Правка раздела «Statement»
en/10.1.2.md
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| @@ -1,6 +1,6 @@ | |||
| ### Statement | |||
| − | $10.1.2.$ [Insert the problem statement] | ||
| + | $10.1.2.$ An electron accelerated by a voltage of 200 V moves in the Earth’s magnetic field, the induction of which is 70 µT. Find the radius of the circle along which the electron moves, if its velocity is perpendicular to the Earth’s magnetic field. | ||
| ### Solution | |||
| Inside the magnetic field, the electron moves following a circular trajectory with constant speed. | |||
| From Newton Second Law: | |||
| $\frac{m_e v^2}{R} = e |\vec{v} \times \vec{B}|$ | |||
| but $\vec{v}$ is perpendicular to $\vec{B}$, so | |||
| $\frac{m_e v^2}{R} = e v B$ | |||
| $R = \frac{m_e v}{eB}$ (1) | |||
| Applying Energy Conservation Law: | |||
| $\frac{m_e v^2}{2} = e U$ | |||
| $v = \sqrt{\frac{2eU}{m_e}}$ (2) | |||
| Substituting (2) into (1) | |||
| $R = \frac{1}{B}\sqrt{\frac{2m_e U}{e}} = 0.68 \rm{m}$ | |||
| #### Answer | |||
| $R = \frac{1}{B}\sqrt{\frac{2m_e U}{e}} = 0.68 \rm{m}$ | |||
| ещё строк без изменений 24 | |||
| @@ -1,6 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $10.1.2.$ [Insert the problem statement] | $10.1.2.$ An electron accelerated by a voltage of 200 V moves in the Earth’s magnetic field, the induction of which is 70 µT. Find the radius of the circle along which the electron moves, if its velocity is perpendicular to the Earth’s magnetic field. | ||
| ### Solution | ### Solution | ||
| Inside the magnetic field, the electron moves following a circular trajectory with constant speed. | Inside the magnetic field, the electron moves following a circular trajectory with constant speed. | ||
| From Newton Second Law: | From Newton Second Law: | ||
| $\frac{m_e v^2}{R} = e |\vec{v} \times \vec{B}|$ | $\frac{m_e v^2}{R} = e |\vec{v} \times \vec{B}|$ | ||
| but $\vec{v}$ is perpendicular to $\vec{B}$, so | but $\vec{v}$ is perpendicular to $\vec{B}$, so | ||
| $\frac{m_e v^2}{R} = e v B$ | $\frac{m_e v^2}{R} = e v B$ | ||
| $R = \frac{m_e v}{eB}$ (1) | $R = \frac{m_e v}{eB}$ (1) | ||
| Applying Energy Conservation Law: | Applying Energy Conservation Law: | ||
| $\frac{m_e v^2}{2} = e U$ | $\frac{m_e v^2}{2} = e U$ | ||
| $v = \sqrt{\frac{2eU}{m_e}}$ (2) | $v = \sqrt{\frac{2eU}{m_e}}$ (2) | ||
| Substituting (2) into (1) | Substituting (2) into (1) | ||
| $R = \frac{1}{B}\sqrt{\frac{2m_e U}{e}} = 0.68 \rm{m}$ | $R = \frac{1}{B}\sqrt{\frac{2m_e U}{e}} = 0.68 \rm{m}$ | ||
| #### Answer | #### Answer | ||
| $R = \frac{1}{B}\sqrt{\frac{2m_e U}{e}} = 0.68 \rm{m}$ | $R = \frac{1}{B}\sqrt{\frac{2m_e U}{e}} = 0.68 \rm{m}$ | ||
| ещё строк без изменений 24 | |||