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Basic Plasma Physics 41 v q Fc B Fig. 3-1. Positively charged particle moving in a uniform vertical magnetic field. mv 2 Fc=qv B= r , (3.3-11) where r is the radius of the cycloidal motion in the magnetic field. Solving for the radius of the circle gives r = rL = mv , (3.3-12) qB which is the Larmor radius. The Larmor radius can be written in a form simple to remember: rL = v = 1 2mV , c B e (3.3-13) using 1 2 mv 2 = eV for the singly charged particle energy in the direction perpendicular to the magnetic field. The direction of particle gyration is always such that the induced magnetic field is opposite in direction to the applied field, which tends to reduce the applied field, an effect called diamagnetism. Any particle motion along the magnetic field is not affected by the field, but causes the particle motion to form a helix along the magnetic field direction with a radius given by the Larmor radius and a pitch given by the ratio of the perpendicular to parallel velocities. Next consider the situation in Fig. 3-1, but with the addition of a finite electric field perpendicular to B. In this case, E is in some direction in the plane of the page. The equation of motion for the charged particle is given by Eq. (3.3-1). Considering the drift to be steady-state, the time derivative is equal to zero, and Eq. (3.3-1) becomes E = v B . (3.3-14) Taking the cross product of both sides with B gives E B=( v B) B=vB2 B(B v). (3.3-15) The dot product is in the direction perpendicular to B, so the last term in Eq. (3.3-15) is equal to zero. Solving for the transverse velocity of the particle givesPDF Image | Fundamentals of Electric Propulsion: Ion and Hall Thrusters
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