in Conducting Sheets and Solid Bodies. 3 



ing to Ohm's law, we shall find that the screening action 

 depends upon the angular velocity multiplied by radius divided 

 by the resistance of the sheet, so that defect of conductivity is 

 fully made up for by a large radius. 



We shall also find expressions for the magnetic moment of 

 a copper sphere or spherical shell rotating in a magnetic field, 

 and for the expenditure of power required to keep up the 

 motion, and also for the rate of damping of the oscillations of 

 such a body. 



Finally, we shall show how solutions may be obtained for 

 the case of a rotating circular disk of conductivity not very 

 large by neglecting the mutual action of the induced currents. 



II. In the case of a conductor rotating steadily in a mag- 

 netic field, which is that with which Jochmann deals, a steady 

 distribution of currents in space will ensue when the conductor 

 is symmetrical about the axis of rotation ; and the electromo- 

 tive force along any line will be given, according to Faraday's 

 rule, by the number of tubes of force of this steady field that 

 are cut through by the line per second. Now when the mag- 

 netic field is symmetrical round the axis of rotation, the 

 number of tubes enclosed in any closed moving circuit in the 

 conductor will not alter at all, so that there will be no current 

 round any circuit, and therefore no induced current whatever: 

 the electromotive force along each open line will accumulate 

 a statical electric charge at one end of it, so that the conductor 

 will become electrified until the induced electromotive force 

 is exactly neutralized by the statical difference of potentials. 

 This conclusion holds whatever be the shape of the body. 



Taking cylindrical coordinates, if a, /3, 7 be the components 

 of the magnetic force at the point rdz in the directions of dr, 

 rd6, dz respectively, the electromotive forces between the ends 

 of these elements of length will be —cor<ydr^ 0, ooradz, where 

 co is the angular velocity, by Faraday's rule ; and thus the 

 electrostatic potential which neutralizes these will be 



yfr = \ (corydr — cor adz) 



= G) \r(ydr — adz). 



From this the electrification of the conductor may be deduced 

 at once. For example, taking a uniform field 7=y parallel 

 to the axis of rotation, we have 



^ = C + ic»7o^ 

 and therefore there is a uniform volume-density of electricity 



4tt V Y 2tt 



B2 



