To demonstrate the similarity of a current carrying solenoid to
a bar magnet, let us calculate axial field of a finite solenoid carrying
current.

Consider:

**= radius of solenoid***a***= length of solenoid with centre O**

*2l***= number of turns per unit length**

*n***= current passing through solenoid**

*I*

*OP = r*

Consider a small element of thickness

**of solenoid at distance***dx***from O. and number of turns in element***x**= n dx.*
We know magnetic field due to n turns coil at axis
of solenoid is given by

It is clear from the above
expression that magnetic moment of a bar magnet is equal to the magnetic moment
of an equivalent solenoid that produce the same magnetic field.

what we have to do to find magnetic field on equitorial line?

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