Edits to “Statement”, “Solution”, “Answer”
en/11.5.24.md
+6 −4
| @@ -1,7 +1,9 @@ | |||
| ### Statement | |||
| − | $11.5.24.$ [Insert the problem statement] | ||
| + | $11.5.24.$ The superconducting box is also divided into two equal parts by a superconducting jumper of thickness $d$. The dimensions of the box are shown in the figure ($h ≪a,l$). A current circulates through the box in the direction perpendicular to the jumper, the linear density of which is $i$. How often will the jumper oscillate if it is given a low speed in the direction shown in the figure? The mass of the jumper is $m$. | ||
| + |  | ||
| + | |||
| ### Solution | |||
| The induction in each of the halves is modulo: | |||
| $$B_0 = \mu_0 i$$ | |||
| Length of each camera: | |||
| $$l_0 = \frac{l-d}{2}$$ | |||
| Let's say we shifted the plate by a small distance $x$. From the law of conservation of flow: | |||
| $$B_1 = \frac{B_0}{1+\frac{x}{l_0}} \approx B_0(1 - \frac{x}{l_0})$$ | |||
| $$B_2 = \frac{B_0}{1- \frac{x}{l_0}} \approx B_0(1 + \frac{x}{l_0})$$ | |||
| Now let's write down the energy density on each side, this will also be the magnetic pressure on the bridge.: | |||
| $$P_1 = \frac{B_1^2}{2 \mu_0} = \frac{B_0^2(1 - \frac{x}{l_0})^2}{2 \mu_0} $$ | |||
| $$P_2 = \frac{B_2^2}{2 \mu_0} = \frac{B_0^2(1 + \frac{x}{l_0})^2}{2 \mu_0} $$ | |||
| Hence the force acting on the partition: | |||
| $$F = ma = (P_2 - P_1)ah$$ | |||
| @@ -30,12 +32,12 @@Solution | |||
| $$F = ma = \frac{4\mu_0 i^2ahx}{l-d}$$ | |||
| − | Solving the | ||
| + | Solving the HOE: | ||
| − | $$ | ||
| + | $$\nu = \frac{i}{\pi} \sqrt{\frac{\mu_0 ah}{m(l-d)}}$$ | ||
| Savchenko has a cyclic frequency, that is, without dividing by $2\pi$ | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $$\nu = \frac{i}{\pi} \sqrt{\frac{\mu_0 ah}{m(l-d)}}$$ | ||
| @@ -1,7 +1,9 @@ | |||
| ### Statement | ### Statement | ||
| $11.5.24.$ [Insert the problem statement] | $11.5.24.$ The superconducting box is also divided into two equal parts by a superconducting jumper of thickness $d$. The dimensions of the box are shown in the figure ($h ≪a,l$). A current circulates through the box in the direction perpendicular to the jumper, the linear density of which is $i$. How often will the jumper oscillate if it is given a low speed in the direction shown in the figure? The mass of the jumper is $m$. | ||
|  | |||
| ### Solution | ### Solution | ||
| The induction in each of the halves is modulo: | The induction in each of the halves is modulo: | ||
| $$B_0 = \mu_0 i$$ | $$B_0 = \mu_0 i$$ | ||
| Length of each camera: | Length of each camera: | ||
| $$l_0 = \frac{l-d}{2}$$ | $$l_0 = \frac{l-d}{2}$$ | ||
| Let's say we shifted the plate by a small distance $x$. From the law of conservation of flow: | Let's say we shifted the plate by a small distance $x$. From the law of conservation of flow: | ||
| $$B_1 = \frac{B_0}{1+\frac{x}{l_0}} \approx B_0(1 - \frac{x}{l_0})$$ | $$B_1 = \frac{B_0}{1+\frac{x}{l_0}} \approx B_0(1 - \frac{x}{l_0})$$ | ||
| $$B_2 = \frac{B_0}{1- \frac{x}{l_0}} \approx B_0(1 + \frac{x}{l_0})$$ | $$B_2 = \frac{B_0}{1- \frac{x}{l_0}} \approx B_0(1 + \frac{x}{l_0})$$ | ||
| Now let's write down the energy density on each side, this will also be the magnetic pressure on the bridge.: | Now let's write down the energy density on each side, this will also be the magnetic pressure on the bridge.: | ||
| $$P_1 = \frac{B_1^2}{2 \mu_0} = \frac{B_0^2(1 - \frac{x}{l_0})^2}{2 \mu_0} $$ | $$P_1 = \frac{B_1^2}{2 \mu_0} = \frac{B_0^2(1 - \frac{x}{l_0})^2}{2 \mu_0} $$ | ||
| $$P_2 = \frac{B_2^2}{2 \mu_0} = \frac{B_0^2(1 + \frac{x}{l_0})^2}{2 \mu_0} $$ | $$P_2 = \frac{B_2^2}{2 \mu_0} = \frac{B_0^2(1 + \frac{x}{l_0})^2}{2 \mu_0} $$ | ||
| Hence the force acting on the partition: | Hence the force acting on the partition: | ||
| $$F = ma = (P_2 - P_1)ah$$ | $$F = ma = (P_2 - P_1)ah$$ | ||
| @@ -30,12 +32,12 @@Solution | |||
| $$F = ma = \frac{4\mu_0 i^2ahx}{l-d}$$ | $$F = ma = \frac{4\mu_0 i^2ahx}{l-d}$$ | ||
| Solving the |
Solving the HOE: | ||
| $$ |
$$\nu = \frac{i}{\pi} \sqrt{\frac{\mu_0 ah}{m(l-d)}}$$ | ||
| Savchenko has a cyclic frequency, that is, without dividing by $2\pi$ | Savchenko has a cyclic frequency, that is, without dividing by $2\pi$ | ||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $$\nu = \frac{i}{\pi} \sqrt{\frac{\mu_0 ah}{m(l-d)}}$$ | ||