422 THE ELECTROMAGNETIC FIELD. [PT. III. CH. XI. 



the direction of ds, the force vanishes, so that the force is per- 

 pendicular to ds. If Bh is perpendicular to the plane of r and ds. 

 we have 



cos (dH, Bh) = 1, sin (ds, Bh) = I, 



. . JTT rdscos(nr) rdssm(r,ds) 



dH = -7*- J- 



Accordingly if we call da- the component of ds perpendicular 

 to r, the magnetic force due to the whole conducting circuit will 

 be obtained if we suppose each element ds to contribute to the 

 field the amount 



(9) ** = , 



which has the direction perpendicular to the element ds and the 

 radius r. 



The total field is the vector sum of all these infinitesimal parts. 

 The proper sign to be chosen may be found by considering the 

 way in which the lines of force are linked with the current, and 

 we find that the direction of the force is given by the rotation 

 of a right-handed screw advancing with the current in the 

 direction of da. The complete specification may be most con- 

 cisely stated by saying that the force due to the element ds is 

 1/r 3 times the vector product of Ids and r, the vector r being 

 drawn from the element ds to the point P, 



(10) dH = ^-VJs.r. 



The resolution of the field into elementary fields is artificial, for 

 the field is of course due to the whole closed circuit. Moreover 

 the resolution may be performed in an infinite number of ways, 

 for it is the integral of the above differential taken around the 

 whole circuit which gives the field. 



We may consequently add to the differential above the differ- 

 ential of any function of the coordinates of the element ds, for in 

 integration around the circuit this function returns to its original 

 value so that the integral vanishes. 



If the coordinates of a point in the current circuit are x lt y l} z l} 

 these of P, x, y, z, since the direction cosines of r and ds l are 



