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The magnetron


This is a valve with a cylindrical anode surrounding a linear cathode (Figure 1). This type of valve is very difficult to buy for schools now but provides a simple way to determine e/m. A solenoid is placed over the valve with its axis coincident with that of the cathode.

With no current in the coil of the solenoid the electrons emitted from the cathode travel in straight lines along a radius and are collected by the anode - a current (IA) flows in the valve. Now if a current is passed through the solenoid a magnetic field will be produced and the emitted electrons will begin to curve in the field - the radii of their paths becoming smaller and smaller as the current, and therefore the magnetic field is increased. Up to this point the anode current remains constant.

Eventually a point will be reached for a solenoid current IC where the diameter of their path is just equal to the radius of the anode, any further increase in the solenoid current and the electrons will not hit the anode. There will then be a sudden drop in the anode current.



From the equations for centripetal force and kinetic energy of a charge in a magnetic field we have:

Bev = mv2/r and ½mv2 = eV

Therefore e/m = 2V/(B2r2) but if R is the radius of the anode (R = 2r) we have:

Magnetron:    Specific charge :  e/m = 8V/[B2R2]

the magnetic field in the solenoid for the critical field is given by the equation:

Bc = μoNI/(d2+L2)1/2

(where d is the diameter of the coils and L their length).

In the magnetron we must assume that the cathode is very narrow, so that the electric field at its surface is high. If we do this then we can also assume that the electrons experience most of their acceleration close to the cathode and so their velocity across the rest of the valve to the anode is uniform.


Example problem
A magnetron has a potential of 200V between the cathode and an anode with a diameter of 24 mm. If the specific charge (e/m) of the electron is 1.76x1011 C/kg calculate the flux density of the field for which the anode current just falls to zero.

The radius of the electron orbit in this condition is a quarter of the diameter of the anode, in this case 6 mm or 6x10-3m.

Therefore:   Critical field (BC) = 8V/[B2R2] = 1.59x10-2 T
 

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© Keith Gibbs 2020