Edits to “Solution”, “Answer”

naz edited
revision #18479 parent #18478 ← older
@@ -15,11 +15,18 @@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: $$B = \frac{\mu_0 (I-I')}{2\pi x}$$
+b) 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 $$B = \frac{\mu_0 I}{2\pi x}$$
+c) 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]
+а.
+Above the plane $B = \frac{\mu_0I}{2πx}$, the magnetic field lines coincide with the field lines of an infinite straight wire; below the plane $B = 0$
+
+b.
+Above the plane $B = \frac{\mu_0I}{2πx}$, below the plane $B = \frac{\mu_0(I − I′)}{2πx}$.
+
+c.
+Inside the cable $B = \frac{\mu_0I}{2πx}$, outside the cable $B = 0$.