Explain briefly how bar magnets act as equivalent solenoids.

Explanation

The magnitude of magnetic field at a point 'P' due to a circular element of thickness 'dx' is given by `dB=((mu_(o))/(4pi))(2pindxIa^(2))/(((r-x)^(2)+a^(2))^(1//2))`
The magnitude of total magnetic field is given by
`B=((mu_(o))/(4pi))2pinIa^(2)int_(-l)^(l)(dx)/([(r-x)^(2)+a^(2)]^(3//2))`
For `r gt gt a ` and r `gt gtl, [(r-x)^(2)+a^(2)]^(3//2)=r^(3)`
`:. " " B = ((mu_(o))/(4pi))(2pinIa^(2))/(r^(3))int_(-l)^(l)dx " " int_(-l)^(l)dx=(x)_(-l)^(l)`
i.e., `" " B = ((mu_(o))/(4pi))(2pinIa^(2))/(r^(3))2l " " = l +l = 2l`
We note that the product n `(2l) I (pia^(2))` = Magnetic moment 'm'
`:. " " B = ((mu_(o))/(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.