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An electron rotates in an orbit of radius 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 , directed along the axis of rotation?

Solutions of Savchenko Problems in Physics
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    <h3 id="back-link"><a href="/#10.1">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
      $10.1.15$
      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? 
    </p>

    <h3>Solution</h3>
    <p>
      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,
    </p>
    <h4>Answer</h4>
    <p>
        $$\omega \approx \omega_0 - \frac{eB}{2m_e}$$
    </p>
    <p>
        Note that $\omega_0$ is just two orders greater than $\frac{eB}{2m_e}$.
    </p>


    <p style="text-align: right; font-style: italic; font-size: 14;">   
      BSc. Luis Daniel Fernández Quintana<br>
      Physics Department (FCNE)<br>
      Universidad de Oriente, Cuba<br>
    </p>



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