Added Luis's English Solution of 10.1.15

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+ <span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span>
+ </div>
+ <p class="author">
+ Solutions&nbsp;of&nbsp;Savchenko Problems&nbsp;in&nbsp;Physics <br>
+ <i><b>knowledge must be free</b></i>
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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, 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 $\omega_0 = v_0 R$,
+ $$\omega_0 = \frac{e^2}{4\pi\varepsilon_0 m_e R^3} \;(1)$$
+ 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>
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