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Taking the scalar product of the Lorentz force equation with the velocity givesThe triple product vanishes because the cross product is perpendicular to its first factor. Thus the kinetic energy is constant, even if the magnetic field varies with position: magnetic forces do no work.

In cylindrical coordinates, the radial, angular and vertical equations areFor a nonstationary circular orbit with constant $r$ and $z$, the first equation gives $-m{\dot{\theta }}^2=qB(r)\dot{\theta }$. Hence its signed angular velocity is $\boxed{\dot{\theta }=-qB(r)/m}$. The sign gives the sense of circulation. The zero root describes a stationary transverse position, not a moving circular orbit. A constant vertical velocity can be superposed to produce a helix.

Define the mechanical angular momentum about the axis by $L=m{r}^2\dot{\theta }$. Multiplying the angular equation by $r$ givesFor the inverse-radius magnetic field this becomes $\dot{L}=-q{B}_0\dot{r}$. Integrating from an initial radius $r_0$ provesLet $c=q{B}_0/m$ and $\ell =(L_0+q{B}_0{r}_0)/m$, so that $L/m=\ell -cr$. Since $\dot{z}$ is constant and the total kinetic energy is constant, the transverse speed squaredis also constant. Substituting the angular momentum relation gives the required radial curve:Only the region $W(r)\ge 0$ is accessible. For $\ell \ne 0$, $W(r)\to -\infty$ as $r\to 0$, and its horizontal asymptote is $v_{\perp }^2-c^2$. Its derivative is $W^{\prime }(r)=2\ell (\ell -cr)/r^3$. Thus, when $\ell c>0$, it has a single maximum $W=v_{\perp }^2$ at $r=\ell /c$. The positive roots of $W$ are the turning points obtained from $r=\ell /(c\pm v_{\perp })$.

For nonzero $c$ and $\ell c>0$, write $r_c=|\ell /c|$. When $0<v_{\perp }<|c|$, two turning points occur atand the radius is confined between them. At $v_{\perp }=|c|$, there is only the finite lower turning point $r_c/2$; $W$ tends to zero from above and an outgoing particle reaches arbitrarily large radius. For $v_{\perp }>|c|$, only the lower turning point remains and the outgoing radial velocity approaches $\sqrt{v_{\perp }^2-c^2}$. The sketch shows these three cases and omits inaccessible negative values of ${\dot{r}}^2$ from the physical interpretation.

For completeness, if $\ell c<0$, $W$ rises monotonically from $-\infty$ to its asymptote; admissible motion then requires $v_{\perp }>|c|$ and is radially unbound after any inward branch turns. If $\ell =0$, $W=v_{\perp }^2-c^2$ is constant; equality gives a circular orbit rather than escape. Above threshold the outgoing branch escapes, whereas an incoming branch reaches the singular axis in finite time and cannot be continued through it without specifying a nonsingular core field. These cases explain why a borderline escape assertion needs an initial-direction qualification.

The transverse escape velocity threshold for radial escape in an inverse-radius axial magnetic field isFor a radial launch at $r_0$, $\ell =c{r}_0$, and equality permits marginal escape; for $\ell \ne 0$, larger transverse speed guarantees radial escape in forward time after any initial inward leg turns. The singular incoming $\ell =0$ branch is the exception described above. More general launches at equality can instead give the circular exception $\ell =0$. A vertical velocity adds to the total speed but cannot pay the transverse escape-energy requirement; if escape merely means unbounded $z$, any nonzero $\dot{z}$ already achieves that. If $q{B}_0=0$, transverse motion is force-free and there is no positive escape-speed threshold.