Правка разделов «Statement», «Answer»
en/10.1.5.md
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| ### Statement | |||
| − | $10.1.5.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| − | ### Statement | ||
| − | |||
| 10.1.5. How much time after their first encounter will two charged particles meet again if they move perpendicular to a magnetic field with induction $B$? The charge of the particles is $q$, and their mass is $m$. Neglect their interaction. | |||
| ### Solution | |||
| Charged particles in a magnetic field move in circular orbits under the action of the Lorentz force, provided that their velocities are perpendicular to the magnetic field. Using the result of problem 10.1.3a, we write the cyclotron frequency: | |||
| $$\omega = \frac{qB}{m} \quad (1)$$ | |||
| Since the charge and mass of the particles are the same, their cyclotron frequencies are also equal: $\omega_1 = \omega_2$. The particles rotate in the same direction, either clockwise or counterclockwise, depending on the sign of the charge. If their trajectories intersected, the next encounter will be after one period. | |||
| The frequency is related to the period of revolution: | |||
| $$\omega = \frac{2\pi}{T} \quad (2)$$ | |||
| Equating (1) and (2): | |||
| $$\frac{qB}{m} = \frac{2\pi}{T}$$ | |||
| $$T = \frac{2\pi m}{qB}$$ | |||
| @@ -25,7 +19,3 @@Solution | |||
| #### Answer | |||
| $$t = \frac{2\pi m}{qB}$$ | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,11 +1,5 @@ | |||
| ### Statement | ### Statement | ||
| $10.1.5.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| 10.1.5. How much time after their first encounter will two charged particles meet again if they move perpendicular to a magnetic field with induction $B$? The charge of the particles is $q$, and their mass is $m$. Neglect their interaction. | 10.1.5. How much time after their first encounter will two charged particles meet again if they move perpendicular to a magnetic field with induction $B$? The charge of the particles is $q$, and their mass is $m$. Neglect their interaction. | ||
| ### Solution | ### Solution | ||
| Charged particles in a magnetic field move in circular orbits under the action of the Lorentz force, provided that their velocities are perpendicular to the magnetic field. Using the result of problem 10.1.3a, we write the cyclotron frequency: | Charged particles in a magnetic field move in circular orbits under the action of the Lorentz force, provided that their velocities are perpendicular to the magnetic field. Using the result of problem 10.1.3a, we write the cyclotron frequency: | ||
| $$\omega = \frac{qB}{m} \quad (1)$$ | $$\omega = \frac{qB}{m} \quad (1)$$ | ||
| Since the charge and mass of the particles are the same, their cyclotron frequencies are also equal: $\omega_1 = \omega_2$. The particles rotate in the same direction, either clockwise or counterclockwise, depending on the sign of the charge. If their trajectories intersected, the next encounter will be after one period. | Since the charge and mass of the particles are the same, their cyclotron frequencies are also equal: $\omega_1 = \omega_2$. The particles rotate in the same direction, either clockwise or counterclockwise, depending on the sign of the charge. If their trajectories intersected, the next encounter will be after one period. | ||
| The frequency is related to the period of revolution: | The frequency is related to the period of revolution: | ||
| $$\omega = \frac{2\pi}{T} \quad (2)$$ | $$\omega = \frac{2\pi}{T} \quad (2)$$ | ||
| Equating (1) and (2): | Equating (1) and (2): | ||
| $$\frac{qB}{m} = \frac{2\pi}{T}$$ | $$\frac{qB}{m} = \frac{2\pi}{T}$$ | ||
| $$T = \frac{2\pi m}{qB}$$ | $$T = \frac{2\pi m}{qB}$$ | ||
| @@ -25,7 +19,3 @@Solution | |||
| #### Answer | #### Answer | ||
| $$t = \frac{2\pi m}{qB}$$ | $$t = \frac{2\pi m}{qB}$$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||