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 ruputure if the tensile strength of the wire is equal to `F_(l i m)`.

Explanation

The magenetic induction `B` is the solenoid is given by `B = mu_(0) nI`. The forces on an element `dI` of the current carrying conductor is,
`dF = (1)/(2) mu_(0) n II dI = (1)/(2) mu_(0) nI^(2) dI`
This is radially outwards. The factor `(1)/(2)` is explained above.
To realte `dF` to teh tenstie strenght `F_(lim)` we proceed as follows. Consider the equlibrium of the element `dI`. The longtitudinal forces `F` have a radial component equal to,
`dF = 2F "sin" (d theta)/(2) = F d theta`
Thus using `dl = R d theta, F = (1)/(2) mu_(0) nl^(2) R`
This equals `F_(lim)` when, `I = I_(lim) = sqrt((2 F_(lim))/(mu_(0) n R))`
Note that `F_(lim)`, here, is actually a force and not a stress.