Новое решение

naz правка от 28 июн. 2026 г., 20:14

правка #19371 позже →

en/11.5.10.md +29 −0

		@@ -0,0 +1,29 @@
		+	### Statement
		+
		+	$11.5.10.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	The superconducting rod displaces the magnetic field from itself, which reduces the inductance of the coil (the area of coverage by the magnetic field decreases), and from the law of conservation of flux this leads to an increase in the current in the coil.:
		+
		+	$$L(x) = \frac{\mu_0 N^2S}{h} - \frac{\mu_0 N^2\sigma}{h}\frac{x}{h}$$
		+
		+	$$L_0 = \frac{\mu_0 N^2S}{h}$$
		+
		+	$$l_0 L_0 = LI$$
		+
		+	$$I(x) = \frac{I_0}{1 - \frac{\sigma}{S}\frac{x}{h}}$$
		+
		+	In the case of point a), the rod can fill the entire solenoid, so:
		+
		+	a)
		+	$$I_{max} = \frac{I_0}{1 - \frac{\sigma}{S}}$$
		+
		+	B) it can't be:
		+
		+	would)
		+	$$I_{max} = $$I_{max} = \frac{I_0}{1 - \frac{\sigma}{S}\frac{l}{h}}$$
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

	@@ -0,0 +1,29 @@
			### Statement

			$11.5.10.$ [Insert the problem statement]

			### Solution

			The superconducting rod displaces the magnetic field from itself, which reduces the inductance of the coil (the area of coverage by the magnetic field decreases), and from the law of conservation of flux this leads to an increase in the current in the coil.:

			$$L(x) = \frac{\mu_0 N^2S}{h} - \frac{\mu_0 N^2\sigma}{h}\frac{x}{h}$$

			$$L_0 = \frac{\mu_0 N^2S}{h}$$

			$$l_0 L_0 = LI$$

			$$I(x) = \frac{I_0}{1 - \frac{\sigma}{S}\frac{x}{h}}$$

			In the case of point a), the rod can fill the entire solenoid, so:

			a)
			$$I_{max} = \frac{I_0}{1 - \frac{\sigma}{S}}$$

			B) it can't be:

			would)
			$$I_{max} = $$I_{max} = \frac{I_0}{1 - \frac{\sigma}{S}\frac{l}{h}}$$

			#### Answer

			[Insert a concise answer or boxed result]