I'm having problems findind the best way to maximize the total area of my solenoid, of course given $l$ as the costant value of the total lenght of my wire.
I can't find it on the internet, so I'm asking for it here.
Let $S = \pi r^2$ be the (approximate) surface of a single winding of my coil and $N = \frac{l}{2 \pi r}$ the number of windings, then $S_{tot} = SN$, so $S_{tot} = \pi r^2 \frac{l}{2 \pi r} = \frac{l r}{2}$ which, given $l$ constant, is a straight line and means that as I get my solenoid bigger in radius, the total area gets the maxinum value, which does not seem right.
Where am I mistaken?
(It may be possible that I'm not evaluating the fact that in magnetic induction happens better for high $\frac{d \phi}{dt}$ so actually the single-winding has to have a maximal value of diameter as the lenght of my magnet, and that's why all solenoids happen to have many many turns with smaller surfaces)
PS: my solenoid is the usual "spring-like" coil/solenoid as shape
Your thinking is correct. Think of a simple case: you have a wire of length $4\pi.$ If you make the radius as big as possible, your radius would be $2.$ The area of this loop is $A = \pi r^{2} = 4\pi.$ If you use two loops, each loop would have radius $1.$ The total area is $A = 2 \cdot \pi r^{2} = 2\pi,$ which is less than the single loop.
The thing you should realize is that the radius cannot be as big as you necessarily want it to be. The largest possible $r,$ which subsequently yields the largest $A$ (as you have proved), is $\frac{l}{2\pi},$ yielding a single loop. The radius cannot be larger than this (assuming the material cannot be stretched).