What pressure does that the laterial surface of a long staraight solenoid with `n` turns per unit length experience when a current `I` flows throguh it?

Explanation

When self-forces are involved, a typical factor of `(1)/(2)` comes into play. For example, the force on a current carrying straight wire in a magnetic induction `B` is `Bil`. If the magnetic induction `B` is due to the current itself then the force can be writtern as,
`F = int_(0)^(I) B (r') dr l`
If `B (r') alpha I'`, then thus becomes, `F = (1)/(2) B(f) I L`
In the present case, `B (f) = mu_(0) n I` and this acts on `nI` ampere turns per unit lenght so, pressure `p = (F)/(Area) = (1)/(2) mu_(0) n (I xx nI xx1 xx l)/(1 xx l) = (1)/(2) mu_(0) n^(2) l^(2)`