= Solution
Taking the <scalar product> of the <Lorentz force> equation with the <velocity> gives
$$
\frac{d}{dt}\left(\frac12m|\dot{\mathbf r}|^2\right)
=q\dot{\mathbf r}\cdot(\dot{\mathbf r}\times\mathbf B)=0.
$$
The triple product vanishes because the <cross product> is perpendicular to its first factor. Thus \b[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 are
$$
m(\ddot r-r\dot\theta^2)=qr\dot\theta B(r),\qquad
m(r\ddot\theta+2\dot r\dot\theta)=-q\dot r B(r),\qquad
m\ddot z=0.
$$
For 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 $L=mr^2\dot\theta$. Multiplying the angular equation by $r$ gives
$$
\dot L=-qrB(r)\dot r.
$$
For the inverse-radius <magnetic field> this becomes $\dot L=-qB_0\dot r$. Integrating from an initial radius $r_0$ proves
$$
\boxed{L=L_0-qB_0(r-r_0).}
$$
Let $c=qB_0/m$ and $\ell=(L_0+qB_0r_0)/m$, so that $L/m=\ell-cr$. Since $\dot z$ is constant and the total <kinetic energy> is constant, the transverse <speed> squared
$$
v_\perp^2=\dot r^2+r^2\dot\theta^2
$$
is also constant. Substituting the <angular momentum> relation gives the required radial curve:
$$
\boxed{\dot r^2=W(r)=v_\perp^2-\left(\frac{\ell}{r}-c\right)^2
=v_\perp^2-c^2+\frac{2\ell c}{r}-\frac{\ell^2}{r^2}.}
$$
Only the region $W(r)\geq0$ is accessible. For $\ell\ne0$, $W(r)\to-\infty$ as $r\to0$, and its horizontal asymptote is $v_\perp^2-c^2$. Its derivative is $W'(r)=2\ell(\ell-cr)/r^3$. 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 at
$$
r_-=\frac{r_c}{1+v_\perp/|c|},\qquad
r_+=\frac{r_c}{1-v_\perp/|c|},
$$
and 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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/ia/paper-4-magnetic-radial-motion.png]
{title=Radial-speed curves and turning points in an inverse-radius axial magnetic field}
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> is
$$
\boxed{v_{\perp,\mathrm{esc}}=\frac{|qB_0|}{m}.}
$$
For a radial launch at $r_0$, $\ell=cr_0$, and equality permits marginal escape; for $\ell\ne0$, 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 $qB_0=0$, transverse motion is force-free and there is no positive escape-speed threshold.
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