A very long straight solenoid carries a current `i`. The cross-sectional area of the solenoid is equal to `S`, the number of turns per unit length is equal to `n`. Find the flux of the vector `B` through the end plane of the solenoid.

Explanation

The magnetic field is given by
`B = (1)/(2) mu_(0) nI (1 - (x)/(sqrt(x^(2) + R^(2))))`
At the end, `B = (1)/(2) mu_(0) n I = (1)/(2) B_(0)`, where `B_(0) = mu_(0) nI`,
is the field deep inside the solement. Thus,
`Phi = (1)/(2) mu_(0) n IS = Phi//2` where `Phi = mu_(0) nl S`
is the flux of the vector `B` through the cross setion deep inside the solenid.