ELEMENTS OF ELECTRICITY AND MAGNETISM. 



(r) 



(20) 



in which mjl is the pole strength per unit length of the shaded 

 area in Fig. 36, and H is the field intensity in gausses at a point 

 r centimeters from the axis of the rod. 



Proof. Imagine a cylindrical surface of radius r to be drawn 

 with its axis coincident with the axis of the rod, as shown by the 

 dotted lines in Fig. 36. The area of this cylindrical surface is equal 

 to 277T/, and, inasmuch as the magnetic field at this cylindrical sur- 

 face is everywhere at right angles to it and everywhere the same 

 value, the magnetic flux through this cylindrical surface is equal to 

 2irrl x H according to equation ( 1 8). This is the total magnetic 

 flux emanating from the pole, and it must be equal to 

 according to equation (19), so that we have 2TrrlH '= 

 whence //= 2m]rl. In this discussion the non-uniformity of 

 the magnetic field near the ends of the long slim pole is ignored ; 

 in fact, the effect of this non-uniformity is negligible if r is small 

 in comparison with the length / of the slim pole. 



The above formula expressing the field intensity at a distance 

 from a long slim pole applies also to the case of the pole which is 

 distributed along the edge of a steel ribbon which is magnetized 

 crosswise as shown in Fig. 37. In this case, however, the 



N 



S 

 side view 



N. 



S 

 end. view 



Fig. 37. 



actual magnetic field at any given point / is the resultant of the 

 fields due to the poles along both edges of the strip. 



