Правка раздела «Statement»
en/8.1.1.md
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| ### Statement | |||
| − | $8.1.1.$ [Insert the problem statement] | ||
| + | $8.1.1.$ a) In a synchrotron, electrons move in an approximately circular orbit of length $\ell = 240\;\rm{m}$. During the acceleration cycle, there are approximately $n = 10^{11}$ electrons in orbit, their speed is almost equal to the speed of light. What is the current equal to?\ | ||
| + | b) Determine the current generated by an electron moving in an orbit of radius $r = 0.5 · 10^{−10}\;\rm{m}$ in a hydrogen atom. | ||
| ### Solution | |||
| The electric current is defined by: | |||
| $i = \frac{dq}{dt}$ | |||
| but $dq = e dn$, where e is the fundamental electrical charge and $dt = ds/c$, where c is the speed of light. | |||
| $i = ce \frac{dn}{ds}$ | |||
| we can approximate $\frac{dn}{ds}$ to $\frac{n}{\ell}$, son | |||
| $i \simeq \frac{cen}{\ell} = 0.02\;\rm{A}$ | |||
| b) Again, electric current can be expressed as | |||
| $i = \frac{dq}{dt} \simeq \frac{e}{t}$ (1) | |||
| Applying Newton Second Law: | |||
| $\frac{e^2}{4\pi\varepsilon r^2} = \frac{m_e v^2}{r}$ | |||
| where $v = \frac{2\pi r}{t}$, | |||
| $t = \sqrt{\frac{16 (\pi r)^3 m_e}{e^2}}$ (2) | |||
| Putting (2) into (1), | |||
| $i = \sqrt{\frac{e^4}{16m_e(\pi r)^3}} = 0.0012\;\rm{A}$ | |||
| #### Answer | |||
| a) $i \simeq \frac{cen}{\ell} = 0.02\;\rm{A}$ | |||
| b) $i = \sqrt{\frac{e^4}{16m_e(\pi r)^3}} = 0.0012\;\rm{A}$ | |||
| [Insert a concise answer or boxed result] | |||
| ещё строк без изменений 34 | |||
| @@ -1,6 +1,7 @@ | |||
| ### Statement | ### Statement | ||
| $8.1.1.$ [Insert the problem statement] | $8.1.1.$ a) In a synchrotron, electrons move in an approximately circular orbit of length $\ell = 240\;\rm{m}$. During the acceleration cycle, there are approximately $n = 10^{11}$ electrons in orbit, their speed is almost equal to the speed of light. What is the current equal to?\ | ||
| b) Determine the current generated by an electron moving in an orbit of radius $r = 0.5 · 10^{−10}\;\rm{m}$ in a hydrogen atom. | |||
| ### Solution | ### Solution | ||
| The electric current is defined by: | The electric current is defined by: | ||
| $i = \frac{dq}{dt}$ | $i = \frac{dq}{dt}$ | ||
| but $dq = e dn$, where e is the fundamental electrical charge and $dt = ds/c$, where c is the speed of light. | but $dq = e dn$, where e is the fundamental electrical charge and $dt = ds/c$, where c is the speed of light. | ||
| $i = ce \frac{dn}{ds}$ | $i = ce \frac{dn}{ds}$ | ||
| we can approximate $\frac{dn}{ds}$ to $\frac{n}{\ell}$, son | we can approximate $\frac{dn}{ds}$ to $\frac{n}{\ell}$, son | ||
| $i \simeq \frac{cen}{\ell} = 0.02\;\rm{A}$ | $i \simeq \frac{cen}{\ell} = 0.02\;\rm{A}$ | ||
| b) Again, electric current can be expressed as | b) Again, electric current can be expressed as | ||
| $i = \frac{dq}{dt} \simeq \frac{e}{t}$ (1) | $i = \frac{dq}{dt} \simeq \frac{e}{t}$ (1) | ||
| Applying Newton Second Law: | Applying Newton Second Law: | ||
| $\frac{e^2}{4\pi\varepsilon r^2} = \frac{m_e v^2}{r}$ | $\frac{e^2}{4\pi\varepsilon r^2} = \frac{m_e v^2}{r}$ | ||
| where $v = \frac{2\pi r}{t}$, | where $v = \frac{2\pi r}{t}$, | ||
| $t = \sqrt{\frac{16 (\pi r)^3 m_e}{e^2}}$ (2) | $t = \sqrt{\frac{16 (\pi r)^3 m_e}{e^2}}$ (2) | ||
| Putting (2) into (1), | Putting (2) into (1), | ||
| $i = \sqrt{\frac{e^4}{16m_e(\pi r)^3}} = 0.0012\;\rm{A}$ | $i = \sqrt{\frac{e^4}{16m_e(\pi r)^3}} = 0.0012\;\rm{A}$ | ||
| #### Answer | #### Answer | ||
| a) $i \simeq \frac{cen}{\ell} = 0.02\;\rm{A}$ | a) $i \simeq \frac{cen}{\ell} = 0.02\;\rm{A}$ | ||
| b) $i = \sqrt{\frac{e^4}{16m_e(\pi r)^3}} = 0.0012\;\rm{A}$ | b) $i = \sqrt{\frac{e^4}{16m_e(\pi r)^3}} = 0.0012\;\rm{A}$ | ||
| [Insert a concise answer or boxed result] | [Insert a concise answer or boxed result] | ||
| ещё строк без изменений 34 | |||