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| <span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span> | | <span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span> |
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| <p class="author"> | | <p class="author"> |
| Solutions of Savchenko Problems in Physics <br> | | Solutions of Savchenko Problems in Physics <br> |
| <i><b>knowledge must be free</b></i> | | <i><b>knowledge must be free</b></i> |
| </p> | | </p> |
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| <h3 id="back-link"><a href="../../#10.1">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#10.1">$\leftarrow$Back</a></h3> |
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| <p> | | <p> |
| Without magnetic field, electrical force is equivalent to centripetal force: | | Without magnetic field, electrical force is equivalent to centripetal force: |
| $$m_e\frac{{v_0}^2}{R} = \frac{e^2}{4\pi\varepsilon_0 R^2}$$ | | $$m_e\frac{{v_0}^2}{R} = \frac{e^2}{4\pi\varepsilon_0 R^2}$$ |
| and cosnidering that $\omega_0 = v_0 R$, | | and cosnidering that $\omega_0 = v_0 R$, |
| $${\omega_0}^2 = \frac{e^2}{4\pi\varepsilon_0 m_e R^3} \;(1)$$ | | $${\omega_0}^2 = \frac{e^2}{4\pi\varepsilon_0 m_e R^3} \;(1)$$ |
| If magnetic force is bigger than electric one, electron abandons the system, so | | If magnetic force is bigger than electric one, electron abandons the system, so |
| $$\frac{e^2}{4\pi\varepsilon_0 R^2}-evB = m_e\frac{v^2}{R}$$ | | $$\frac{e^2}{4\pi\varepsilon_0 R^2}-evB = m_e\frac{v^2}{R}$$ |
| where $v = \omega R$, then | | where $v = \omega R$, then |
| $$\omega^2 = \frac{e^2}{4\pi\varepsilon_0 m_e R^3} - \frac{e\omega B}{m_e}$$ | | $$\omega^2 = \frac{e^2}{4\pi\varepsilon_0 m_e R^3} - \frac{e\omega B}{m_e}$$ |
| Taking in account (1), | | Taking in account (1), |
| $$\omega^2 + \frac{e\omega B}{m_e} - {\omega_0}^2 = 0$$ | | $$\omega^2 + \frac{e\omega B}{m_e} - {\omega_0}^2 = 0$$ |
| Applying the general formula of solution for quadratic equations, and considering $\omega > 0$ | | Applying the general formula of solution for quadratic equations, and considering $\omega > 0$ |
| $$\omega = -\frac{eB}{2m_e} + \sqrt{\frac{e^2B^2}{4{m_e}^2}+{\omega_0}^2}$$ | | $$\omega = -\frac{eB}{2m_e} + \sqrt{\frac{e^2B^2}{4{m_e}^2}+{\omega_0}^2}$$ |
| 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, | | 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, |
| $${\omega_0}^2 = 10^{32} {\rm{s^{-2}}}$$ | | $${\omega_0}^2 = 10^{32} {\rm{s^{-2}}}$$ |
| and | | and |
| $$\frac{e^2B^2}{4{m_e}^2} \approx 10^{28} {\rm{s^{-2}}}$$ | | $$\frac{e^2B^2}{4{m_e}^2} \approx 10^{28} {\rm{s^{-2}}}$$ |
| For that, | | For that, |
| $${\omega_0}^2 >> \frac{e^2B^2}{4{m_e}^2}$$ | | $${\omega_0}^2 >> \frac{e^2B^2}{4{m_e}^2}$$ |
| by four orders. Finally, | | by four orders. Finally, |
| </p> | | </p> |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$\omega \approx \omega_0 - \frac{eB}{2m_e}$$ | | $$\omega \approx \omega_0 - \frac{eB}{2m_e}$$ |
| </p> | | </p> |
| <p> | | <p> |
| Note that $\omega_0$ is just two orders greater than $\frac{eB}{2m_e}$. | | Note that $\omega_0$ is just two orders greater than $\frac{eB}{2m_e}$. |
| </p> | | </p> |
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| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p style="text-align: right; font-style: italic; font-size: 14;"> |
| BSc. Luis Daniel Fernández Quintana<br> | | BSc. Luis Daniel Fernández Quintana<br> |
| Physics Department (FCNE)<br> | | Physics Department (FCNE)<br> |
| Universidad de Oriente, Cuba<br> | | Universidad de Oriente, Cuba<br> |
| </p> | | </p> |
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