( 255 ) 



Then for an arbitrary gradient distribution holds : 



Jx = v/J 



- ^ ^/ ° FAr)dr. . . , . . . (/) 



4jr 



V. The field E^ of a circular current according to the equator 

 plane in the origin, is identical outside the origin to the above field 

 E^ ; but now each force line is closed, and has a line integral of 

 4:ji along itself. 



According to the method of A § IX we find of this field E^ the 

 planivector potential H in the meridian plane and independent of the 

 azimuth. 



We find when writing n — r = ^ : 



So 



2 = - sin ^(p(l -\- ^ cot r) dd: 



jt 



1 . I -\- ^cotr 



H =. — sin<p ; , 



n sin r 



vanishing along all principal circles in the opposite point. 



From which we deduce for the force of an element of current 



with unity-intensity in the origin directed according to the axis of 



the spherical system of coordinates : 



1 , 1 -\- ^cotr 



— sin (p ; , 



n smr 



directed normally to the meridianplane. 



VI. From this we deduce as in A ^ XI a vector potential V of 

 an element of current parallel to that element of current and a 

 function of r only. For the scalar value U of that vector potential 

 we have the differential equation : 



d I , , 1 d I ) 



—\U sin w sin r dcp] dr — ^-- \U cos (pdr \ d(p z=z 



Or \ ) 0(p { ) 



1 , \ -\- ^cotr 



= — sin<p ; . dr . sm r d<p. 



n sin r 



Or: 



d ) 1 



U — — Usinr\——i\-{-^cotr), 

 Or 1 ^ 



of which the solution is 



c 1 I i|3' 



cos^ \r n \ cos* \r sinr\ 



