Новое решение

Alexphysics правка от 25 июн. 2026 г., 17:51

правка #19236 позже →

en/11.3.21.md +20 −0

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		+	### Statement
		+
		+	$11.3.21.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	The transformer hums due to magnetostriction: the ferromagnetic core slightly contracts and expands when magnetized by the alternating current.
		+	The deformation$ \epsilon $is proportional to the square of the magnetic induction B
		+
		+	$\epsilon(t) \propto B(t)^2 = B_0^2 \sin^2(\omega t) = \frac{B_0^2}{2}\bigl[1 - \cos(2\omega t)\bigr], \qquad \omega = 2\pi f_{\text{grid}}$
		+
		+	The oscillatory component has twice the frequency of the mains.
		+	For an industrial mains frequency of $f_{\text{grid}} = 50\ \text{Hz}$
		+
		+	$\boxed{\nu = 2 \times 50\ \text{Hz} = 100\ \text{Hz}}$
		+
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$11.3.21.$ [Insert the problem statement]

			### Solution

			The transformer hums due to magnetostriction: the ferromagnetic core slightly contracts and expands when magnetized by the alternating current.
			The deformation$ \epsilon $is proportional to the square of the magnetic induction B

			$\epsilon(t) \propto B(t)^2 = B_0^2 \sin^2(\omega t) = \frac{B_0^2}{2}\bigl[1 - \cos(2\omega t)\bigr], \qquad \omega = 2\pi f_{\text{grid}}$

			The oscillatory component has twice the frequency of the mains.
			For an industrial mains frequency of $f_{\text{grid}} = 50\ \text{Hz}$

			$\boxed{\nu = 2 \times 50\ \text{Hz} = 100\ \text{Hz}}$


			#### Answer

			[Insert a concise answer or boxed result]