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# Force due to a Solenoidal Magnet

To obtain this we must remember that the work done on a unit pole by the forces of any system in going from a point P1 to a second point P2, V1, V2 being the potentials at P1 and P2, is V1 - V2. Let a be the distance between these two points, and let F be the average value of the magnetic force acting from P1 to P2 resolved along the line P1P2. Then the work done by the force F in moving the pole is Fa.

Hence

Fa = V1 - V2

and if the distance a be sufficiently small, F, the average value of the force between P1 and P2 may be taken as the force in the direction P1P2 at either P1 or P2.

Denoting it by F we have

when a is very small

Let us suppose that P1, P2 are two points on the same radius from O, that OP1=r and OP2=r+δ.

Then θ is the same for the two points, and we have

neglecting (δ/r)2 and higher powers (see p. 42).

Also, in this case, a=δ. Thus

We shall denote this by R, so that R is the force outwards, in the direction of the radius-vector, on a unit pole at a distance r from the centre of a small solenoidal magnet of moment M. If the radius-vector make an angle θ with the axis of the magnet, we have

Again, let us suppose that P1P2 (fig, 46) is a small arc of a circle with O as centre, so that

OP1=OP2=r
let P1ON = θ
and P2ON = θ+φ.

Thus

a=P1P2 = OP1 x P1OP2 = rφ.

The force, in this case, will be that at right angles to the radius vector, tending to increase θ; if we call it T we have

These two expressions are approximately true if the magnet NS be very small and solenoidal. We may dispense with the latter condition if the magnet be sufficiently small; for, as we have said, any carefully and regularly magnetised bar behaves approximately like a solenoid with its poles not quite coincident with its ends. In such a case 2l will be the distance between the poles, not the real length of the magnet, and 2ml will still be the magnetic moment.

Last Update: 2011-03-16