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en/11.3.21.md
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| + | ### Statement | ||
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| + | $11.3.21.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | 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 | ||
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| + | $\epsilon(t) \propto B(t)^2 = B_0^2 \sin^2(\omega t) = \frac{B_0^2}{2}\bigl[1 - \cos(2\omega t)\bigr], \qquad \omega = 2\pi f_{\text{grid}}$ | ||
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| + | The oscillatory component has twice the frequency of the mains. | ||
| + | For an industrial mains frequency of $f_{\text{grid}} = 50\ \text{Hz}$ | ||
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| + | $\boxed{\nu = 2 \times 50\ \text{Hz} = 100\ \text{Hz}}$ | ||
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| + | #### Answer | ||
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| + | [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] | |||