1898 - 99 .] Prof. Knott on Magnetic Twist in Nickel Tubes. 591 
the axis of the tube; then Q is perpendicular to P in the tangent 
plane to the tube, and E, is perpendicular to both, that is, radial. 
The tangential stress on the surface at right angles to the axis 
of the tube is 
J (P - Q) sin 2(9 = n (X - y) sin 2 6. 
But if r is the twist and r the radius of the tube, this tangential 
stress is also equal to n r r. Hence 
r r = (A - /x) sin 20. 
How introduce the assumption that the direction of P is the 
direction of the resultant magnetizing force. Then 
jj/ 
tan 6 — -y- 
and 
sin 
2 <9 = 
2HH' 
H 2 + H' 2 * 
Thus finally the twist is 
2(X-/x) HH' 
T “ r ’ H 2 + H' 2 
where A. and g are the elongations in field ^/H 2 4- H' 2 . 
This is exactly the same formula I obtained formerly as the 
result of a much more complicated piece of analysis (see Trans. 
Roy. Soc. Edin ., vol. xxxii. p. 388). 
Since H' is generally much smaller than H, the last factor may, 
without serious error in most cases, he written H7H ; and in this 
form we see at once why, with (A. - y) for nickel increasing in 
numerical value rapidly at first and then very slowly in higher 
fields, the value of r should pass through a maximum in moderate 
fields. 
In the case of iron, A. has itself a maximum value ; but the form 
of the expression above shows that the maximum twist will occur 
in lower fields than the maximum elongation. 
The maximum twist was observed by me both in iron and 
nickel, and its existence in the latter case in which there is no 
maximum elongation is explained by the formula just given. 
