= Solution
In classical <charged particle in a uniform magnetic field> motion, the <Lorentz force> and <Newton's second law> give
$$
m\dot{\mathbf v}=q\mathbf v\times\mathbf B.
$$
Assume $\mathbf B\ne0$ and choose an orthonormal coordinate system with $\mathbf B=B\mathbf e_z$, $B=|\mathbf B|$. Write $\Omega=qB/m$, retaining the sign of the <electric charge>. The <velocity> equations are
$$
\dot v_x=\Omega v_y,\qquad\dot v_y=-\Omega v_x,\qquad\dot v_z=0.
$$
Thus the parallel <velocity> is constant, and the perpendicular <velocity> rotates at constant <angular frequency>. Choosing the time origin and perpendicular axes conveniently gives
$$
v_x=v_\perp\cos\Omega t,\qquad v_y=-v_\perp\sin\Omega t,\qquad v_z=v_\parallel.
$$
Integrating the <velocity>, one obtains
$$
x=x_c+\frac{v_\perp}{\Omega}\sin\Omega t,\qquad y=y_c+\frac{v_\perp}{\Omega}\cos\Omega t,\qquad z=z_0+v_\parallel t.
$$
The first two coordinates describe a <circle> of radius $mv_\perp/(|q|B)$, while the third advances uniformly. This is <helical motion in a uniform magnetic field>: the <helix> has \b[axis parallel to $\mathbf B$] through the <guiding centre>. Its <cyclotron frequency> is
$$
\boxed{\omega_c=\frac{|q|B}{m}},
$$
and its signed <angular velocity> vector about that axis is $-q\mathbf B/m$. The minus sign follows from the ordering $\mathbf v\times\mathbf B$ in the <Lorentz force>. The <helix> degenerates to a <circle> when $v_\parallel=0$ and to a straight line when $v_\perp=0$. If $q=0$ or $B=0$, there is only straight-line motion. In a relativistic extension with fixed <Lorentz factor> $\gamma$, the frequency becomes $|q|B/(\gamma m)$; the classical result above uses $m\mathbf v$ as <momentum>.
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