EQUIVALENCE BETWEEN A BAR MAGNET AND A SOLENOID

The magnetic field produced by a solenoid is due to passage of electric current through it. How a magnetic field is produced by a bar magnet without any apparent observable current can be explained as under.

A bar magnet is made up of atoms in which definite number of electrons move in various possible orbits. This constitutes electric current around a closed path. This current and the spin of electrons result in magnetic dipole moment. If the vector sum of magnetic dipole moments of all electrons is zero, then such a substance will NOT act as a bar magnet.

Despite the fact that the atoms of iron possess magnetic dipole moment, an ordinary piece of iron does not behave as a magnet. It can be converted into a magnet by keeping in a strong magnetic field for some time and reducing the magnetic field slowly to zero. When a piece of iron is kept in a strong magnetic field, the elemental atomic currents get redistributed in the iron piece and do not return to the original current distribution on removal of the externally applied magnetic field. The following figures represent this process. 

The rectangular piece of iron is enclosed in a solenoid carrying current If . This produces the applied magnetic field. The randomly distributed atomic currents shown in part ( a ) of the figure gets redistributed as current Ib as shown in part ( b ) of the figure.

This new distribution of currents remains even on removal of the external current If . It is because of this that the piece of iron behaves as a bar magnet.

 Now consider a solenoid shown in the following figure. It has a large number of closely wound turns of current carrying conducting coil around a soft iron core. For clarity, a few turns separated from one another are shown. Each turn can be treated as a closed current loop possessing magnetic dipole moment. Thus each turn can be treated as a tiny magnet with north and south poles. On looking normally at the plane of the loop, if the current appears to flow in the clockwise direction, then the side of the loop towards the eye behaves as the south pole and the other side behaves as the north pole. Similarly, if the current appears to flow in the anticlockwise direction, then the side of the loop towards the eye will behave as the north pole. This is shown in the two figures above on the right side. Thus, in the figure of the solenoid above, current in the circuit in front of the eye being in the anticlockwise direction, the side of the first turn towards the eye behaves as a North Pole, while the other side of the turn is the South Pole for that turn. For the second turn, the side towards the eye is again North Pole and so on for all the turns. Thus, magnetic dipole moment of each turn is in the same direction and hence the magnetic dipole moment of the solenoid is the vector sum of dipole moments of all the turns. For current I  through a solenoid with total number of turns N and with cross-sectional area A, the magnetic dipole moment of the solenoid is given by

M_S = N I A             ... ... ... ( 1 )

The magnetic dipole moment of a bar magnet of pole strength, m, and length, 2l, is given by

M_b = 2ml             ... ... ... ( 2 )

By analogy between solenoid and bar magnet, the pole strength, ms, of the solenoid can be obtained using the above two equations as under.

2msl = N I A

or     ms = N I A / 2 l = n I A, where, n = N / 2l = no. of turns per unit length of solenoid.

Thus Pole-strength of solenoid =(Number of turns per unit length)*(electric current)*(cross-sectional area of solenoid). 

The unit of pole-strength is Am ( Ampere-meter ).