Правка разделов «Statement», «Solution», «Answer»
en/10.1.13.md
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| ### Statement | |||
| − | $10.1.13.$ [Insert the problem statement] | ||
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| − | ### Solution | ||
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| − | ### Statement | ||
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| 10.1.13. <b>a.</b> A vacuum device consists of a coaxial cylinder of radius $R$ and a wire placed in a longitudinal magnetic field of induction $B$. When the wire is heated, electrons with kinetic energy $K$ are emitted from its surface; in this case, a current $I$ flows in the external circuit between the cylinder and the wire. Draw the dependence of $I$ on $B$. Find the values of $B$ at which the current in vacuum is zero. | |||
| <b>b.</b> The figure shows two dependencies of $I$ on $B$ at different residual gas pressures $P_1$ and $P_2$. Which pressure is higher? | |||
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| ### Solution | |||
| <b>a.</b> We neglect the transverse dimension of the wire. We will assume that electrons are emitted from the wire radially and their velocities $v$ are strictly perpendicular to the field lines. The Lorentz force acts on the electron, twisting it into a circle. According to Newton's second law: | |||
| $$eBv = \frac{m_ev^2}{r} \implies r = \frac{m_ev}{eB} \quad (1)$$ | |||
| From the definition of kinetic energy: | |||
| $$\frac{1}{2}m_ev^2 = K \implies v = \sqrt{\frac{2K}{m_e}} \quad (2)$$ | |||
| Substitute (2) into (1): | |||
| $$r = \frac{m_e}{eB}\sqrt{\frac{2K}{m_e}} = \frac{\sqrt{2m_eK}}{eB}$$ | |||
| We see that the larger $B$ is, the smaller $r$ is. The condition for the electron to reach the cylinder wall of radius $R$: | |||
| $$2r \ge R$$ | |||
| $$2 \cdot \frac{\sqrt{2m_eK}}{eB} \ge R$$ | |||
| Hence the boundary value of the magnetic induction: | |||
| $$B_0 = \frac{2\sqrt{2m_eK}}{eR}$$ | |||
| @@ -30,15 +26,18 @@Solution | |||
| At $B \le B_0$ in vacuum, all electrons will reach the cylinder and some constant current $I_0$ will flow. At $B > B_0$, the electrons will not reach the cylinder wall and the current will be absent. The graph of the dependence $I(B)$ in vacuum will be in the form of a step. | |||
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| <b>b.</b> In the presence of residual gas, electrons can collide with gas molecules. Such collisions change the directions of velocities — this gives the electron a chance to reach the cylinder wall even if initially $2r < R$. Therefore, at a higher pressure, more collisions occur and the current is maintained at larger values of $B$. From the graph in the problem statement, it is seen that the boundary value of the magnetic induction: | |||
| $$B_{0_2} > B_{0_1}$$ | |||
| Therefore, $P_2 > P_1$. | |||
| #### Answer | |||
| − | <b>a.</b> | ||
| − | <b>b.</b> $P_2 > P_1$ | ||
| + | <b>a.</b> | ||
| − | #### Answer | ||
| + |  | ||
| − | [Insert a concise answer or boxed result] | ||
| + | $B \ge B_0 = \frac{2\sqrt{2m_eK}}{eR}$ | ||
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| + | <b>b.</b> $P_2 > P_1$ | ||
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| ### Statement | ### Statement | ||
| $10.1.13.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| 10.1.13. <b>a.</b> A vacuum device consists of a coaxial cylinder of radius $R$ and a wire placed in a longitudinal magnetic field of induction $B$. When the wire is heated, electrons with kinetic energy $K$ are emitted from its surface; in this case, a current $I$ flows in the external circuit between the cylinder and the wire. Draw the dependence of $I$ on $B$. Find the values of $B$ at which the current in vacuum is zero. | 10.1.13. <b>a.</b> A vacuum device consists of a coaxial cylinder of radius $R$ and a wire placed in a longitudinal magnetic field of induction $B$. When the wire is heated, electrons with kinetic energy $K$ are emitted from its surface; in this case, a current $I$ flows in the external circuit between the cylinder and the wire. Draw the dependence of $I$ on $B$. Find the values of $B$ at which the current in vacuum is zero. | ||
| <b>b.</b> The figure shows two dependencies of $I$ on $B$ at different residual gas pressures $P_1$ and $P_2$. Which pressure is higher? | <b>b.</b> The figure shows two dependencies of $I$ on $B$ at different residual gas pressures $P_1$ and $P_2$. Which pressure is higher? | ||
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| ### Solution | ### Solution | ||
| <b>a.</b> We neglect the transverse dimension of the wire. We will assume that electrons are emitted from the wire radially and their velocities $v$ are strictly perpendicular to the field lines. The Lorentz force acts on the electron, twisting it into a circle. According to Newton's second law: | <b>a.</b> We neglect the transverse dimension of the wire. We will assume that electrons are emitted from the wire radially and their velocities $v$ are strictly perpendicular to the field lines. The Lorentz force acts on the electron, twisting it into a circle. According to Newton's second law: | ||
| $$eBv = \frac{m_ev^2}{r} \implies r = \frac{m_ev}{eB} \quad (1)$$ | $$eBv = \frac{m_ev^2}{r} \implies r = \frac{m_ev}{eB} \quad (1)$$ | ||
| From the definition of kinetic energy: | From the definition of kinetic energy: | ||
| $$\frac{1}{2}m_ev^2 = K \implies v = \sqrt{\frac{2K}{m_e}} \quad (2)$$ | $$\frac{1}{2}m_ev^2 = K \implies v = \sqrt{\frac{2K}{m_e}} \quad (2)$$ | ||
| Substitute (2) into (1): | Substitute (2) into (1): | ||
| $$r = \frac{m_e}{eB}\sqrt{\frac{2K}{m_e}} = \frac{\sqrt{2m_eK}}{eB}$$ | $$r = \frac{m_e}{eB}\sqrt{\frac{2K}{m_e}} = \frac{\sqrt{2m_eK}}{eB}$$ | ||
| We see that the larger $B$ is, the smaller $r$ is. The condition for the electron to reach the cylinder wall of radius $R$: | We see that the larger $B$ is, the smaller $r$ is. The condition for the electron to reach the cylinder wall of radius $R$: | ||
| $$2r \ge R$$ | $$2r \ge R$$ | ||
| $$2 \cdot \frac{\sqrt{2m_eK}}{eB} \ge R$$ | $$2 \cdot \frac{\sqrt{2m_eK}}{eB} \ge R$$ | ||
| Hence the boundary value of the magnetic induction: | Hence the boundary value of the magnetic induction: | ||
| $$B_0 = \frac{2\sqrt{2m_eK}}{eR}$$ | $$B_0 = \frac{2\sqrt{2m_eK}}{eR}$$ | ||
| @@ -30,15 +26,18 @@Solution | |||
| At $B \le B_0$ in vacuum, all electrons will reach the cylinder and some constant current $I_0$ will flow. At $B > B_0$, the electrons will not reach the cylinder wall and the current will be absent. The graph of the dependence $I(B)$ in vacuum will be in the form of a step. | At $B \le B_0$ in vacuum, all electrons will reach the cylinder and some constant current $I_0$ will flow. At $B > B_0$, the electrons will not reach the cylinder wall and the current will be absent. The graph of the dependence $I(B)$ in vacuum will be in the form of a step. | ||
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| <b>b.</b> In the presence of residual gas, electrons can collide with gas molecules. Such collisions change the directions of velocities — this gives the electron a chance to reach the cylinder wall even if initially $2r < R$. Therefore, at a higher pressure, more collisions occur and the current is maintained at larger values of $B$. From the graph in the problem statement, it is seen that the boundary value of the magnetic induction: | <b>b.</b> In the presence of residual gas, electrons can collide with gas molecules. Such collisions change the directions of velocities — this gives the electron a chance to reach the cylinder wall even if initially $2r < R$. Therefore, at a higher pressure, more collisions occur and the current is maintained at larger values of $B$. From the graph in the problem statement, it is seen that the boundary value of the magnetic induction: | ||
| $$B_{0_2} > B_{0_1}$$ | $$B_{0_2} > B_{0_1}$$ | ||
| Therefore, $P_2 > P_1$. | Therefore, $P_2 > P_1$. | ||
| #### Answer | #### Answer | ||
| <b>a.</b> |
<b>a.</b> | ||
| <b>b.</b> $P_2 > P_1$ | |||
| #### Answer |  | ||
| [Insert a concise answer or boxed result] | $B \ge B_0 = \frac{2\sqrt{2m_eK}}{eR}$ | ||
| <b>b.</b> $P_2 > P_1$ | |||