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xii. 3] MAGNETIC TUBES IN ROTATION 163 
The method of evaluating the integral is given in the paper 
cited where it is shown that 
f Hp 2 ds = . . . . 12: 1 
J jU 
Substituting this value in our expression for the angular momen 
tum we find that the total angular momentum for the single 
quantum tube is 
K/u h 4M KM h 
—— . — . CO . —— = . - . CO . . . 12:2 
4 n e ¡x n e 
And so we see that the magnetic tube which we are considering 
has an equivalent moment of inertia 
I = KMh 
Tie 
Remembering that the angular velocity, co, can be expressed 
in terms of the frequency by writing co = 2nv, we find that the 
Angular Momentum = —. . . 12 : 3 
We can now write down the kinetic energy of the quantum 
tube, for it is found by multiplying the angular momentum by 
2co, and the Kinetic Energy = f^^co 2 or flco 2 
Tie 
2nKM.h „ 
= v 2 . . . . 12:4 
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3. The Equivalent Charge of a Rotating Tube 
When a magnetic tube is in motion, it produces an electric 
field. In the case of the rotating quantum tube already con 
sidered the electric field set up at a point D in the equatorial 
plane is equal to By where v is the velocity of the rotating tube 
at that point. If the distance OD is a, the velocity v = aco, and 
the electric field is Baco, or 27iaBv. 
Let us assume that the electric field at D is the same as would 
be produced by an electrostatic charge E at the point O, and 
then determine the value of that charge. As the two fields are 
assumed to be equal, 
E 
——= By = Baw — 2naBv . . . 12 : 5 
K a 2 
Now the magnetic field at D is supposed to be due to an