328 BELL SYSTEM TECHNICAL JOURNAL 



pole-piece, and there is a vertical gradient of field-strength which may be 

 made fairly constant over most of the interspace. Rabi at Columbia 

 achieves the same result more efficiently by peculiar arrangements of 

 current-carrying wires instead of iron magnets. 



Now consider (thinking classically!) a few of the atomic magnets as 

 they shoot across this non-uniform field. Think of one which originally 

 is oriented vertically, with north pole down and south pole up — the 

 south pole will be in a stronger part of the field than its mate; it will 

 be drawn upward harder than the north pole is pushed downward; 

 the atom will sweep in a parabolic arc upward. Think of another 

 which originally is oriented vertically with north pole up and south 

 pole down — it will be swept in a parabolic arc downward. Think of 

 another which originally has its axis pointing horizontally — ^it will 

 shoot in a straight line across the field, as though the pole-pieces were 

 not there.^ Now think of all those which are oblique to the vertical ; 

 they will describe parabolic arcs of intermediate curvatures, upward 

 or downward as the case may be. One infers that the beam must be 

 spread out into a continuous fan, making a continuous vertical band 

 upon the photographic plate. 



Moreover, from the upper edge or from the lower edge of this 

 continuous band, it should be possible to determine the magnetic 

 moment /x of the atoms. For let us consider one of the vertically- 

 oriented atoms, and call its pole-strength M and the length of its 

 magnet r. The upward force on the north pole is MH, — H standing 

 for the field-strength at the point where the north pole is. The 

 downward force on the south pole is M{H + dH/dz • r) . The net force 

 is Mr-dll/dz, which is iJ.(dH/dz), because Mr is the magnetic moment 

 fx of the magnet by definition. The acceleration of the little atom is 

 equal to this force divided by nia, the mass of the sodium atom. The 

 deflection is equal to half the acceleration by the square of the time 

 during which the atom is exposed to the force. This time" is equal to 

 the distance D which the atoms traverse across the field, divided by v 

 the speed which they have when they come out of the furnace. So 

 finally, we have: 



Deflection = -7:— — - (D v)-. 



2 nia 



We ascertain the deflection by looking at the end of the band on the 

 photographic plate, and we can ascertain all the other things in the 



^ When thinking classically, we must not expect the atomic magnets to turn their 

 axes toward the field-direction as they shoot across the field; the gyroscopic quality 

 of these magnets, due to their angular momentum, inhibits this. 



