Правка разделов «Statement», «Answer»
en/14.4.30.md
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| ### Statement | |||
| − | $14.4.30 | ||
| + | $14.4.30 $ | ||
| + | a high voltage is applied between the flat anode and the cathode. The systemis located in a magnetic field of induction B = 10 T, which is parallel to the electrodes. The distance between the anode and cathode is h = 10 cm. At what minimum voltage will the electrons reach the anode | ||
| + | |||
| ### Solution | |||
| The electron leaves the cathode from rest and is accelerated toward the anode by the voltage U. Simultaneously, the magnetic field B (parallel to the plates) acts perpendicularly to its velocity and curves the trajectory. For the electron to reach the anode, the radius of gyration in the magnetic field must be at least equal to the plate separation h. We use the limiting condition: exactly when the electron touches the anode, its trajectory is tangent to it and the local radius of curvature is h. | |||
| The Lorentz force gives us the relationship between momentum and radius of curvature. In CGS units (where the magnetic field is denoted by H and is equal to B in vacuum): | |||
| $\frac{d\mathbf{p}}{dt} = \frac{e}{c}\,\mathbf{v} \times \mathbf{H}$ | |||
| For uniform circular motion in a perpendicular field, the magnitude of the force is $p v / r = (e/c) v H$ Canceling v: | |||
| $p = \frac{e}{c} H r$ | |||
| Multiplying by c we obtain the energy–momentum invariant: | |||
| $pc = e H r$ | |||
| With the limiting condition$ r = h $we have $ pc = e H h$ | |||
| The total energy of the electron after being accelerated is: | |||
| $\mathcal{E} = \sqrt{(pc)^2 + (m_e c^2)^2} = \sqrt{(e H h)^2 + (m_e c^2)^2}$ | |||
| The kinetic energy comes entirely from the electrical work: | |||
| $eU = \mathcal{E} - m_e c^2$ | |||
| Substituting the expression for $\mathcal{E}$ we arrive at: | |||
| $\boxed{eU = \sqrt{(m_e c^2)^2 + (hH)^2} - m_e c^2}$ | |||
| @@ -33,4 +35,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $\boxed{eU = \sqrt{(m_e c^2)^2 + (hH)^2} - m_e c^2}$ | ||
| @@ -1,7 +1,9 @@ | |||
| ### Statement | ### Statement | ||
| $14.4.30 |
$14.4.30 $ | ||
| a high voltage is applied between the flat anode and the cathode. The systemis located in a magnetic field of induction B = 10 T, which is parallel to the electrodes. The distance between the anode and cathode is h = 10 cm. At what minimum voltage will the electrons reach the anode | |||
| ### Solution | ### Solution | ||
| The electron leaves the cathode from rest and is accelerated toward the anode by the voltage U. Simultaneously, the magnetic field B (parallel to the plates) acts perpendicularly to its velocity and curves the trajectory. For the electron to reach the anode, the radius of gyration in the magnetic field must be at least equal to the plate separation h. We use the limiting condition: exactly when the electron touches the anode, its trajectory is tangent to it and the local radius of curvature is h. | The electron leaves the cathode from rest and is accelerated toward the anode by the voltage U. Simultaneously, the magnetic field B (parallel to the plates) acts perpendicularly to its velocity and curves the trajectory. For the electron to reach the anode, the radius of gyration in the magnetic field must be at least equal to the plate separation h. We use the limiting condition: exactly when the electron touches the anode, its trajectory is tangent to it and the local radius of curvature is h. | ||
| The Lorentz force gives us the relationship between momentum and radius of curvature. In CGS units (where the magnetic field is denoted by H and is equal to B in vacuum): | The Lorentz force gives us the relationship between momentum and radius of curvature. In CGS units (where the magnetic field is denoted by H and is equal to B in vacuum): | ||
| $\frac{d\mathbf{p}}{dt} = \frac{e}{c}\,\mathbf{v} \times \mathbf{H}$ | $\frac{d\mathbf{p}}{dt} = \frac{e}{c}\,\mathbf{v} \times \mathbf{H}$ | ||
| For uniform circular motion in a perpendicular field, the magnitude of the force is $p v / r = (e/c) v H$ Canceling v: | For uniform circular motion in a perpendicular field, the magnitude of the force is $p v / r = (e/c) v H$ Canceling v: | ||
| $p = \frac{e}{c} H r$ | $p = \frac{e}{c} H r$ | ||
| Multiplying by c we obtain the energy–momentum invariant: | Multiplying by c we obtain the energy–momentum invariant: | ||
| $pc = e H r$ | $pc = e H r$ | ||
| With the limiting condition$ r = h $we have $ pc = e H h$ | With the limiting condition$ r = h $we have $ pc = e H h$ | ||
| The total energy of the electron after being accelerated is: | The total energy of the electron after being accelerated is: | ||
| $\mathcal{E} = \sqrt{(pc)^2 + (m_e c^2)^2} = \sqrt{(e H h)^2 + (m_e c^2)^2}$ | $\mathcal{E} = \sqrt{(pc)^2 + (m_e c^2)^2} = \sqrt{(e H h)^2 + (m_e c^2)^2}$ | ||
| The kinetic energy comes entirely from the electrical work: | The kinetic energy comes entirely from the electrical work: | ||
| $eU = \mathcal{E} - m_e c^2$ | $eU = \mathcal{E} - m_e c^2$ | ||
| Substituting the expression for $\mathcal{E}$ we arrive at: | Substituting the expression for $\mathcal{E}$ we arrive at: | ||
| $\boxed{eU = \sqrt{(m_e c^2)^2 + (hH)^2} - m_e c^2}$ | $\boxed{eU = \sqrt{(m_e c^2)^2 + (hH)^2} - m_e c^2}$ | ||
| @@ -33,4 +35,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $\boxed{eU = \sqrt{(m_e c^2)^2 + (hH)^2} - m_e c^2}$ | ||