Bar magnet is equivalent

to the solenoid

 



1) consider a Solenoid of length 2l, and O is the center of the solenoid as shown in the above



2)

n = no of turns per unit length of the solenoid.

I = current flowing through the solenoid.

a = radius of the solenoid.



3) Now we have to determine magnetic field strength (B) at P, a distance r from the center of the solenoid.



4) Consider a small thickness dx of the solenoid

At a distance x from the center O of the solenoid.



5) We know the magnetic strength on the axis of a circular current loop , So for  dx  thickness, magnetic field at P, is

 

 


6) If is located at a large distance then

r ≫ a,  r ≫ x,

 

 


 

So, we get





7) Now total magnetic field B at P from

x = -l  to  x = l  is














Now we know that magnetic moment of the solenoid is


 



therefore




 

This is also same as the axial magnetic field due to a Bar magnet, Thus you can say that a Bar magnet and a Solenoid is equivalent.