70 [ELECTKOAIAGNETIbM. Denote the angles between the positive directions of da and ds and the direction of D from da to ds by and 6, then we have dD -T- da- ., cose -j , cos0 = - -T- ds - dD T , - - T -- D Ry means of these we get T a< dsdff ., /V = tz. // JJ ^- a -^lads, D da ds (16). The part which is a complete differential has been left out, because it disappears when the integration is carried round closed circuits, as we always suppose it to be. Consider now the work done in a small displacement which alters D and S, da not da; we have
ds V ff l =- nJJ - , and ds, but D <ZP , , - dads - f d<r efc rfs ^D d) dSs , , -, -3 -- j- dl ? ds da- ds ds The parts containing 8,s disappear in this expression, and if the rest be arranged by integration by parts as usual, we get SM-y/RSD^vofo-^O ..... (17), where R-n 2c 3 *- 3 c s e C03e> . Hence the electrodynamical action of the two circuits is completely accounted for by supposing every element c/o- to attract every element Js with a force ii dsdaf jy a I 2 cos e - 3 cos cos (18). ction m nagnetic jole. We may therefore use this elementary formula whenever it suits our convenience to do so. It is very easy to obtain a similar elementary formula, which is very often useful, for the action of an element of a circuit on a unit north pole. "We have seen above how to find the action on an element PQ (ds) of a circuit in a given magnetic field. Let the field be that due to a unit north pole N (fig. 32). Then the magnetic induction at P is in the direction NPK, and is equal to : , if NP = D. Hence by (6) the force R on PQ 1 is perpendicular to NP and PQ, is in the direction PM shown in the figure, and is equal to S1 . Now, by the princip of "action and reaction," the force on N is R in the Fig. 32. direction PM opposite to PM, i.e. is equal to a force R acting at N in a direction NM"parallel to I M , together with a couple whose moment is Rx PN, and whose axis is perpendicular to NP and in the plane XPQ. Now a simple calculation, which we leave to the reader, will show that for any closed circuit the resultant of all the couples thus introduced is nil; hence, since we deal with closed circuits only, we may neglect the couple. The force exerted by a closed circuit on a unit north pole may therefore be found by supposing each element ds to act on the pole with a force equal to ids sine (19), D* whose direction is perpendicular to the plane containing the pole and the element, and such that it tends to cause rotation round the element related to the direction of the current in it by the right-handed screw relation. PQ is supposed to be drawn from the reader. Agree- meilt of Straight current. Comparison of Theory with Experiment. The best veri fication of the theory which has just been laid down con sists in its uniform accordance with experience. We pro ceed to give a few instances of its application, adopting now one, now another, of the equivalent principles deduced from it. We have already remarked that the lines of magnetic force in an electric field due to an infinite straight current are circles having the current for axis. It is easy to deduce from the fact that there is a magnetic potential that the force must vary inversely as the distance from the current. This may also be proved by means of the formula (19) ; in fact, the resultant force at P is given by H = i I . ds = i I ,, -- cosec 6d9 = , (20), J D- J (Z 2 cosec a d (j d being the distance of P from the current. Let AB (fig. 33) be a very long straight current, and Parallel poq an element ds of a parallel cur- B rent, having the same direction as AB. If we draw the line of force (a circle with C as centre) though O, the -* tangent OR is the direction of the force at O; hence by (6) and (20), the force on poq is - ds, and acts in the direction 00 ; poq is therefore attracted. If the current in poq be reversed, the force will have the same numerical value, but will act in the direction CO. Hence two parallel straight conductors attract or repel each other according as the currents in them have the same or opposite directions. Let AB (fig. 34) be an infinitely long (or very long) Inclined
- * currents
current, CD a portion of a current inclined to it, and passing very near it at 0. If the plane of the paper contain AB and CD, then at every point in OD the magnetic force is perpendicular to -^ the plane of the paper and towards the reader, at every point in OC perpendicular to the plane of the paper and from the reader; hence Fig. 34. at the elements P and Q the forces acting will be in the direction of the arrows in the figure, and CD will tend to place itself parallel to AB. If both the currents be re versed, the action will be unaltered; but if the current in CD alone be reversed, it will move so that the acute angle DOB increases. Hence it is often said that cur rents that meet at an angle at tract each other, when both flow to or both flow from the angle, but repel when one flows to and the other flows from the angle. These actions may be demon strated in a great Pig. 35. variety of ways. Figure 35 shows an arrangement for demonstrating the attraction or repulsion of parallel currents, wliuih is essentially that first used by Ampere. A is an upright consisting of a tube in good metal lic connection with one of the binding screws t, and with a little cup p, containing a drop of mercury. A stout wire passes up the centre of the tube, and is insulated from it, but in metallic cormec- Am pere s ap
paratu.s.