554 
Proceedings of the Royal Society 
distance. If the strip be moved in the opposite direction, the same 
thing will happen with the opposite sign. Hence if we make the 
frame on which the coils are wound such that the values of the 
deviating forces due to the two strips EBcF and E'B'c'E' are 
numerically equal, the former being on the nearer side of the 
positive maximum and the latter on the further side of the negative 
maximum, the deviating force on the magnetometer will he insensible. 
This is illustrated by fig. 6, where the curve shows the deviating 
force caused by a strip in various positions along the line NS. 
To find the action of each magnetic strip — 
Let NS in the annexed figure represent the magnetic meridian. 
Let the centre of the magnetic strip be on the line ns^ parallel to 
N S,and let the magnetometer be placed at m. From the preced- 
ing figure it will be evident that the breadth of the strip is 28a, its 
thickness Sc, its length 25, and the distance of its centre from N S 
the point m ; differentiate it with respect to a ] multiply the result 
by Me, where ^ is the strength of the current, and n the number 
of turns of wire per unit of length. 
Since the strip is supposed to be very narrow, we can, without 
sensible error, project the area perpendicular to the direction of r, 
and take w subtended by this projected area in finding the potential 
at the point m due to the strip. To do this we have only to find 
the solid angle subtended by a rectangle whose height is 25 and 
breadth 2Sa^, namely, 
N 
n 
is a. Further, let c be 
the distance of its centre 
from WE, and r the dis- 
tance from m. 
Fig. 7. 
To find the force in the 
direction of a, which is 
the deviating force, we 
have to find the solid 
angle o> subtended by 
the area of the strip at 
w ~ 4 cos 
