Question

A current $I$ flows in a long single-layer solenoid with cross-sectional radius $R$. The number of turns per unit length of the solenoid equals $n$. Find the limiting current at which the winding may rupture if the tensile strength of the wire is equal to $F_{lim}$.

Answer 1

The magnetic induction $B$ in the solenoid is given by $B_0={\mu }_{0}nI$. The force on an element $dl$ of the current carrying conductor is,
$dF=\frac{1}{2}{\mu }_{0}nIIdl=\frac{1}{2}{\mu }_{0}n{I}^{2}dl$
This is radially outwards. The factor $\frac{1}{2}$ is explained above.
To relate $dF$ to the tensile strength $F_{lim}$ we proceed as follows. Consider the equilibrium of the element $dl$. The longitudinal forces $F$ have a radial component equal to,
$dF=2F\sin \frac{d\theta }{2}=Fd\theta$
Thus using $dl=Rd\theta$, $F=\frac{1}{2}{\mu }_{0}n{I}^{2}R$
This equals $F_{lim}$ when, $I=I_{lim}=\sqrt{\frac{2F_{lim}}{{\mu }_{0}nR}}$
Note that $F_{lim}$, here, is actually a force and not a stress.