Правка разделов «Statement», «Answer»
en/11.5.27.md
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| @@ -1,6 +1,8 @@ | |||
| ### Statement | |||
| − | $11.5.27.$ | ||
| + | $11.5.27.$ The external magnetic field of induction $B$, in which there is a long perfectly conducting tube, is not completely shielded by the tube walls | ||
| + | due to the fact that the mass of electrons is finite. The field partially penetrates into the tube. The axis of the tube is directed along the magnetic field, its radius $r$ is much greater than the thickness of the walls $h$. The number of conduction electrons per unit volume of the tube material $n_e$. Calculate the induction of the field penetrated into the tube in the case | ||
| + | $B = 10 T$, $r = 1 mm$, $h = 0.1 mm$, $n_e = 10^{20} cm^{−3}$. | ||
| ### Solution | |||
| Before equilibrium is established, the flow changes according to the following law: | |||
| $$\frac{d\Phi}{d t} =\frac{d(B - B_e)}{dt} \pi r^2$$ | |||
| Where $B_e$ is the interference generated by the motion of electrons. | |||
| The vortex electric field is determined by the circulation theorem: | |||
| $$E 2\pi r = \frac{d(B - B_e)}{dt} \pi r^2$$ | |||
| $$E = \frac{d(B - B_e)}{dt} \frac{r}{2}$$ | |||
| Let's write 2nd law: | |||
| $$m\frac{d \upsilon}{dt} = \frac{d(B - B_e)}{dt} \frac{re}{2}$$ | |||
| From here: | |||
| $$\upsilon = (B - B_e) \frac{re}{2m}$$ | |||
| Let's find the current density: | |||
| $$j = \frac{n_e \upsilon dt Lh e}{Lh dt}$$ | |||
| $$j = n_e \upsilon e$$ | |||
| The equivalent linear current density is: | |||
| $$i = n_e \upsilon e h$$ | |||
| Therefore: | |||
| $$B_e = \mu_0 n_e \upsilon e h$$ | |||
| Substitute the velocity and find $B_e$: | |||
| $$B_e = \frac{B}{1 + \frac{2m}{\mu_0 n_e e^2 h r}}$$ | |||
| From here, the desired induction inside: | |||
| $$B' = B - B_e = \frac{Bm}{m + \frac{\mu_0 n_e e^2 h r}{2}} \approx 5.7 \cdot 10^{-5} Т$$ | |||
| @@ -28,4 +30,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $$B' = B - B_e = \frac{Bm}{m + \frac{\mu_0 n_e e^2 h r}{2}} \approx 5.7 \cdot 10^{-5} Т$$ | ||
| @@ -1,6 +1,8 @@ | |||
| ### Statement | ### Statement | ||
| $11.5.27.$ |
$11.5.27.$ The external magnetic field of induction $B$, in which there is a long perfectly conducting tube, is not completely shielded by the tube walls | ||
| due to the fact that the mass of electrons is finite. The field partially penetrates into the tube. The axis of the tube is directed along the magnetic field, its radius $r$ is much greater than the thickness of the walls $h$. The number of conduction electrons per unit volume of the tube material $n_e$. Calculate the induction of the field penetrated into the tube in the case | |||
| $B = 10 T$, $r = 1 mm$, $h = 0.1 mm$, $n_e = 10^{20} cm^{−3}$. | |||
| ### Solution | ### Solution | ||
| Before equilibrium is established, the flow changes according to the following law: | Before equilibrium is established, the flow changes according to the following law: | ||
| $$\frac{d\Phi}{d t} =\frac{d(B - B_e)}{dt} \pi r^2$$ | $$\frac{d\Phi}{d t} =\frac{d(B - B_e)}{dt} \pi r^2$$ | ||
| Where $B_e$ is the interference generated by the motion of electrons. | Where $B_e$ is the interference generated by the motion of electrons. | ||
| The vortex electric field is determined by the circulation theorem: | The vortex electric field is determined by the circulation theorem: | ||
| $$E 2\pi r = \frac{d(B - B_e)}{dt} \pi r^2$$ | $$E 2\pi r = \frac{d(B - B_e)}{dt} \pi r^2$$ | ||
| $$E = \frac{d(B - B_e)}{dt} \frac{r}{2}$$ | $$E = \frac{d(B - B_e)}{dt} \frac{r}{2}$$ | ||
| Let's write 2nd law: | Let's write 2nd law: | ||
| $$m\frac{d \upsilon}{dt} = \frac{d(B - B_e)}{dt} \frac{re}{2}$$ | $$m\frac{d \upsilon}{dt} = \frac{d(B - B_e)}{dt} \frac{re}{2}$$ | ||
| From here: | From here: | ||
| $$\upsilon = (B - B_e) \frac{re}{2m}$$ | $$\upsilon = (B - B_e) \frac{re}{2m}$$ | ||
| Let's find the current density: | Let's find the current density: | ||
| $$j = \frac{n_e \upsilon dt Lh e}{Lh dt}$$ | $$j = \frac{n_e \upsilon dt Lh e}{Lh dt}$$ | ||
| $$j = n_e \upsilon e$$ | $$j = n_e \upsilon e$$ | ||
| The equivalent linear current density is: | The equivalent linear current density is: | ||
| $$i = n_e \upsilon e h$$ | $$i = n_e \upsilon e h$$ | ||
| Therefore: | Therefore: | ||
| $$B_e = \mu_0 n_e \upsilon e h$$ | $$B_e = \mu_0 n_e \upsilon e h$$ | ||
| Substitute the velocity and find $B_e$: | Substitute the velocity and find $B_e$: | ||
| $$B_e = \frac{B}{1 + \frac{2m}{\mu_0 n_e e^2 h r}}$$ | $$B_e = \frac{B}{1 + \frac{2m}{\mu_0 n_e e^2 h r}}$$ | ||
| From here, the desired induction inside: | From here, the desired induction inside: | ||
| $$B' = B - B_e = \frac{Bm}{m + \frac{\mu_0 n_e e^2 h r}{2}} \approx 5.7 \cdot 10^{-5} Т$$ | $$B' = B - B_e = \frac{Bm}{m + \frac{\mu_0 n_e e^2 h r}{2}} \approx 5.7 \cdot 10^{-5} Т$$ | ||
| @@ -28,4 +30,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $$B' = B - B_e = \frac{Bm}{m + \frac{\mu_0 n_e e^2 h r}{2}} \approx 5.7 \cdot 10^{-5} Т$$ | ||