Consider a long thin conducting wire carrying a uniform current $I$. A particle having mass $M$ and charge $q$ is released at a distance $a$ from the wire with a speed $v_0$ along the direction of current in the wire. The particle gets attracted to the wire due to the magnetic force. The particle turns round when it is at distance $x$ from the wire. The value of $x$ is [$\mu_0$ is vacuum permeability]
Answer: (A) $a\,e^{-\frac{4\pi Mv_0}{q\mu_0I}}$
Take the wire along $z$ with current in $+z$, and the particle at distance $r$ along $x$. There $\vec B = \dfrac{\mu_0I}{2\pi r}\hat y$ and
$$\vec F = q\vec v\times\vec B \Rightarrow M\frac{dv_z}{dt} = qv_rB = \frac{q\mu_0I}{2\pi r}\frac{dr}{dt}$$
Integrating from $r = a$ ($v_z = v_0$):
$$M(v_z - v_0) = \frac{q\mu_0I}{2\pi}\ln\frac{r}{a}$$
The magnetic force does no work, so the speed stays $v_0$. The particle turns round where its radial velocity is zero, i.e. its velocity is purely along $z$; having been turned back, $v_z = -v_0$:
$$-2Mv_0 = \frac{q\mu_0I}{2\pi}\ln\frac{x}{a} \Rightarrow x = a\,e^{-\frac{4\pi Mv_0}{q\mu_0I}}$$
Solution by Sreeraj P, M.Sc Physics