Новое решение

Alexphysics правка от
Посмотреть решение История
правка #19178 позже →
en/11.3.10.md +61 −0
@@ -0,0 +1,61 @@
+### Statement
+
+$11.3.10.$ [Insert the problem statement]
+
+### Solution
+
+Magnetic field inside the wire
+
+Enclosed current at radius r:
+
+$I_{\text{enc}} = I \frac{r^2}{r_2^2}$
+
+Ampère's law:
+
+$H_1 \cdot 2\pi r = I \frac{r^2}{r_2^2} \;\Longrightarrow\; H_1 = \frac{I r}{2\pi r_2^2}$
+
+Magnetic flux density:
+
+$B_1 = \mu_1 \mu_0 H_1 = \frac{\mu_1 \mu_0 I r}{2\pi r_2^2}$
+
+Magnetic field between the wire and the casing
+
+The enclosed current is the total current of the wire (I)
+
+$H_2 \cdot 2\pi r = I \;\Longrightarrow\; H_2 = \frac{I}{2\pi r}$
+
+$B_2 = \mu_2 \mu_0 H_2 = \frac{\mu_2 \mu_0 I}{2\pi r}$
+
+For$ r > r_1$, the net current is zero and the field is zero
+
+Magnetic flux per unit length
+
+Internal flux$ (\Phi'_{\text{int}})$
+According to the convention adopted in the problem, $B_1 $ is integrated directly over the entire cross-section of the wire:
+
+$\Phi'_{\text{int}} = \int_0^{r_2} B_1 \, dr
+= \frac{\mu_1 \mu_0 I}{2\pi r_2^2} \int_0^{r_2} r \, dr
+= \frac{\mu_1 \mu_0 I}{2\pi r_2^2} \cdot \frac{r_2^2}{2}
+= \frac{\mu_1 \mu_0 I}{4\pi}$
+
+External flux $(\Phi'_{\text{ext}})$
+Between the two conductors:
+
+$\Phi'_{\text{ext}} = \int_{r_2}^{r_1} B_2 \, dr
+= \frac{\mu_2 \mu_0 I}{2\pi} \int_{r_2}^{r_1} \frac{dr}{r}
+= \frac{\mu_2 \mu_0 I}{2\pi} \ln \frac{r_1}{r_2}$
+
+Total flux per unit length:
+
+$\Phi' = \Phi'_{\text{int}} + \Phi'_{\text{ext}}
+= \frac{\mu_0 I}{4\pi} \left( \mu_1 + 2\mu_2 \ln \frac{r_1}{r_2} \right)$
+
+Inductance per unit length
+
+By definition have $ L' = \Phi'/I$
+
+$\boxed{L' = \frac{\mu_0}{4\pi} \left( \mu_1 + 2\mu_2 \ln \frac{r_1}{r_2} \right)}$
+
+#### Answer
+
+[Insert a concise answer or boxed result]