Правка разделов «Statement», «Solution», «Answer»
en/9.2.23.md
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| @@ -1,15 +1,15 @@ | |||
| ### Statement | |||
| − | $9.2.23.$ [Insert the problem statement] | ||
| + | $9.2.23.$ The interaction force of two thin magnetized square plates located at a distance $H$ above each other is equal to $F$. The dimensions of the plates are $a × a × h$. Estimate the magnetic moment per unit volume of the plate if the thickness of the plate is $h ≪ H$ and $H ≪ a$. | ||
| ### Solution | |||
| − | Since $h<<H we will consider the interaction of two frames with a current I | ||
| + | Since $h<<H$ we will consider the interaction of two frames with a current $I = Mh$. So, seeing that $a>>H$, we will take into account the interaction of only the nearest beams, the force of interaction between them is calculated assuming that the field of rotation is close to the field of infinite filaments: | ||
| $$B = \frac{\mu_0 Mh}{2\pi H}$$ | |||
| $$F_1 = \frac{\mu_0 (Mh)^2a}{2\pi H}$$ | |||
| $$F = 4F_1 = \frac{2\mu_0 (Mh)^2a}{\pi H}$$ | |||
| − | |||
| + | $$M = \sqrt{\frac{F\pi H}{2\mu_0 a h^2}}$$ | ||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $$M = \sqrt{\frac{F\pi H}{2\mu_0 a h^2}}$$ | ||
| @@ -1,15 +1,15 @@ | |||
| ### Statement | ### Statement | ||
| $9.2.23.$ [Insert the problem statement] | $9.2.23.$ The interaction force of two thin magnetized square plates located at a distance $H$ above each other is equal to $F$. The dimensions of the plates are $a × a × h$. Estimate the magnetic moment per unit volume of the plate if the thickness of the plate is $h ≪ H$ and $H ≪ a$. | ||
| ### Solution | ### Solution | ||
| Since $h<<H we will consider the interaction of two frames with a current I |
Since $h<<H$ we will consider the interaction of two frames with a current $I = Mh$. So, seeing that $a>>H$, we will take into account the interaction of only the nearest beams, the force of interaction between them is calculated assuming that the field of rotation is close to the field of infinite filaments: | ||
| $$B = \frac{\mu_0 Mh}{2\pi H}$$ | $$B = \frac{\mu_0 Mh}{2\pi H}$$ | ||
| $$F_1 = \frac{\mu_0 (Mh)^2a}{2\pi H}$$ | $$F_1 = \frac{\mu_0 (Mh)^2a}{2\pi H}$$ | ||
| $$F = 4F_1 = \frac{2\mu_0 (Mh)^2a}{\pi H}$$ | $$F = 4F_1 = \frac{2\mu_0 (Mh)^2a}{\pi H}$$ | ||
| $$M = \sqrt{\frac{F\pi H}{2\mu_0 a h^2}}$$ | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $$M = \sqrt{\frac{F\pi H}{2\mu_0 a h^2}}$$ | ||