Правка разделов «Statement», «Answer»
en/10.1.25.md
+2 −2
| @@ -1,6 +1,6 @@ | |||
| ### Statement | |||
| − | $10.1.25.$ [Insert the problem statement] | ||
| + | $10.1.25.$ Regions $I$ and $II$ of two homogeneous unidirectional magnetic fields with induction $B_1$ and $B_2$ have an axisymmetric thin junction $AA'$, in which the magnetic field has a large radial component. The electron in region I moves along the magnetic field at a distance $R$ from the axis of symmetry of the transition. What angular momentum relative to the axis of symmetry does an electron acquire during the transition from region $I$ to region $II$? Does the sum $M+(1/2π)eΦ$ remain constant during the motion of this particle (see the notation in problem $10.1.24$)? | ||
| ### Solution | |||
| To begin with, we will show the invariance of the sum in this case as well. The force acting on the electron inside the transition zone : | |||
| $$F = -B_r e \upsilon_z$$ | |||
| Where $B_r$ is the radial component of induction, $\upsilon_z$ is the component of velocity along the axis of symmetry. | |||
| $$\frac{dM}{dt} = -B_r e r \upsilon_z $$ | |||
| $$dM = - B_r e r dz$$ | |||
| Now let's consider the flow change through the cross section of a ring of arbitrary radius R: | |||
| $$d\Phi = 2 \pi R B_r dz$$ | |||
| From here: | |||
| $$dM = -\frac{1}{2\pi}ed\Phi$$ | |||
| $$M+\frac{1}{2\pi} e \Phi = const$$ | |||
| Well, now, it's very easy to find the desired moment, at first the moment of momentum is zero, because the electron does not rotate: | |||
| $$\frac{1}{2}eB_1 R^2 = M + \frac{1}{2}eB_2 R^2$$ | |||
| $$M = \frac{(B_2 - B_1)R^2e}{2}$$ | |||
| @@ -32,4 +32,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $$M = \frac{(B_2 - B_1)R^2e}{2}$$ | ||
| @@ -1,6 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $10.1.25.$ [Insert the problem statement] | $10.1.25.$ Regions $I$ and $II$ of two homogeneous unidirectional magnetic fields with induction $B_1$ and $B_2$ have an axisymmetric thin junction $AA'$, in which the magnetic field has a large radial component. The electron in region I moves along the magnetic field at a distance $R$ from the axis of symmetry of the transition. What angular momentum relative to the axis of symmetry does an electron acquire during the transition from region $I$ to region $II$? Does the sum $M+(1/2π)eΦ$ remain constant during the motion of this particle (see the notation in problem $10.1.24$)? | ||
| ### Solution | ### Solution | ||
| To begin with, we will show the invariance of the sum in this case as well. The force acting on the electron inside the transition zone : | To begin with, we will show the invariance of the sum in this case as well. The force acting on the electron inside the transition zone : | ||
| $$F = -B_r e \upsilon_z$$ | $$F = -B_r e \upsilon_z$$ | ||
| Where $B_r$ is the radial component of induction, $\upsilon_z$ is the component of velocity along the axis of symmetry. | Where $B_r$ is the radial component of induction, $\upsilon_z$ is the component of velocity along the axis of symmetry. | ||
| $$\frac{dM}{dt} = -B_r e r \upsilon_z $$ | $$\frac{dM}{dt} = -B_r e r \upsilon_z $$ | ||
| $$dM = - B_r e r dz$$ | $$dM = - B_r e r dz$$ | ||
| Now let's consider the flow change through the cross section of a ring of arbitrary radius R: | Now let's consider the flow change through the cross section of a ring of arbitrary radius R: | ||
| $$d\Phi = 2 \pi R B_r dz$$ | $$d\Phi = 2 \pi R B_r dz$$ | ||
| From here: | From here: | ||
| $$dM = -\frac{1}{2\pi}ed\Phi$$ | $$dM = -\frac{1}{2\pi}ed\Phi$$ | ||
| $$M+\frac{1}{2\pi} e \Phi = const$$ | $$M+\frac{1}{2\pi} e \Phi = const$$ | ||
| Well, now, it's very easy to find the desired moment, at first the moment of momentum is zero, because the electron does not rotate: | Well, now, it's very easy to find the desired moment, at first the moment of momentum is zero, because the electron does not rotate: | ||
| $$\frac{1}{2}eB_1 R^2 = M + \frac{1}{2}eB_2 R^2$$ | $$\frac{1}{2}eB_1 R^2 = M + \frac{1}{2}eB_2 R^2$$ | ||
| $$M = \frac{(B_2 - B_1)R^2e}{2}$$ | $$M = \frac{(B_2 - B_1)R^2e}{2}$$ | ||
| @@ -32,4 +32,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $$M = \frac{(B_2 - B_1)R^2e}{2}$$ | ||