Bar Magnet as an Equivalent Solenoid
Ampere's hypothesis that all magnetic phenomena can be explained in terms of circulating currents. Recall that the magnetic dipole moment `m` associated with a current loop was defined to be
`m = NI A`
where `N` is the number of turns in the loop, `I` the current and `A` the area vector.
The resemblance of magnetic field lines for a bar magnet and a solenoid suggest that a bar magnet may be thought of as a large number of circulating currents in analogy with a solenoid. Cutting a bar magnet in half is like cutting a solenoid. We get two smaller solenoids with weaker magnetic properties. The field lines remain continuous, emerging from one face of the solenoid and entering into the other face. One can test this analogy by moving a small compass needle in the neighborhood of a bar magnet and a current-carrying finite solenoid and noting that the deflections of the needle are similar in both cases.
To make this analogy more firm we calculate the axial field of a finite solenoid depicted in Fig(a). We shall demonstrate that at large distances this axial field resembles that of a bar magnet.
Let the solenoid of Fig.(a) consists of `n` turns per unit length. Let its length be `2l` and radius `a`. We can evaluate the axial field at a point P, at a distance r from the centre O of the solenoid. To do this, consider a circular element of thickness dx of the solenoid at a distance x from its centre. It consists of ndx turns. Let `I` be the current in the solenoid. The magnetic field on the axis of a circular current loop.
`dB=(mu_0ndxIa^2)/(2[(r-x)^2+a^2]^(3//2)`
The magnitude of the total field is obtained by summing over all the elements - in other words by integrating from `x = - l` to `x = + l`. Thus,
`B=(mu_0nIa^2)/2 int_(-l)^l(dx)/[(r-x)^2+a^2]^(3//2)`
Consider the far axial field of the solenoid, i.e., `r` >> `a` and `r` >> `l` . Then the denominator is approximated by
`[(r-x)^2+a^2]^(3//2)approxr^3`
and `B=(mu_0nIa^2)/(2r^3)int_(-l)^ldx`
`=(mu_0nI)/2quad(2la^2)/r^3`
Note that the magnitude of the magnetic moment of the solenoid is, `m = n (2l) I (πa^2)` = (total number of turns x current x cross sectional area). Thus,
`B=mu_0/(4pi)quad(2m)/r^3`
This is also the far axial magnetic field of a bar magnet which one may obtain experimentally. Thus, a bar magnet and a solenoid produce similar magnetic fields. The magnetic moment of a bar magnet is thus equal to the magnetic moment of an equivalent solenoid that produces the same magnetic field.
`m = NI A`
where `N` is the number of turns in the loop, `I` the current and `A` the area vector.
The resemblance of magnetic field lines for a bar magnet and a solenoid suggest that a bar magnet may be thought of as a large number of circulating currents in analogy with a solenoid. Cutting a bar magnet in half is like cutting a solenoid. We get two smaller solenoids with weaker magnetic properties. The field lines remain continuous, emerging from one face of the solenoid and entering into the other face. One can test this analogy by moving a small compass needle in the neighborhood of a bar magnet and a current-carrying finite solenoid and noting that the deflections of the needle are similar in both cases.
To make this analogy more firm we calculate the axial field of a finite solenoid depicted in Fig(a). We shall demonstrate that at large distances this axial field resembles that of a bar magnet.
Let the solenoid of Fig.(a) consists of `n` turns per unit length. Let its length be `2l` and radius `a`. We can evaluate the axial field at a point P, at a distance r from the centre O of the solenoid. To do this, consider a circular element of thickness dx of the solenoid at a distance x from its centre. It consists of ndx turns. Let `I` be the current in the solenoid. The magnetic field on the axis of a circular current loop.
`dB=(mu_0ndxIa^2)/(2[(r-x)^2+a^2]^(3//2)`
The magnitude of the total field is obtained by summing over all the elements - in other words by integrating from `x = - l` to `x = + l`. Thus,
`B=(mu_0nIa^2)/2 int_(-l)^l(dx)/[(r-x)^2+a^2]^(3//2)`
Consider the far axial field of the solenoid, i.e., `r` >> `a` and `r` >> `l` . Then the denominator is approximated by
`[(r-x)^2+a^2]^(3//2)approxr^3`
and `B=(mu_0nIa^2)/(2r^3)int_(-l)^ldx`
`=(mu_0nI)/2quad(2la^2)/r^3`
Note that the magnitude of the magnetic moment of the solenoid is, `m = n (2l) I (πa^2)` = (total number of turns x current x cross sectional area). Thus,
`B=mu_0/(4pi)quad(2m)/r^3`
This is also the far axial magnetic field of a bar magnet which one may obtain experimentally. Thus, a bar magnet and a solenoid produce similar magnetic fields. The magnetic moment of a bar magnet is thus equal to the magnetic moment of an equivalent solenoid that produces the same magnetic field.