Edits to “Statement”, “Solution”, “Answer”
en/11.5.25.md
+3 −3
| @@ -1,6 +1,6 @@ | |||
| ### Statement | |||
| − | $11.5.25.$ [Insert the problem statement] | ||
| + | $11.5.25.$ Between two superconducting busbars, a piston moves at a constant velocity $v_0$ towards a jumper of mass $m$, forming a superconducting circuit. Find the maximum velocity of the jumper if it was initially at rest, the current in the circuit was $I_0$, and the distance between the piston and the jumper was $x$. The inductance per unit length of the busbars is $L$. Friction can be neglected. | ||
| ### Solution | |||
| From the law of conservation of magnetic flux: | |||
| $$LxI_0 = Ll I$$ | |||
| $$I = I_0 \frac{x}{l}$$ | |||
| The force acting on the plate: | |||
| $$m \frac{d \upsilon'}{dt} = \frac{\partial U}{\partial l}$$ | |||
| The derivative of the magnetic field energy with respect to distance is equal to: | |||
| $$\frac{\partial U}{\partial l} = \frac{1}{2} I^2 \mathcal {L}$$ | |||
| $$\mathcal {L} = Ll$$ | |||
| $$\frac{\partial U}{\partial l} = \frac{1}{2} I^2 L$$ | |||
| Substitute the expressions for the current strength: | |||
| $$\frac{\partial U}{\partial l} = \frac{1}{2} I_0^2 x^2 L\frac{1}{l^2}$$ | |||
| Now, let's return to the left-hand side of Newton's Second Law: | |||
| $$m \frac {d \upsilon'}{d t} = m \frac {d \upsilon'}{d l}\frac {d l}{d t}$$ | |||
| $$m \frac {d \upsilon'}{d t} = m \frac {d \upsilon'}{d l}(\upsilon' - \upsilon)$$ | |||
| @@ -21,10 +21,10 @@Solution | |||
| Let's go back to the beginning, substitute all the values, and integrate: | |||
| $$\int_0^{\upsilon_{max}}(\upsilon' - \upsilon)d \upsilon' = \int_0^\infty \frac{1}{2} I_0^2 x^2 L\frac{dl}{l^2}$$ | |||
| We get a quadratic equation: | |||
| − | $$\frac{1}{2}\upsilon_{max} - \upsilon \upsilon_{max} - \frac{1}{2} I_0^2 x | ||
| + | $$\frac{1}{2}\upsilon_{max}^2 - \upsilon \upsilon_{max} - \frac{1}{2} I_0^2 xL = 0$$ | ||
| And we find the answer: | |||
| $$\upsilon_{max} = \upsilon(1 + \sqrt{1 + \frac{I_0^2 x L}{m \upsilon^2}})$$ | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $$\upsilon_{max} = \upsilon(1 + \sqrt{1 + \frac{I_0^2 x L}{m \upsilon^2}})$$ | ||
| @@ -1,6 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $11.5.25.$ [Insert the problem statement] | $11.5.25.$ Between two superconducting busbars, a piston moves at a constant velocity $v_0$ towards a jumper of mass $m$, forming a superconducting circuit. Find the maximum velocity of the jumper if it was initially at rest, the current in the circuit was $I_0$, and the distance between the piston and the jumper was $x$. The inductance per unit length of the busbars is $L$. Friction can be neglected. | ||
| ### Solution | ### Solution | ||
| From the law of conservation of magnetic flux: | From the law of conservation of magnetic flux: | ||
| $$LxI_0 = Ll I$$ | $$LxI_0 = Ll I$$ | ||
| $$I = I_0 \frac{x}{l}$$ | $$I = I_0 \frac{x}{l}$$ | ||
| The force acting on the plate: | The force acting on the plate: | ||
| $$m \frac{d \upsilon'}{dt} = \frac{\partial U}{\partial l}$$ | $$m \frac{d \upsilon'}{dt} = \frac{\partial U}{\partial l}$$ | ||
| The derivative of the magnetic field energy with respect to distance is equal to: | The derivative of the magnetic field energy with respect to distance is equal to: | ||
| $$\frac{\partial U}{\partial l} = \frac{1}{2} I^2 \mathcal {L}$$ | $$\frac{\partial U}{\partial l} = \frac{1}{2} I^2 \mathcal {L}$$ | ||
| $$\mathcal {L} = Ll$$ | $$\mathcal {L} = Ll$$ | ||
| $$\frac{\partial U}{\partial l} = \frac{1}{2} I^2 L$$ | $$\frac{\partial U}{\partial l} = \frac{1}{2} I^2 L$$ | ||
| Substitute the expressions for the current strength: | Substitute the expressions for the current strength: | ||
| $$\frac{\partial U}{\partial l} = \frac{1}{2} I_0^2 x^2 L\frac{1}{l^2}$$ | $$\frac{\partial U}{\partial l} = \frac{1}{2} I_0^2 x^2 L\frac{1}{l^2}$$ | ||
| Now, let's return to the left-hand side of Newton's Second Law: | Now, let's return to the left-hand side of Newton's Second Law: | ||
| $$m \frac {d \upsilon'}{d t} = m \frac {d \upsilon'}{d l}\frac {d l}{d t}$$ | $$m \frac {d \upsilon'}{d t} = m \frac {d \upsilon'}{d l}\frac {d l}{d t}$$ | ||
| $$m \frac {d \upsilon'}{d t} = m \frac {d \upsilon'}{d l}(\upsilon' - \upsilon)$$ | $$m \frac {d \upsilon'}{d t} = m \frac {d \upsilon'}{d l}(\upsilon' - \upsilon)$$ | ||
| @@ -21,10 +21,10 @@Solution | |||
| Let's go back to the beginning, substitute all the values, and integrate: | Let's go back to the beginning, substitute all the values, and integrate: | ||
| $$\int_0^{\upsilon_{max}}(\upsilon' - \upsilon)d \upsilon' = \int_0^\infty \frac{1}{2} I_0^2 x^2 L\frac{dl}{l^2}$$ | $$\int_0^{\upsilon_{max}}(\upsilon' - \upsilon)d \upsilon' = \int_0^\infty \frac{1}{2} I_0^2 x^2 L\frac{dl}{l^2}$$ | ||
| We get a quadratic equation: | We get a quadratic equation: | ||
| $$\frac{1}{2}\upsilon_{max} - \upsilon \upsilon_{max} - \frac{1}{2} I_0^2 x |
$$\frac{1}{2}\upsilon_{max}^2 - \upsilon \upsilon_{max} - \frac{1}{2} I_0^2 xL = 0$$ | ||
| And we find the answer: | And we find the answer: | ||
| $$\upsilon_{max} = \upsilon(1 + \sqrt{1 + \frac{I_0^2 x L}{m \upsilon^2}})$$ | $$\upsilon_{max} = \upsilon(1 + \sqrt{1 + \frac{I_0^2 x L}{m \upsilon^2}})$$ | ||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $$\upsilon_{max} = \upsilon(1 + \sqrt{1 + \frac{I_0^2 x L}{m \upsilon^2}})$$ | ||