Download Electron Beams and Microwave Vacuum Electronics by Shulim E. Tsimring PDF

By Shulim E. Tsimring

This ebook specializes in a primary function of vacuum electronics: the robust interplay of the physics of electron beams and vacuum microwave electronics, together with millimeter-wave electronics. the writer publications readers from the roots of classical vacuum electronics to the newest achievements within the box. distinctive cognizance is dedicated to the physics and thought of relativistic beams and microwave units, in addition to the speculation and functions of particular units.

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Usually, these conditions are written without a factor 2p. This means that changes in field magnitudes on the Larmor radius scale must be much smaller than full field magnitudes. The alternative condition takes place when j vB j ) j v v j þ j v t j and hence In this case, we obtain from Eq. 72)   dF h  ¼ 1j F j dr j v j % vB (1:80) (1:81) where h ¼ vB Tg is the pitch of the helical trajectory. Hence, Eq. 79) must be fulfilled as a condition of small field changes on the helical trajectory pitch.

This formula is valid in a nonrelativistic approximation and for w(z)=kT ) 1. Substituting Eq. 54) for small a(z)½sin a(z) % a(z)Š into Eq. 52), we obtain the maximum brightness: Bmax ¼ jc e0 w(z) pkT (I:55) Correspondingly, the invariance maximum brightness [Eq. 53)] is Binv, max ¼ jc e0 w pkT (I:56) According to Eq. 54), for small a the maximum current density is jmax ¼ jc e0 w a(z)2 kT (I:57) PART I ELECTRON BEAMS Electron beams are flows of free electrons moving in the direction chosen. This is called the axis of the beam.

87) is the Larmor circle of the electron gyrating in the static electromagnetic field if E=cB , 1. Then, taking into account the fact that the momentum on the circle is P ¼ P? ¼ const:, we obtain þ þ P dl ¼ P? dl ¼ 2pp? r? À e0 2pr? A? (1:88) Application of Stokes’ theorem to Eq. 3) yields ð ð þ Bn dS ¼ curln A dS ¼ A? dl ¼ 2pr? A? 89) is taken over by the Larmor circle, where normal components inhomogeneous magnetic field Bn ¼ B ¼ const: Therefore, Ðof a weakly 2 B dS ¼pr B. By substituting this in Eqs.

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