$10.1.15$ An electron rotates in an orbit of radius $R$ around a proton. How will the frequency of rotation of an electron in the same orbit change if the system is placed in a weak magnetic field of induction $B$, directed along the axis of rotation?
Without magnetic field, electrical force is equivalent to centripetal force: $$m_e\frac{{v_0}^2}{R} = \frac{e^2}{4\pi\varepsilon_0 R^2}$$ and cosnidering that $\omega_0 = v_0 R$, $${\omega_0}^2 = \frac{e^2}{4\pi\varepsilon_0 m_e R^3} \quad(1)$$ If magnetic force is bigger than electric one, electron abandons the system, so $$\frac{e^2}{4\pi\varepsilon_0 R^2}-evB = m_e\frac{v^2}{R}$$ where $v = \omega R$, then $$\omega^2 = \frac{e^2}{4\pi\varepsilon_0 m_e R^3} - \frac{e\omega B}{m_e}$$ Taking in account (1), $$\omega^2 + \frac{e\omega B}{m_e} - {\omega_0}^2 = 0$$ Applying the general formula of solution for quadratic equations, and considering $\omega > 0$ $$\omega = -\frac{eB}{2m_e} + \sqrt{\frac{e^2B^2}{4{m_e}^2}+{\omega_0}^2}$$ Now, it's time to check the orders of each magnitude. Let's see the terms under square root. Let's consider $e \approx 10^{-19} {\rm{C}}$, $B \approx 10^{-6} T$ (due to problem refers to weak magnetic field, we take in account the order of Earth's magnetic field from Problem 10.1.2), $m_e \approx 10^{-31} {\rm{kg}}$ and $R \approx 10^{-10} {\rm{m}}$ (in the order of Å). Hence, $${\omega_0}^2 = 10^{32} {\rm{s^{-2}}}$$ and $$\frac{e^2B^2}{4{m_e}^2} \approx 10^{28} {\rm{s^{-2}}}$$ For that, $${\omega_0}^2 >> \frac{e^2B^2}{4{m_e}^2}$$ by four orders. Finally,
$$\omega\approx \omega_0 - \frac{eB}{2m_e}$$
Note that $\omega_0$ is just two orders greater than $\frac{eB}{2m_e}$.