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naz edited Jul 1, 2026, 9:26 AM

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			### 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:
		@@ -19,9 +19,9 @@Solution
			$$m \frac {d \upsilon'}{d t} = m \frac {d \upsilon'}{d l}\frac {d l}{d t}$$
			$$m \frac {d \upsilon'}{d t} = m \frac {d \upsilon'}{d l}(\upsilon' - \upsilon)$$
			Let's go back to the beginning, substitute all the values, and integrate:
		−	$$\int_0^{\upsilon_{max}}(\upsilon' - \upsilon)d \upsilon' = \int_0^\infty \frac{1}{2} I_0^2 x^2 L\frac{dl}{l^2}$$
		+	$$m\int_0^{\upsilon_{max}}(\upsilon' - \upsilon)d \upsilon' = \int_x^\infty \frac{1}{2} I_0^2 x^2 L\frac{dl}{l^2}$$
			We get a quadratic equation:
		−	$$\frac{1}{2}\upsilon_{max}^2 - \upsilon \upsilon_{max} - \frac{1}{2} I_0^2 xL = 0$$
		+	$$\frac{1}{2}\upsilon_{max}^2 - \upsilon \upsilon_{max} - \frac{1}{2m} I_0^2 xL = 0$$
			And we find the answer:
			$$\upsilon_{max} = \upsilon(1 + \sqrt{1 + \frac{I_0^2 x L}{m \upsilon^2}})$$

			#### Answer

			$$\upsilon_{max} = \upsilon(1 + \sqrt{1 + \frac{I_0^2 x L}{m \upsilon^2}})$$
		unchanged lines 3

	### Statement		### 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.		$11.5.25.$ 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:
	@@ -19,9 +19,9 @@Solution
	$$m \frac {d \upsilon'}{d t} = m \frac {d \upsilon'}{d l}\frac {d l}{d t}$$		$$m \frac {d \upsilon'}{d t} = m \frac {d \upsilon'}{d l}\frac {d l}{d t}$$
	$$m \frac {d \upsilon'}{d t} = m \frac {d \upsilon'}{d l}(\upsilon' - \upsilon)$$		$$m \frac {d \upsilon'}{d t} = m \frac {d \upsilon'}{d l}(\upsilon' - \upsilon)$$
	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:
	$$\int_0^{\upsilon_{max}}(\upsilon' - \upsilon)d \upsilon' = \int_0^\infty \frac{1}{2} I_0^2 x^2 L\frac{dl}{l^2}$$		$$m\int_0^{\upsilon_{max}}(\upsilon' - \upsilon)d \upsilon' = \int_x^\infty \frac{1}{2} I_0^2 x^2 L\frac{dl}{l^2}$$
	We get a quadratic equation:		We get a quadratic equation:
	$$\frac{1}{2}\upsilon_{max}^2 - \upsilon \upsilon_{max} - \frac{1}{2} I_0^2 xL = 0$$		$$\frac{1}{2}\upsilon_{max}^2 - \upsilon \upsilon_{max} - \frac{1}{2m} I_0^2 xL = 0$$
	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

	$$\upsilon_{max} = \upsilon(1 + \sqrt{1 + \frac{I_0^2 x L}{m \upsilon^2}})$$		$$\upsilon_{max} = \upsilon(1 + \sqrt{1 + \frac{I_0^2 x L}{m \upsilon^2}})$$
	unchanged lines 3