Irodov 3.252: The Field That Bursts a Coil – Amit & Deepali Sinha Foundation

Problem Irodov 3.252

A coil carrying a current $I = 10$ mA sits in a uniform magnetic field with its axis along the field. The coil has a single layer of turns of radius $R = 30$ mm, wound from copper wire of diameter $d = 0.10$ mm (tensile strength of copper $\sigma_m = 3.0\times 10^8$ Pa). How strong must the external field be for the winding to break?

The single-layer coil with turns of radius R carrying the current I, its axis along the uniform external field B; the current is drawn in the sense that makes the coil's own field along B (assumed). — The single-layer coil with turns of radius $R$ carrying the current $I$ , its axis along the uniform external field $\vect B$ ; the current is drawn in the sense that makes the coil’s own field along $\vect B$ (assumed).

Given

$I = 10$ mA
$d = 0.10$ mm
$R = 30$ mm
copper, $\sigma_m = 3.0\times 10^8$ Pa (table value)
the coil’s own field along $\vect B$ , so the forces stretch the turns (assumed)

Find

the external

B

at which the winding breaks

Solution

Step 1The magnetic load on a turn

Each element $d\vect l$ of a turn lies in the plane perpendicular to $\vect B$ , so the Ampere force $I\,d\vect l\times\vect B$ is radial, of magnitude $IB\,dl$ . We take the coil’s own field along the external one (the stretching case, assumed): then, seen from the tip of $\vect B$ , the current circulates counter-clockwise and the force on every element points outward. The load per unit length of wire is

f = IB \tag{1}

Step 2The tension in the wire

Half a turn seen along the axis, B out of the page: the magnetic load d F pushes every element outward, and the tension T at the two cut ends holds the half turn in place. — Half a turn seen along the axis, $\vect B$ out of the page: the magnetic load $d\vect F$ pushes every element outward, and the tension $T$ at the two cut ends holds the half turn in place.

Cut a turn along a diameter and balance the half turn along the symmetry line. The element at the angle $\varphi$ from the cut, of length $R\,d\varphi$ , carries the outward force $fR\,d\varphi$ , whose component along the symmetry line is $fR\sin\varphi\,d\varphi$ ; the two cut ends each pull back with the tension $T$ :

2T = \int_0^{\pi} fR\sin\varphi\,d\varphi = 2fR

so with (1)

T = fR = IBR

(2)

Step 3The breaking field

The wire breaks when the tension reaches the strength of copper times its cross-section $\pi d^2/4$ :

IBR = \sigma_m\,\frac{\pi d^2}{4}

Solving for $B$ and substituting,

B = \frac{\pi\,(1.0\times 10^{-4}\text{ m})^2\,(3.0\times 10^{8}\text{ Pa})}{4\,(1.0\times 10^{-2}\text{ A})\,(3.0\times 10^{-2}\text{ m})}

B = \frac{\pi d^2\sigma_m}{4IR} \approx 7.9\text{ kT}

(3)

Check

Units: $\text{m}^2\cdot\text{N/m}^2$ over $\text{A}\cdot\text{m}$ is $\text{N}/(\text{A}\cdot\text{m})$ , a tesla. The tension (2) also follows by virtual work: let the turn’s radius grow by $dR$ ; the outward load does the work $f\cdot 2\pi R\,dR$ , and the wire, stretched by $2\pi\,dR$ , takes up $T\cdot 2\pi\,dR$ . Equating the two gives $T = fR$ again.

Remark

The answer is thousands of tesla, far beyond the strongest steady laboratory fields (about $45$ T), so a fine copper coil at $10$ mA is never in danger. Real high-field magnets reverse the balance: there the current is huge, the hoop tension $IBR$ is the design limit, and the windings are reinforced with steel or wound inside a strong outer shell for exactly this reason.