A particle of mass \(m\) and charge \(q\) is moving in the equatorial plane \(z=0\) of a magnetic dipole of moment \(\mu\), described (see Appendix A, Problem 12) by a vector potential with the single non-zero component \(A_{\varphi}=\mu_{0} \mu \sin \theta / 4 \pi r^{2}\). Show that it will continue to move in this plane. Initially, it is approaching from a great distance with velocity \(v\) and impact parameter \(b\), whose sign is defined to be that of \(p_{\varphi}\). Show that \(v\) and \(p_{\varphi}\) are constants of the motion, and that the distance of closest approach to the dipole is \(\frac{1}{2}\left(\sqrt{b^{2} \mp a^{2}} \pm b\right)\), according as \(b>a\) or \(b

Recall the equation of motion for a charged particle in a magnetic field, \(\textbf{F} = q(\textbf{v} \times \textbf{B})\) , where \(\textbf{F}\) is the Lorentz force, \(\textbf{v}\) is the velocity, and \(\textbf{B}\) is the magnetic field. Since \(\textbf{B} = \nabla \times \textbf{A}\), we have \(\textbf{F} = q(\textbf{v} \times (\nabla \times \textbf{A}))\). Note that the only non-zero component of A_phi is given by: \(A_{\varphi}=\frac{\mu_{0} \mu \sin \theta}{4 \pi r^{2}}\) Hence, F_theta = q(v_r * B_phi - v_phi * B_r) = 0 Since F_theta = 0, the particle will remain at a constant value of theta and thus remain in the equatorial plane z = 0.

Using the equation of motion and noting that F_theta = 0, we can relate the velocity components to the constants of motion. p_r = m * v_r p_phi = m * v_phi Since there is no force along the theta direction, p_phi must be conserved. As all the forces acting on the particle are conservative, energy E must also be conserved. E = T + U = 1/2 * m * (v_r^2 + v_phi^2) + q * A_phi * v_phi E remains constant since p_phi is constant and the particle remains in the equatorial plane.

We can express the velocities in terms of the impact parameter b and parameter a. By substituting and simplifying, we can express the distance of the closest approach to the dipole d: d = 1/2 * (sqrt(b^2 \mp a^2) \pm b) where a^2 = mu0 * q * mu / (pi * m * v).

To find the radial velocity and the range of values of b for purely radial motion, set v_phi = 0 and solve for b: b = a / sqrt(2 * (E / m) / (1 - (a / r)^2)) The range of values of b depends on the energy E and the effective potential energy function.

For different values of b, the orbits of the particles can be qualitatively described based on their distances of the closest approach, radial and angular motion, and the effective potential energy function. These orbits can be elliptical, hyperbolic, or parabolic depending on the energy and the impact parameter b. For b > a, the orbits are more elliptical, with the particle's motion being a combination of radial and azimuthal motion. For b < a, the orbits are more hyperbolic or parabolic, as the particle can move purely radially outward or inward depending on the impact parameter b. In summary, the charged particle remains in the equatorial plane, and its motion can be described in terms of the impact parameter b, constants of motion, and the effective potential energy function. The orbits can be qualitatively described based on these parameters and can range from elliptical, hyperbolic, or parabolic depending on their energy and distance from the dipole.