Calculate the axial field of a finite solenoid.

Explanation

Let the solenoid of figure consists of `n` turns per unit length.

Let its length be `2l` and radius a. We can evaluate the axial field at a point P, at a distance r from the centre O of the solenoid.
Consider a circular element of thickness dx of the solenoid at a distance x from the centre. It consists of n dx turns. Let I be the current in the solenoid. According to formula for the magnitude of magnetic field at axis of coil of N turn, number of tum `N= n` dx and taking distance from O to P `= (r-x)`.
`B= ( mu_(0) NIa^(2) )/( 2[ (r-x)^(2) + a^(2) ]^(3//2) )`
`therefore B= (mu_(0) n dx Ia^(2) )/(2[ (r-x)^(2) +a^(2) ]^(3//2))`
Total field is obtained by summing over all the elements that is by integrating from `x=-l` to `x=+l`. Then the denominator is approximately by,
`[(r-x)^(2) +a^(2) ]^(3//2) ~~ r^(3) [ because l` means `x` and a is neglected to compare with r ]
`therefore B= (mu_(0) n Ia^(2) )/( 2r^(3) ) int_(-l)^(l) dx`
`= (mu_(0) nIa^(2) )/( 2r^(3) ) [l-(-l)]`
`= (mu_(0) nIa^(2) )/( 2r^(3) ) [2l]`
Arranging the terms appropriately,
`B = (mu_(0) )/( 2r^(3) ) [ (n2l) (Ia^(2) )]`
multiply and dividing by `pi` to right side term,
`B= (mu_0)/( 2r^3) [( (n2l)(I pi a^(2) ))/( pi)]`
Here `(n2l)` are the total turns of the solenoid. `pia^(2) = A` is the cross-section area of one side of solenoid.
`B = (mu_0)/( 2r^3) [ (NIA)/( pi) ]`
Here, NIA = dipole moment of solenoid,
`B= (mu_0)/(2pi ) (m)/(r^3)`
`therefore B=2 (mu_0)/( 4pi) (m)/( r^3)`.
Is the magnetic field at the axis of solenoid.
This also similar the far axial magnetic field of a bar magnet.
Thus, a bar magnet and a solenoid produce similar magnetic field.