= Solution
The <Lorentz force> and <Newton's second law> give
$$
m\ddot{\mathbf x}=q\dot{\mathbf x}\times\mathbf B.
$$
Writing $\Omega=qB/m$, the component equations are
$$
\ddot x=\Omega\dot y,\qquad \ddot y=-\Omega\dot x,\qquad \ddot z=0.
$$
The initial <velocity> therefore gives
$$
\dot x=v\sin(\Omega t),\qquad \dot y=v\cos(\Omega t),\qquad \dot z=v,
$$
and integration with the initial <position> gives
$$
\boxed{\mathbf x(t)=\left(1+\frac v\Omega[1-\cos(\Omega t)],\ \frac v\Omega\sin(\Omega t),\ vt\right)}.
$$
This is <helical motion in a uniform magnetic field>: a <helix> of radius $v/\Omega$ around the line $x=1+v/\Omega$, $y=0$, with its axis parallel to the <magnetic field>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-4-magnetic-helix.png]
{title=The helical trajectory of a charged particle in a magnetic field}
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