Saturday, January 12, 2013

Bar Magnet as an Equivalent Solenoid

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:           a = radius of solenoid
                        2l = length of solenoid with centre O
                        n = number of turns per unit length
                        I = current passing through solenoid
                        OP = r

Consider a small element of thickness dx of solenoid at distance x from O. and number of turns in element = 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.

1 comment:

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