A very long straight solenoid has a cross section radius `R`. A number of turns per unit length is equal to `n`. Find the magnetic induction at the centre of the coil when a current `I` flows thougth it.

Explanation

On the surface of the solenoid there is a surface current density
`vec(j) n = n I hat(e)_(varphi)`
Then, `vec(B) = (mu_(0))/(4pi) n I int R d varphi dz (hat(e)_(varphi) xx vec(r_(0)))/(r_(0)^(3))`
where `vec(r_(0))` is the vector fromt the current element to the field point, which is the centre of the solenoid,
Now, `-hat(e)_(varphi) xx vec(r_(0)) = R hat(e)_(x)`
`r_(0) = (x^(2) + R^(2))^(1//2)`
Thus, `B = B_(x) = (mu_(0) n I)/(4pi)xx 2pi R^(2) int_(-1//2)^(-1//2) (dz)/((R^(2) + x^(2))^(3//2))`
`= (1)/(2) mu_(0) n I int_(-"tan"^(-1)(l)/(2R))^(+"tan"^(-1)(l)/(2R)) cos alpha d alpha` (on putting `z = R tan alpha`)
`= mu_(0) n I sin alpha = mu_(0) nI (l//2)/(sqrt((1//2)^(2) + R^(2))) = mu_(0) n I//sqrt(1 + ((2R)/(l))^(2))`