Edits to “Statement”, “Answer”
en/11.3.10.md
+10 −2
| @@ -1,7 +1,14 @@ | |||
| ### Statement | |||
| − | $11.3.10.$ | ||
| + | $11.3.10.$ | ||
| + | On the axis of a thin conducting cylindrical shell of radius r1is a wire of ra- | ||
| + | dius r2, the magnetic permeability of which is µ1. The space between them is | ||
| + | filled with a substance with magnetic permeability µ2. Find the line induc- | ||
| + | tance per unit length. The current in the wire is evenly distributed over the | ||
| + | cross-section, equal in modulus and opposite in direction to the current of the | ||
| + | cylindrical shell. | ||
| + | |||
| ### 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$ | |||
| $\boxed{L' = \frac{\mu_0}{4\pi} \left( \mu_1 + 2\mu_2 \ln \frac{r_1}{r_2} \right)}$ | |||
| @@ -58,4 +65,5 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | |||
| + | $\boxed{L' = \frac{\mu_0}{4\pi} \left( \mu_1 + 2\mu_2 \ln \frac{r_1}{r_2} \right)}$ | ||
| @@ -1,7 +1,14 @@ | |||
| ### Statement | ### Statement | ||
| $11.3.10.$ |
$11.3.10.$ | ||
| On the axis of a thin conducting cylindrical shell of radius r1is a wire of ra- | |||
| dius r2, the magnetic permeability of which is µ1. The space between them is | |||
| filled with a substance with magnetic permeability µ2. Find the line induc- | |||
| tance per unit length. The current in the wire is evenly distributed over the | |||
| cross-section, equal in modulus and opposite in direction to the current of the | |||
| cylindrical shell. | |||
| ### Solution | ### Solution | ||
| Magnetic field inside the wire | Magnetic field inside the wire | ||
| Enclosed current at radius r: | Enclosed current at radius r: | ||
| $I_{\text{enc}} = I \frac{r^2}{r_2^2}$ | $I_{\text{enc}} = I \frac{r^2}{r_2^2}$ | ||
| Ampère's law: | Ampère's law: | ||
| $H_1 \cdot 2\pi r = I \frac{r^2}{r_2^2} \;\Longrightarrow\; H_1 = \frac{I r}{2\pi r_2^2}$ | $H_1 \cdot 2\pi r = I \frac{r^2}{r_2^2} \;\Longrightarrow\; H_1 = \frac{I r}{2\pi r_2^2}$ | ||
| Magnetic flux density: | Magnetic flux density: | ||
| $B_1 = \mu_1 \mu_0 H_1 = \frac{\mu_1 \mu_0 I r}{2\pi r_2^2}$ | $B_1 = \mu_1 \mu_0 H_1 = \frac{\mu_1 \mu_0 I r}{2\pi r_2^2}$ | ||
| Magnetic field between the wire and the casing | Magnetic field between the wire and the casing | ||
| The enclosed current is the total current of the wire (I) | The enclosed current is the total current of the wire (I) | ||
| $H_2 \cdot 2\pi r = I \;\Longrightarrow\; H_2 = \frac{I}{2\pi r}$ | $H_2 \cdot 2\pi r = I \;\Longrightarrow\; H_2 = \frac{I}{2\pi r}$ | ||
| $B_2 = \mu_2 \mu_0 H_2 = \frac{\mu_2 \mu_0 I}{2\pi r}$ | $B_2 = \mu_2 \mu_0 H_2 = \frac{\mu_2 \mu_0 I}{2\pi r}$ | ||
| For$ r > r_1$, the net current is zero and the field is zero | For$ r > r_1$, the net current is zero and the field is zero | ||
| Magnetic flux per unit length | Magnetic flux per unit length | ||
| Internal flux$ (\Phi'_{\text{int}})$ | 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: | 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 | $\Phi'_{\text{int}} = \int_0^{r_2} B_1 \, dr | ||
| = \frac{\mu_1 \mu_0 I}{2\pi r_2^2} \int_0^{r_2} r \, dr | = \frac{\mu_1 \mu_0 I}{2\pi r_2^2} \int_0^{r_2} r \, dr | ||
| = \frac{\mu_1 \mu_0 I}{2\pi r_2^2} \cdot \frac{r_2^2}{2} | = \frac{\mu_1 \mu_0 I}{2\pi r_2^2} \cdot \frac{r_2^2}{2} | ||
| = \frac{\mu_1 \mu_0 I}{4\pi}$ | = \frac{\mu_1 \mu_0 I}{4\pi}$ | ||
| External flux $(\Phi'_{\text{ext}})$ | External flux $(\Phi'_{\text{ext}})$ | ||
| Between the two conductors: | Between the two conductors: | ||
| $\Phi'_{\text{ext}} = \int_{r_2}^{r_1} B_2 \, dr | $\Phi'_{\text{ext}} = \int_{r_2}^{r_1} B_2 \, dr | ||
| = \frac{\mu_2 \mu_0 I}{2\pi} \int_{r_2}^{r_1} \frac{dr}{r} | = \frac{\mu_2 \mu_0 I}{2\pi} \int_{r_2}^{r_1} \frac{dr}{r} | ||
| = \frac{\mu_2 \mu_0 I}{2\pi} \ln \frac{r_1}{r_2}$ | = \frac{\mu_2 \mu_0 I}{2\pi} \ln \frac{r_1}{r_2}$ | ||
| Total flux per unit length: | Total flux per unit length: | ||
| $\Phi' = \Phi'_{\text{int}} + \Phi'_{\text{ext}} | $\Phi' = \Phi'_{\text{int}} + \Phi'_{\text{ext}} | ||
| = \frac{\mu_0 I}{4\pi} \left( \mu_1 + 2\mu_2 \ln \frac{r_1}{r_2} \right)$ | = \frac{\mu_0 I}{4\pi} \left( \mu_1 + 2\mu_2 \ln \frac{r_1}{r_2} \right)$ | ||
| Inductance per unit length | Inductance per unit length | ||
| By definition have $ L' = \Phi'/I$ | By definition have $ L' = \Phi'/I$ | ||
| $\boxed{L' = \frac{\mu_0}{4\pi} \left( \mu_1 + 2\mu_2 \ln \frac{r_1}{r_2} \right)}$ | $\boxed{L' = \frac{\mu_0}{4\pi} \left( \mu_1 + 2\mu_2 \ln \frac{r_1}{r_2} \right)}$ | ||
| @@ -58,4 +65,5 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | |||
| $\boxed{L' = \frac{\mu_0}{4\pi} \left( \mu_1 + 2\mu_2 \ln \frac{r_1}{r_2} \right)}$ | |||