Irodov 3.263: The Field Inside Pushes the Winding Out

Problem Irodov 3.263

A long straight solenoid is wound with $n$ turns per unit length and carries a current $I$ . What pressure acts on its lateral (side) surface?

A long straight solenoid with n turns per unit length carrying the current I. — A long straight solenoid with $n$ turns per unit length carrying the current $I$ .

Given

$n$ turns per unit length
current $I$
long straight solenoid

Find

the pressure on the lateral surface

Solution

Step 1The field inside and outside

The same solenoid with its field: B runs along the axis inside, in the right-hand sense of the current, and vanishes outside. — The same solenoid with its field: $\vect B$ runs along the axis inside, in the right-hand sense of the current, and vanishes outside.

Far from the ends the field of a long solenoid is uniform inside and zero outside. A rectangular Amperian loop with one long side inside along the axis and the other outside encloses $n$ turns per unit length of its side, so Ampere’s law gives the familiar

B = \mu_0 n I \tag{1}

The winding is a cylindrical current sheet: per unit length of the solenoid it carries the current of $n$ turns, so its surface current is

i = n I \tag{2}

Step 2The field that acts on the winding

The field jumps from $B$ inside to zero outside across the winding. A small patch of the winding is a plane current sheet on its own scale, whose own field has magnitude $\mu_0 i/2$ and reverses across it; the rest of the winding supplies a field $B'$ that is smooth across the patch. Inside and outside, by (2),

\begin{aligned} B &= B' + \frac{\mu_0 nI}{2}\\ 0 &= B' - \frac{\mu_0 nI}{2} \end{aligned}

Either line gives, with (1),

B' = \frac{B}{2} = \frac{\mu_0 n I}{2}

(3)

Step 3The pressure

The current in the winding runs round the axis and $\vect B'$ runs along it, so they are perpendicular and the force per unit area is $i\,B'$ . Its sense is that of $\vect i\times\vect B'$ , which by the right-hand rule points away from the axis: the winding is pushed outward. Substituting (2) and (3):

p = i\,B' = nI\cdot\frac{\mu_0 n I}{2}

p = \frac{\mu_0 n^2 I^2}{2}

(4)

The pressure acts outward, stretching the winding.

Check

Virtual work at fixed current gives the same value. The field energy $B^2/2\mu_0$ fills the volume $\pi R^2\ell$ of a length $\ell$ of a solenoid of radius $R$ . Widening the radius by $dR$ at fixed $I$ keeps $B$ fixed, and at fixed current the work done by the magnetic forces on the winding equals the increase of the field energy (the source supplies twice that):

p\cdot 2\pi R\ell\,dR = \frac{B^2}{2\mu_0}\,2\pi R\ell\,dR,

so $p = B^2/2\mu_0$ , which with (1) is $\mu_0 n^2I^2/2$ , the result (4). The outward sense also agrees with the rule that antiparallel currents repel: the two sides of each turn carry opposite currents.

Remark

The magnetic pressure equals the field energy density, $B^2/2\mu_0$ , and it grows as the square of the field: at $B = 1$ T it is about $4\times 10^5$ Pa, four atmospheres, and at $10$ T it is a hundred times that, which is why the windings of strong electromagnets need a steel jacket or a self-supporting design. Compare the tube with an axial current (3.262): there the field is outside, and the same pressure $B^2/2\mu_0$ pushes the wall inward, toward the field-free side. In both cases the field pushes the current sheet toward the side where there is no field.