Новое решение

naz правка от
Посмотреть решение История
правка #19455 предыдущая #19454 ← раньше позже →
en/11.5.25.md +30 −0
@@ -0,0 +1,30 @@
+### Statement
+
+$11.5.25.$ [Insert the problem statement]
+
+### 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:
+$$m \frac {d \upsilon'}{d t} = m \frac {d \upsilon'}{d l}\frac {d l}{d t}$$
+$$m \frac {d \upsilon'}{d t} = m \frac {d \upsilon'}{d l}(\upsilon' - \upsilon)$$
+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:
+$$\upsilon_{max} = \upsilon(1 + \sqrt{1 + \frac{I_0^2 x L}{m \upsilon^2}})$$
+
+#### Answer
+
+[Insert a concise answer or boxed result]