Chapter 4 SINGLE PARTICLE MOTIONS 4.1 Introduction

4.4 Inhomogeneous Fields 115Comparing with Eq. (4.98) we obtainv r = − mqB θ˙ v ‖which integrates to giver − r 0 = − m v ‖ (4.102)qB θwhere r 0 is the turning point at which v ‖ = 0. Equation (4.102) describesthe poloidal projection of the banana orbit. To obtain the orbit width, we useEq. (4.86) with K ≈ mv⊥/2 2 to obtain for the change in v ‖ around the orbitδv ‖ =[ 2m (K − µB) ] 1/2≈ v ⊥ ɛ 1/2 (4.103)where we have substituted from Eq. (4.92) for B. Alsonotethatso that Eq. (4.102) becomesω cθ = qB θm ≈ɛ qB 0q(r) m= ɛω cq(r)(4.104)δr = δv ‖ω cθ≈ q(r)r L /ɛ 1/2 . (4.105)The width of the banana orbit is bigger again by the factor ɛ −1/2 than thepassing particles. This has profound consequences, with neoclassical diffusionincreased again over the cylindrical value.ProblemsProblem 4.1 The polarization drift v P can also be derived from energy conservation.If E is oscillating, the E×B drift also oscillates, and there is an energy 1 2 mv2 Eassociated with the guiding centre motion. Since energy can be gained from an Efield only be motion along E, there must be a drift v P in the E direction. By equatingthe rate of change of energy gain from v P .E, find the required value of v P .HINT: v P and E are in quadrature.Problem 4.2 A1keVprotonwithv ‖ =0in a uniform magnetic field B =0.1Tis accelerated as B is slowly increased to 1T. It then makes an elastic collision witha heavy particle and changes direction so that v ‖ = v ⊥ . The B field is then slowlydecreased back to 0.1 T. What is the proton energy now?