= Solution
After the <electric field> is removed, the uniform <magnetic field> preserves the speed. Decompose the instantaneous initial <velocity> as $\mathbf v_0=\mathbf v_\parallel+\mathbf v_\perp$, with $\hat{\mathbf b}=\mathbf B/B$ and signed $\Omega=qB/m$. Then
$$
\mathbf v(t)=\mathbf v_\parallel+\mathbf v_\perp\cos\Omega t+(\mathbf v_\perp\times\hat{\mathbf b})\sin\Omega t,
$$
and, integrating,
$$
\boxed{\mathbf r(t)=\mathbf r_0+\mathbf v_\parallel t+\frac{\mathbf v_\perp}{\Omega}\sin\Omega t+\frac{\mathbf v_\perp\times\hat{\mathbf b}}{\Omega}(1-\cos\Omega t).}
$$
This is a <helical motion in a uniform magnetic field> of radius $m|\mathbf v_\perp|/(|q|B)$ and angular frequency $|q|B/m$. It is a circle if $\mathbf v_\parallel=0$, and a straight line if $\mathbf v_\perp=0$. The formula assumes a uniform, nonzero field and nonrelativistic motion.
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