Irodov 3.264: A Solenoid Bursts Like a Boiler – Amit & Deepali Sinha Foundation

Problem Irodov 3.264

A long solenoid is wound in a single layer, with $n$ turns per unit length on a cross-section of radius $R$ , and carries a current $I$ . The wire breaks when the tension in it reaches $F_{lim}$ . At what current may the winding rupture?

A long single-layer solenoid of radius R with n turns per unit length carrying the current I. — A long single-layer solenoid of radius $R$ with $n$ turns per unit length carrying the current $I$ .

Given

radius $R$
$n$ turns per unit length
single layer
wire breaks at tension $F_{lim}$

Find

the current

I_{lim}

at which the winding ruptures

Solution

Step 1The magnetic pressure on the winding

By symmetry, far from the ends the field is axial, uniform inside and zero outside; Ampere’s law round a rectangle with a side of length $\ell$ inside along the axis and the opposite side outside gives

B\ell = \mu_0 n\ell I

so $B = \mu_0 nI$ . The winding is a current sheet carrying $nI$ per unit length, and the field acting on it is the average of the fields on its two faces, $B/2$ , because a patch of the sheet exerts no force on itself. Their product is the outward force per unit area:

p = nI\cdot\frac{B}{2} = nI\cdot\frac{\mu_0 nI}{2}

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

(1)

Step 2The tension in one turn

Half of one turn as a free body (the wire drawn exaggerated): the magnetic pressure p pushes it outward all along, and the other half pulls on it with the wire's tension F at each cut. — Half of one turn as a free body (the wire drawn exaggerated): the magnetic pressure $p$ pushes it outward all along, and the other half pulls on it with the wire’s tension $F$ at each cut.

Cut one turn along a diameter and keep the upper half. The turn occupies an axial length $1/n$ of the winding, so an arc element $R\,d\theta$ of it, at angle $\theta$ from the cut plane, receives the outward force $p\,R\,d\theta/n$ . By symmetry only the component normal to the cut plane survives, and it is balanced by the tension $F$ pulling at each of the two cut ends:

2F = \int_0^{\pi}\frac{pR}{n}\,\sin\theta\,d\theta = \frac{pR}{n}\Big[-\cos\theta\Big]_0^{\pi} = \frac{2pR}{n}

With (1) this is

F = \frac{pR}{n} = \frac{\mu_0 n^2 I^2}{2}\cdot\frac{R}{n}

F = \frac{\mu_0 n R I^2}{2}

(2)

Step 3The current that breaks the wire

The wire ruptures when this tension reaches $F_{lim}$ . Setting $F = F_{lim}$ in (2):

\frac{\mu_0 n R I_{lim}^2}{2} = F_{lim}

and solving for the current,

I_{lim} = \sqrt{\frac{2F_{lim}}{\mu_0 n R}}

(3)

Check

Units: $\mu_0$ is in N/A $^2$ , $n$ in 1/m and $R$ in m, so $\mu_0 nR$ is in N/A $^2$ and $F_{lim}/\mu_0 nR$ is in A $^2$ : the root is a current. Trends: a wider coil ( $R$ larger) has more tension per turn at the same field, so it bursts at a smaller current; a stronger wire takes a larger one, growing only as $F_{lim}^{1/2}$ because the pressure grows as $I^2$ .

Remark

The relation $F = pR/n$ is the hoop formula of a thin-walled boiler: the tension per unit length of wall, $nF$ , equals the pressure times the radius. The only difference here is that the pressure comes from the field, $B^2/2\mu_0$ , and the pull $2F$ on the half turn is the force on a straight wire along its chord, of length $2R$ , carrying $I$ across the average field $B/2$ , which gives $2F = I\cdot 2R\cdot B/2$ once more.