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en/10.1.4.md
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| ### Statement | |||
| − | $10.1.4.$ [Insert the problem statement] | ||
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| − | ### Solution | ||
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| − | ### Statement | ||
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| 10.1.4. What is the ratio of the trajectory radii of two electrons with kinetic energies $K_1$ and $K_2$ if a uniform magnetic field is perpendicular to their velocities? | |||
| ### Solution | |||
| When an electron moves in a magnetic field, it experiences the Lorentz force, which causes it to move in a circular trajectory. The Lorentz force for an electron moving with velocity $v$ perpendicular to the magnetic field $B$ is: | |||
| $$F = evB$$ | |||
| The centripetal acceleration for circular motion is described by the expression: | |||
| $$a = \frac{v^2}{R}$$ | |||
| Let us write Newton's second law: | |||
| $$evB = \frac{m_e v^2}{R}$$ | |||
| From this equation, we can express the trajectory radius $R$: | |||
| $$R = \frac{m_e v}{eB} \quad (1)$$ | |||
| The kinetic energy of the first electron: | |||
| $$K_1 = \frac{1}{2}m_e v^2$$ | |||
| From here, we express the velocity: | |||
| $$v = \sqrt{\frac{2K_1}{m_e}}$$ | |||
| Substitute into (1): | |||
| $$R_1 = \frac{m_e \sqrt{\frac{2K_1}{m_e}}}{eB} = \frac{\sqrt{2K_1 m_e}}{eB} \quad (2)$$ | |||
| Similarly, for the electron with kinetic energy $K_2$: | |||
| $$R_2 = \frac{\sqrt{2K_2 m_e}}{eB} \quad (3)$$ | |||
| Let us take the ratio of the radii: | |||
| $$\frac{R_1}{R_2} = \sqrt{\frac{K_1}{K_2}}$$ | |||
| @@ -40,7 +34,3 @@Solution | |||
| #### Answer | |||
| $$\frac{R_1}{R_2} = \sqrt{\frac{K_1}{K_2}}$$ | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,11 +1,5 @@ | |||
| ### Statement | ### Statement | ||
| $10.1.4.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| 10.1.4. What is the ratio of the trajectory radii of two electrons with kinetic energies $K_1$ and $K_2$ if a uniform magnetic field is perpendicular to their velocities? | 10.1.4. What is the ratio of the trajectory radii of two electrons with kinetic energies $K_1$ and $K_2$ if a uniform magnetic field is perpendicular to their velocities? | ||
| ### Solution | ### Solution | ||
| When an electron moves in a magnetic field, it experiences the Lorentz force, which causes it to move in a circular trajectory. The Lorentz force for an electron moving with velocity $v$ perpendicular to the magnetic field $B$ is: | When an electron moves in a magnetic field, it experiences the Lorentz force, which causes it to move in a circular trajectory. The Lorentz force for an electron moving with velocity $v$ perpendicular to the magnetic field $B$ is: | ||
| $$F = evB$$ | $$F = evB$$ | ||
| The centripetal acceleration for circular motion is described by the expression: | The centripetal acceleration for circular motion is described by the expression: | ||
| $$a = \frac{v^2}{R}$$ | $$a = \frac{v^2}{R}$$ | ||
| Let us write Newton's second law: | Let us write Newton's second law: | ||
| $$evB = \frac{m_e v^2}{R}$$ | $$evB = \frac{m_e v^2}{R}$$ | ||
| From this equation, we can express the trajectory radius $R$: | From this equation, we can express the trajectory radius $R$: | ||
| $$R = \frac{m_e v}{eB} \quad (1)$$ | $$R = \frac{m_e v}{eB} \quad (1)$$ | ||
| The kinetic energy of the first electron: | The kinetic energy of the first electron: | ||
| $$K_1 = \frac{1}{2}m_e v^2$$ | $$K_1 = \frac{1}{2}m_e v^2$$ | ||
| From here, we express the velocity: | From here, we express the velocity: | ||
| $$v = \sqrt{\frac{2K_1}{m_e}}$$ | $$v = \sqrt{\frac{2K_1}{m_e}}$$ | ||
| Substitute into (1): | Substitute into (1): | ||
| $$R_1 = \frac{m_e \sqrt{\frac{2K_1}{m_e}}}{eB} = \frac{\sqrt{2K_1 m_e}}{eB} \quad (2)$$ | $$R_1 = \frac{m_e \sqrt{\frac{2K_1}{m_e}}}{eB} = \frac{\sqrt{2K_1 m_e}}{eB} \quad (2)$$ | ||
| Similarly, for the electron with kinetic energy $K_2$: | Similarly, for the electron with kinetic energy $K_2$: | ||
| $$R_2 = \frac{\sqrt{2K_2 m_e}}{eB} \quad (3)$$ | $$R_2 = \frac{\sqrt{2K_2 m_e}}{eB} \quad (3)$$ | ||
| Let us take the ratio of the radii: | Let us take the ratio of the radii: | ||
| $$\frac{R_1}{R_2} = \sqrt{\frac{K_1}{K_2}}$$ | $$\frac{R_1}{R_2} = \sqrt{\frac{K_1}{K_2}}$$ | ||
| @@ -40,7 +34,3 @@Solution | |||
| #### Answer | #### Answer | ||
| $$\frac{R_1}{R_2} = \sqrt{\frac{K_1}{K_2}}$$ | $$\frac{R_1}{R_2} = \sqrt{\frac{K_1}{K_2}}$$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||