Новое решение
en/9.2.23.md
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| + | ### Statement | ||
| + | |||
| + | $9.2.23.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | Since $h<<H we will consider the interaction of two frames with a current I 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: | ||
| + | |||
| + | $$B = \frac{\mu_0 Mh}{2\pi H}$$ | ||
| + | $$F_1 = \frac{\mu_0 (Mh)^2a}{2\pi H}$$ | ||
| + | $$F = 4F_1 = \frac{2\mu_0 (Mh)^2a}{\pi H}$$ | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $9.2.23.$ [Insert the problem statement] | |||
| ### Solution | |||
| Since $h<<H we will consider the interaction of two frames with a current I 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: | |||
| $$B = \frac{\mu_0 Mh}{2\pi H}$$ | |||
| $$F_1 = \frac{\mu_0 (Mh)^2a}{2\pi H}$$ | |||
| $$F = 4F_1 = \frac{2\mu_0 (Mh)^2a}{\pi H}$$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||