Edits to “Statement”, “Answer”
en/11.3.13.md
+7 −2
| @@ -1,6 +1,11 @@ | |||
| ### Statement | |||
| − | $11.3.13.$ | ||
| + | $11.3.13.$ | ||
| + | What is the inductance of two long solenoids of radius r1and r2connected as | ||
| + | shown in the figure? The internal solenoid has a length of l1, the external l2. | ||
| + | The number of turns per unit length of the internal solenoid n1, external n2. | ||
| + | Consider cases where the current directions in the turns of both solenoids are | ||
| + | the same and opposite. | ||
| ### Solution | |||
| Self‑inductances | |||
| For a long solenoid, the total self‑inductance is $L = \mu_0 n^2 \pi r^2 l $ | |||
| Therefore: | |||
| $L_1 = \mu_0 n_1^2 \pi r_1^2 l_1, \qquad | |||
| L_2 = \mu_0 n_2^2 \pi r_2^2 l_2$ | |||
| Mutual inductance | |||
| The mutual inductance M is defined from the flux that one coil sends through the other. | |||
| The magnetic field created by the outer solenoid (of length$ l_2$) in its interior is uniform and is given by$ B_2 = \mu_0 n_2 I_2$ This field passes through the cross‑section of the inner solenoid ($\pi r_1^2$) along the common length, which is $l_2$ | |||
| The flux linked by the inner solenoid due to current $I_2 $is: | |||
| $\Phi_{12} = (B_2)(\pi r_1^2)(n_1 l_2) = \mu_0 n_1 n_2 \pi r_1^2 l_2 I_2$ | |||
| By definition, $M = \Phi_{12}/I_2$: | |||
| $M = \mu_0 n_1 n_2 \pi r_1^2 l_2$ | |||
| Total inductance according to the direction of the currents | |||
| In a series connection the total inductance is: | |||
| $L = L_1 + L_2 \pm 2M$ | |||
| where the + sign corresponds to currents in the same direction (fields reinforcing each other) and the - sign to currents in opposite directions (fields opposing each other). | |||
| Substituting the expressions: | |||
| $\boxed{L_{\pm} = \mu_0 \pi \Bigl( n_1^2 r_1^2 l_1 + n_2^2 r_2^2 l_2 \pm 2 n_1 n_2 r_1^2 l_2 \Bigr)}$ | |||
| @@ -38,4 +43,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $\boxed{L_{\pm} = \mu_0 \pi \Bigl( n_1^2 r_1^2 l_1 + n_2^2 r_2^2 l_2 \pm 2 n_1 n_2 r_1^2 l_2 \Bigr)}$ | ||
| @@ -1,6 +1,11 @@ | |||
| ### Statement | ### Statement | ||
| $11.3.13.$ |
$11.3.13.$ | ||
| What is the inductance of two long solenoids of radius r1and r2connected as | |||
| shown in the figure? The internal solenoid has a length of l1, the external l2. | |||
| The number of turns per unit length of the internal solenoid n1, external n2. | |||
| Consider cases where the current directions in the turns of both solenoids are | |||
| the same and opposite. | |||
| ### Solution | ### Solution | ||
| Self‑inductances | Self‑inductances | ||
| For a long solenoid, the total self‑inductance is $L = \mu_0 n^2 \pi r^2 l $ | For a long solenoid, the total self‑inductance is $L = \mu_0 n^2 \pi r^2 l $ | ||
| Therefore: | Therefore: | ||
| $L_1 = \mu_0 n_1^2 \pi r_1^2 l_1, \qquad | $L_1 = \mu_0 n_1^2 \pi r_1^2 l_1, \qquad | ||
| L_2 = \mu_0 n_2^2 \pi r_2^2 l_2$ | L_2 = \mu_0 n_2^2 \pi r_2^2 l_2$ | ||
| Mutual inductance | Mutual inductance | ||
| The mutual inductance M is defined from the flux that one coil sends through the other. | The mutual inductance M is defined from the flux that one coil sends through the other. | ||
| The magnetic field created by the outer solenoid (of length$ l_2$) in its interior is uniform and is given by$ B_2 = \mu_0 n_2 I_2$ This field passes through the cross‑section of the inner solenoid ($\pi r_1^2$) along the common length, which is $l_2$ | The magnetic field created by the outer solenoid (of length$ l_2$) in its interior is uniform and is given by$ B_2 = \mu_0 n_2 I_2$ This field passes through the cross‑section of the inner solenoid ($\pi r_1^2$) along the common length, which is $l_2$ | ||
| The flux linked by the inner solenoid due to current $I_2 $is: | The flux linked by the inner solenoid due to current $I_2 $is: | ||
| $\Phi_{12} = (B_2)(\pi r_1^2)(n_1 l_2) = \mu_0 n_1 n_2 \pi r_1^2 l_2 I_2$ | $\Phi_{12} = (B_2)(\pi r_1^2)(n_1 l_2) = \mu_0 n_1 n_2 \pi r_1^2 l_2 I_2$ | ||
| By definition, $M = \Phi_{12}/I_2$: | By definition, $M = \Phi_{12}/I_2$: | ||
| $M = \mu_0 n_1 n_2 \pi r_1^2 l_2$ | $M = \mu_0 n_1 n_2 \pi r_1^2 l_2$ | ||
| Total inductance according to the direction of the currents | Total inductance according to the direction of the currents | ||
| In a series connection the total inductance is: | In a series connection the total inductance is: | ||
| $L = L_1 + L_2 \pm 2M$ | $L = L_1 + L_2 \pm 2M$ | ||
| where the + sign corresponds to currents in the same direction (fields reinforcing each other) and the - sign to currents in opposite directions (fields opposing each other). | where the + sign corresponds to currents in the same direction (fields reinforcing each other) and the - sign to currents in opposite directions (fields opposing each other). | ||
| Substituting the expressions: | Substituting the expressions: | ||
| $\boxed{L_{\pm} = \mu_0 \pi \Bigl( n_1^2 r_1^2 l_1 + n_2^2 r_2^2 l_2 \pm 2 n_1 n_2 r_1^2 l_2 \Bigr)}$ | $\boxed{L_{\pm} = \mu_0 \pi \Bigl( n_1^2 r_1^2 l_1 + n_2^2 r_2^2 l_2 \pm 2 n_1 n_2 r_1^2 l_2 \Bigr)}$ | ||
| @@ -38,4 +43,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $\boxed{L_{\pm} = \mu_0 \pi \Bigl( n_1^2 r_1^2 l_1 + n_2^2 r_2^2 l_2 \pm 2 n_1 n_2 r_1^2 l_2 \Bigr)}$ | ||