= Solution
Apply the rotating-vector derivative rule first to the position <vector> and then to its <velocity>. Since $\boldsymbol\omega$ is constant,
$$
\mathbf v=\mathbf v'+\boldsymbol\omega\times\mathbf x,\qquad \mathbf a=\mathbf a'+2\boldsymbol\omega\times\mathbf v'+\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf x).
$$
For example, the derivative of $\mathbf v'$ contributes $\mathbf a'+\boldsymbol\omega\times\mathbf v'$, while that of $\boldsymbol\omega\times\mathbf x$ contributes the other two terms. This is the <equation of motion in a rotating frame>. The mutual <central forces> are unchanged in form because rotations preserve lengths and differences of position vectors. <Newton's second law> for particle $i$ becomes
$$
m\mathbf a_i'=\sum_{j\ne i}\mathbf F_{ij}+e(\mathbf v_i'+\boldsymbol\omega\times\mathbf x_i)\times\mathbf B-2m\boldsymbol\omega\times\mathbf v_i'-m\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf x_i).
$$
Set $\boldsymbol\omega=-e\mathbf B/(2m)$. The velocity-dependent <Lorentz force> is $e\mathbf v_i'\times\mathbf B=-e\mathbf B\times\mathbf v_i'$, and it cancels the <Coriolis force> $-2m\boldsymbol\omega\times\mathbf v_i'=e\mathbf B\times\mathbf v_i'$. The remaining terms combine to $m\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf x_i)$ and are quadratic in the <magnetic field>. Neglecting them gives
$$
\boxed{m\mathbf a_i'=\sum_{j\ne i}\mathbf F_{ij}+O(B^2)},
$$
identical to the zero-field equations to first order. The <Larmor theorem> requires the common charge-to-mass ratio; a mixture with different ratios cannot use one rotating frame to cancel every magnetic term.
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