Новое решение

naz правка от 9 июн. 2026 г., 19:24

правка #18513 позже →

en/9.3.20.md +23 −0

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		+	### Statement
		+
		+	$9.3.20.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	In problem $9.3.15$, we obtained formulas for the field inside and outside a cylindrical rod and plate. Replacing $I=j \pi (h/2)^2$ in the formulas yields the formulas for induction, respectively:
		+	$$B^{in}_{full} = \mu_0 j x$$
		+	$$B^{out}_{full}=\frac{\mu_0 j h}{2}$$
		+	$$B^{in}_{cyl} = \frac{\mu_0 j x}{2}$$
		+	$$B^{out}_{cyl}=\frac{\mu_0 j h^2}{8x}$$
		+
		+	Now, we represent the desired field as the difference between the field of the plate and the cylindrical tube. We obtain the answers:
		+
		+	For $0<x<h/2$:
		+	$$B = \frac{\mu_0 j x}{2}$$
		+
		+	For $x > h/2$:
		+	$$B = \frac{\mu_0 j h}{2}(1-\frac{h}{4x})$$
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$9.3.20.$ [Insert the problem statement]

			### Solution

			In problem $9.3.15$, we obtained formulas for the field inside and outside a cylindrical rod and plate. Replacing $I=j \pi (h/2)^2$ in the formulas yields the formulas for induction, respectively:
			$$B^{in}_{full} = \mu_0 j x$$
			$$B^{out}_{full}=\frac{\mu_0 j h}{2}$$
			$$B^{in}_{cyl} = \frac{\mu_0 j x}{2}$$
			$$B^{out}_{cyl}=\frac{\mu_0 j h^2}{8x}$$

			Now, we represent the desired field as the difference between the field of the plate and the cylindrical tube. We obtain the answers:

			For $0<x<h/2$:
			$$B = \frac{\mu_0 j x}{2}$$

			For $x > h/2$:
			$$B = \frac{\mu_0 j h}{2}(1-\frac{h}{4x})$$

			#### Answer

			[Insert a concise answer or boxed result]