Field of Cylindrical Currents. 437 



deduced the formula 



27T 8 N 1 N,& 2 / a 2 4/ <2 2 -36 2 

 M ~ ^ "V 1 iW 4 



a 2 (4/y-3a 2 ) 8V-20/< 2 2 6 2 + 56 4 _ \ 

 Self ' 8 ■•/ , 



where d 1 2 = h 1 2 -\-a 2 . The value of M found by this formula 

 in the case considered is 0*0011999 henry. 



We shall find a formula in the next section by means of 

 which M can always be evaluated to any specified degree of 

 accuracy. 



8. Formulas for the mutual inductance between a helix and 

 a coaxial cylindrical current sheet. 



We shall first find Jones's formula * for the mutual in- 

 ductance between a circle and a coaxial helix. 



From (19) we see that the mutual inductance M between 

 two coaxial circular currents is given by 



.V 2 cos20d0 



= -Sirab\' 

 Jo 



{(a + b) 2 +■ z 2 - 4ab sin 2 6. }V»» 



where z is the distance between their planes, and a and b are 

 their radii. Hence the mutual inductance dM. between a 

 sector of the circle whose radius is b and the circle whose 

 radius is a is given by 



«, A „.. C^ 2 cos20dd 



where flf-v/r is the angle of the sector. 



Let p be the pitch of the helix, and let the coordinates of 

 points on it be given by #=&cos yjr, y — b sin yjr, and z=pyfr. 



We shall first find the mutual inductance between this 

 helix and a coaxial circle in the plane XY (fig. 3). Let us 

 find the linkages of the helix with the flux due to unit 

 current in the circular filament. Consider two contiguous 

 points P x and P 2 on the helix and draw PiN x and P 2 N 2 per- 

 pendicular to OZ. Also draw P 2 P parallel to OZ to meet 

 the plane passing through PiN x and perpendicular to OZ 

 at P. Join PNi and PPi. Now every line of force due to 

 the circular current must lie in a plane which passes through 

 OZ. Hence, since PiP 2 are infinitely close together all the 



* Proc. Roy. Soc. vol. lxii. p. 247 (1897). 



