584 EXPLORATION GEOPHYSICS 



field at P due to the current element dl is given by the equation 



,,, . dl sin (dl.r) . dl cos B 

 dH — I \ '- = I j-^ 



or 



jrr I dwdl COS B 



dH= — • z — ^ 



Ztt r^ 



Also, the field dH is perpendicular to the plane passing through dl and r. 



Due to the fact that the cones radiate from the source into the earth 

 in all directions, the upward and downward vertical components of the 

 fields produced by the cones cancel each other; that is, the resultant field 

 at any point on the earth's surface (xy plane) is parallel to the xy plane. 

 Also, the resultant field is perpendicular to any radius through the current 

 source ; that is, at a point such as P the resultant field is perpendicular to 

 X and parallel to the 3; axis.* Hence, the calculation of the total field 

 at P reduces to a calculation of the y component. 



The y component of the field due to the element dl is 



,rj _ / dw dl cos /3 cos ^ 



where (f> is the angle between the y axis and the perpendicular to the plane 

 passing through dl and r.** The last equation may be written in a more 

 convenient form by making use of the following relations : 



dw = sin 6 dd d<f> 

 r=R sec JS 

 R= X sin 6 

 1= Rtan/3 



dl= R sec^/? dp = —^ = . 



Whence 



or 



But 



/7r/2 



) e—n- 



e—ir/2 



R X sin 6 

 dHy = - — do cos <f> d^ cos ^ dp 



l/kX 



//y = :r— \ dQ \ cos^d^K COS p dp 



^TtXJ } —k/2. /e-v/2 



COS p dp =\- sin {6 — ir/2) = 1 + cos ^ 



* At a point such as P', the field is parallel to the x axis. 

 ** It is evident from elementary geometry that <p is also the angle between the plane 

 passing through dl and r and the xz plane. 



