Правка разделов «Statement», «Solution»
en/10.1.10.md
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| ### Statement | |||
| 10.1.10*. Particles of mass $m$ and charge $q$ fly out from point A with velocity $v$, having a small angular spread $\delta\alpha$, and then move in a uniform magnetic field of induction $B$ perpendicular to it. Determine at what distance from point A the beam will converge, and estimate its transverse dimension at this location. | |||
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| ### Solution | |||
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| The particle velocity is perpendicular to the magnetic field induction vector, which means they will move not along helical paths, but along circles lying in the same plane. Since their velocities are the same, the radii of all trajectories are equal: | |||
| $$R = \frac{mv}{qB}$$ | |||
| The position of the circle's center depends on the emission angle $\beta$. Let us place the system in a rectangular coordinate system so that the $Oy$ axis passes through the center of the beam. The focusing of the beam will occur exactly near $Oy$. Note that the coordinate of the intersection point of the trajectory and $Oy$ is determined as | |||
| $$y = 2R\cos\beta$$ | |||
| Moreover, $y_{\max} = 2R$ corresponds to $\beta = 0$, and $y_{\min} = 2R\cos(\delta\alpha/2)$ corresponds to the maximum angle $\beta = \delta\alpha/2$. The difference between these values will be the required transverse dimension. The angle $\delta\alpha$ is small, which means we can assume that $\delta\alpha/2$ is small, and $\cos(\delta\alpha/2) \approx 1 - \frac{(\delta\alpha/2)^2}{2}$. | |||
| Then | |||
| $$\Delta y = y_{\max} - y_{\min} = 2R - 2R\left(1 - \frac{(\delta\alpha/2)^2}{2}\right) = \frac{mv(\delta\alpha)^2}{4qB}$$ | |||
| The edge of the region $\Delta y$ will be located at a distance $l = y_{\max} = 2R = \frac{2mv}{qB}$ from point A. | |||
| #### Answer | |||
| $$l = \frac{2mv}{qB}; \quad \Delta y = \frac{mv(\delta\alpha)^2}{4qB}$$ | |||
| ещё строк без изменений 13 | |||
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| ### Statement | ### Statement | ||
| 10.1.10*. Particles of mass $m$ and charge $q$ fly out from point A with velocity $v$, having a small angular spread $\delta\alpha$, and then move in a uniform magnetic field of induction $B$ perpendicular to it. Determine at what distance from point A the beam will converge, and estimate its transverse dimension at this location. | 10.1.10*. Particles of mass $m$ and charge $q$ fly out from point A with velocity $v$, having a small angular spread $\delta\alpha$, and then move in a uniform magnetic field of induction $B$ perpendicular to it. Determine at what distance from point A the beam will converge, and estimate its transverse dimension at this location. | ||
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| ### Solution | ### Solution | ||
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| The particle velocity is perpendicular to the magnetic field induction vector, which means they will move not along helical paths, but along circles lying in the same plane. Since their velocities are the same, the radii of all trajectories are equal: | The particle velocity is perpendicular to the magnetic field induction vector, which means they will move not along helical paths, but along circles lying in the same plane. Since their velocities are the same, the radii of all trajectories are equal: | ||
| $$R = \frac{mv}{qB}$$ | $$R = \frac{mv}{qB}$$ | ||
| The position of the circle's center depends on the emission angle $\beta$. Let us place the system in a rectangular coordinate system so that the $Oy$ axis passes through the center of the beam. The focusing of the beam will occur exactly near $Oy$. Note that the coordinate of the intersection point of the trajectory and $Oy$ is determined as | The position of the circle's center depends on the emission angle $\beta$. Let us place the system in a rectangular coordinate system so that the $Oy$ axis passes through the center of the beam. The focusing of the beam will occur exactly near $Oy$. Note that the coordinate of the intersection point of the trajectory and $Oy$ is determined as | ||
| $$y = 2R\cos\beta$$ | $$y = 2R\cos\beta$$ | ||
| Moreover, $y_{\max} = 2R$ corresponds to $\beta = 0$, and $y_{\min} = 2R\cos(\delta\alpha/2)$ corresponds to the maximum angle $\beta = \delta\alpha/2$. The difference between these values will be the required transverse dimension. The angle $\delta\alpha$ is small, which means we can assume that $\delta\alpha/2$ is small, and $\cos(\delta\alpha/2) \approx 1 - \frac{(\delta\alpha/2)^2}{2}$. | Moreover, $y_{\max} = 2R$ corresponds to $\beta = 0$, and $y_{\min} = 2R\cos(\delta\alpha/2)$ corresponds to the maximum angle $\beta = \delta\alpha/2$. The difference between these values will be the required transverse dimension. The angle $\delta\alpha$ is small, which means we can assume that $\delta\alpha/2$ is small, and $\cos(\delta\alpha/2) \approx 1 - \frac{(\delta\alpha/2)^2}{2}$. | ||
| Then | Then | ||
| $$\Delta y = y_{\max} - y_{\min} = 2R - 2R\left(1 - \frac{(\delta\alpha/2)^2}{2}\right) = \frac{mv(\delta\alpha)^2}{4qB}$$ | $$\Delta y = y_{\max} - y_{\min} = 2R - 2R\left(1 - \frac{(\delta\alpha/2)^2}{2}\right) = \frac{mv(\delta\alpha)^2}{4qB}$$ | ||
| The edge of the region $\Delta y$ will be located at a distance $l = y_{\max} = 2R = \frac{2mv}{qB}$ from point A. | The edge of the region $\Delta y$ will be located at a distance $l = y_{\max} = 2R = \frac{2mv}{qB}$ from point A. | ||
| #### Answer | #### Answer | ||
| $$l = \frac{2mv}{qB}; \quad \Delta y = \frac{mv(\delta\alpha)^2}{4qB}$$ | $$l = \frac{2mv}{qB}; \quad \Delta y = \frac{mv(\delta\alpha)^2}{4qB}$$ | ||
| ещё строк без изменений 13 | |||