Added Luis's English solution of problems 10.2.11 and 10.2.12

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+ <h2>Solutions of Savchenko Physics Textbook</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../#10.2">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $10.2.12.$ Find the drift velocity of the electron and proton in the gravity field and the Earth's magnetic field, the induction of which is equal to 0.7 · 10$^{-4}$ T. The magnetic field is perpendicular to the gravity field.
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ Drift velocity is achieved when particle's path is unaffected respect to force fields that acting over it. So, absolute values of these forces are equal and their directions are opposed. Hence,
+ </p>
+ <p style="text-align: center;">
+ $mg = q~v_d~B$
+ </p>
+ <p style="text-align: center;">
+ $v_d = \frac{mg}{qB}$
+ </p>
+ <p>
+ Now, considering the following data, calculations for each particle are possible: mass of electron $m_e = 9.1 \dot 10^{-31}$ kg, mass of proton $m_p = 1.672 \dot 10^{-27}$ kg, absolute value of charge $q = 1.6 \dot 10^{-19}$ C and acceleration due to gravity $g = 9.8$ m/s$^2$. Calculating,
+ </p>
+ <p style="text-align: center;">
+ $v_e = 7.9625 \dot 10^{-7}$ m/s
+ </p>
+ <p style="text-align: center;">
+ $v_p = 1.423625 \dot 10^{-3}$ m/s
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$v_e \simeq 8 \dot 10^{-7} m/s, v_p \simeq 1.4 \dot 10^{-3} m/s$$
+ </p>
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