Show that a current carrying solenoid is equivalent to a bar magnet.

Explanation

The magnitude of magnetic field at a point .P. due to a circular element of thickness .dx. is given by
`dB=(mu_0/(4pi)) (2pindxla^2)/(((r-x)^2 + a^2)^(3/2))`
The magnitude of total magnetic field is given by
`B=(mu_0/(4pi)) 2pinIa^2 int_(-1)^1 (dx)/([(r-x)^2 +a^2]^(3/2))`
For r > > a and r > > l , `[(r-x)^2 + a^2]^(3//2) ~~ r^3`
`therefore B =(mu_0/(4pi)) (2pinIa^2)/r^3 int_(-l)^l dx " " int_(-l)^ldx = (x)^(-l)^l`
i.e., `B=(mu_0/(4pi)) (2pinIa^2)/r^3 2l`
=l + l =2l
We note that the product n(2l) I `(pi a^2)` = Magnetic moment .m.
`therefore B=(mu_0/(4pi)) (2m)/r^3` --(1)
This expression is similar to the magnetic field at a point on the axis of a short magnet. Therefore a bar magnet and a solenoid produce similar magnetic fields.