B_{\mathrm{out}}$ in $iB$ would double the answer; the field acting on any surface current is always the average of the fields on its two faces, which is the same factor $1/2$ that appears in the force on a charged conductor’s surface.