Новое решение
en/9.3.17.md
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| + | ### Statement | ||
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| + | $9.3.17.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | а) | ||
| + | Let's surround the wire with rings of radius $2\pi x$, above the plane they carry current I and induction $$B = \frac{\mu_0 I}{2\pi x}$$, below the plane there is no current, therefore the induction is zero. | ||
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| + | б) Similarly, over the plane the induction $$B = \frac{\mu_0 I}{2\pi x}$$ under: $$B = \frac{\mu_0 (I-I')}{2\pi x}$$ | ||
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| + | в) Using the same type of circuit, we'll analyze it. Inside the circuit, the current only enters. The answer is $$B = \frac{\mu_0 I}{2\pi x}$$ | ||
| + | Let's take a wider circuit. The current enters it and the same amount exits $\Longrightarrow$ outside $$B = 0$$ | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $9.3.17.$ [Insert the problem statement] | |||
| ### Solution | |||
| а) | |||
| Let's surround the wire with rings of radius $2\pi x$, above the plane they carry current I and induction $$B = \frac{\mu_0 I}{2\pi x}$$, below the plane there is no current, therefore the induction is zero. | |||
| б) Similarly, over the plane the induction $$B = \frac{\mu_0 I}{2\pi x}$$ under: $$B = \frac{\mu_0 (I-I')}{2\pi x}$$ | |||
| в) Using the same type of circuit, we'll analyze it. Inside the circuit, the current only enters. The answer is $$B = \frac{\mu_0 I}{2\pi x}$$ | |||
| Let's take a wider circuit. The current enters it and the same amount exits $\Longrightarrow$ outside $$B = 0$$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||