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: $$m \frac {d \upsilon'}{d t} = m \frac {d \upsilon'}{d l}\frac {d l}{d t}$$ $$m \frac {d \upsilon'}{d t} = m \frac {d \upsilon'}{d l}(\upsilon' - \upsilon)$$ 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: $$\frac{1}{2}\upsilon_{max} - \upsilon \upsilon_{max} - \frac{1}{2} I_0^2 x^2 L\frac{1}{l^2} = 0$$ And we find the answer: $$\upsilon_{max} = \upsilon(1 + \sqrt{1 + \frac{I_0^2 x L}{m \upsilon^2}})$$
