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XXX. Magnetic Shielding by a Hollow Iron Cylinder. Simplest 

 Case. By John Perry, F.R.S* 



CONSIDER going and return electric conductors each at 



i 



a distance a from the axis in a diametral plane of the 

 hollow cylinder, the current in each heing C. 



As to the inducing magnetic potential, all over one half of 

 the cylindric surface of radius a the potential is constant, 7rC, 

 and all over the other half it is constant, being — 7rC. At a 

 point P whose distance from the axis is ?% the angle made by 

 the axial plane through P and the plane through the con- 

 ductors being 6, it is evident that V, the inducing potential, 



\r 6r 6 or / 



as this becomes +7tC or — 7rC where r=a, and also satisfies 



v 2 v=o. 



Let the coefficient of sin n6 in V be denoted by A? ~ n : let 

 the corresponding coefficients be Aj? ,w + Ar _M for the total 

 (induced plus inducing) potential inside the tube ; A 2 r n + B 2 r~ n 

 total in the iron and B%r~ n in outside space. Let the inside 

 and outside radii of the iron cylinder be a 1 and a 2 . Then, on 

 stating the equality of the potentials just inside and just 

 outside the two surfaces, and also that the normal induction 

 is the same on both sides, we have four equations, which 

 enable the coefficients A ln A 2 , B 2 , and B§ to be calculated. 

 The two important ones are 



B 3 =4M/{0" + l) 2 -^-l) 2 }, 

 where e stands for a 1 2n /a 2 2 '\ 



and A^-A/Kl-^/a/V-'), 



where p = (fi + l)(fi — 1) . 



If we take a 2 =a l -f t, and t is small, since /x is large in iron, 

 we have approximately 



b,=a/(i+^1). 



The higher terms are unimportant. Taking the first : the 

 unshielded potential was 4Car -1 sin 0, and it is converted by 

 shielding into 4Ca sin 0/r{ 1 + \^tja x \ . - 



Thus, taking /i=1200, a! = 150 cm., t=5 cm., the first and 

 most important term is reduced to the 1/2 1st part of its 



* Communicated by the Physical Society : read June 22, 1894. 





