Edits to “Statement”, “Answer”
en/11.3.2.md
+8 −2
| @@ -1,6 +1,12 @@ | |||
| ### Statement | |||
| − | $11.3.2.$ | ||
| + | $11.3.2.$ | ||
| + | A coil of radius r was bent along its diameter at a right angle and placed | ||
| + | inside the long solenoid so that one of the planes was located to the axis of the solenoid at an angle α, and the other at an angle π | ||
| + | The number of turns | ||
| + | 2−α | ||
| + | per unit length of the solenoid n. What is the mutual inductance of the bent | ||
| + | coil and the solenoid? | ||
| ### Solution | |||
| The mutual inductance between the solenoid and the coil is obtained by calculating the magnetic flux of the solenoid through the coil. | |||
| The magnetic field inside the long solenoid is uniform and axial: $B = \mu_0 n I$, where I is the current in the solenoid and n is the number of turns per unit length. | |||
| The bent coil is formed by two half-turns of radius r, each with area$ S = \pi r^2/2$. | |||
| Their planes make angles $\alpha $and $\frac{\pi}{2}-\alpha$ with the solenoid axis, so their normals make angles $\frac{\pi}{2}-\alpha $and $\alpha$, respectively, with the direction of the field. The flux through each half-turn is $B\,S\cos\theta$, where $\theta$ is the angle between the normal and the field. Thus, the total flux is: | |||
| $\Phi = B\frac{\pi r^2}{2}\left[\cos\!\left(\frac{\pi}{2}-\alpha\right) + \cos\alpha\right] = \mu_0 n I\frac{\pi r^2}{2}(\sin\alpha + \cos\alpha)$ | |||
| The mutual inductance is $M = \Phi/I$ So | |||
| $\boxed{M = \frac{1}{2}\,\mu_0 n \pi r^2 (\sin\alpha + \cos\alpha)}$ | |||
| @@ -20,4 +26,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $\boxed{M = \frac{1}{2}\,\mu_0 n \pi r^2 (\sin\alpha + \cos\alpha)}$ | ||
| @@ -1,6 +1,12 @@ | |||
| ### Statement | ### Statement | ||
| $11.3.2.$ |
$11.3.2.$ | ||
| A coil of radius r was bent along its diameter at a right angle and placed | |||
| inside the long solenoid so that one of the planes was located to the axis of the solenoid at an angle α, and the other at an angle π | |||
| The number of turns | |||
| 2−α | |||
| per unit length of the solenoid n. What is the mutual inductance of the bent | |||
| coil and the solenoid? | |||
| ### Solution | ### Solution | ||
| The mutual inductance between the solenoid and the coil is obtained by calculating the magnetic flux of the solenoid through the coil. | The mutual inductance between the solenoid and the coil is obtained by calculating the magnetic flux of the solenoid through the coil. | ||
| The magnetic field inside the long solenoid is uniform and axial: $B = \mu_0 n I$, where I is the current in the solenoid and n is the number of turns per unit length. | The magnetic field inside the long solenoid is uniform and axial: $B = \mu_0 n I$, where I is the current in the solenoid and n is the number of turns per unit length. | ||
| The bent coil is formed by two half-turns of radius r, each with area$ S = \pi r^2/2$. | The bent coil is formed by two half-turns of radius r, each with area$ S = \pi r^2/2$. | ||
| Their planes make angles $\alpha $and $\frac{\pi}{2}-\alpha$ with the solenoid axis, so their normals make angles $\frac{\pi}{2}-\alpha $and $\alpha$, respectively, with the direction of the field. The flux through each half-turn is $B\,S\cos\theta$, where $\theta$ is the angle between the normal and the field. Thus, the total flux is: | Their planes make angles $\alpha $and $\frac{\pi}{2}-\alpha$ with the solenoid axis, so their normals make angles $\frac{\pi}{2}-\alpha $and $\alpha$, respectively, with the direction of the field. The flux through each half-turn is $B\,S\cos\theta$, where $\theta$ is the angle between the normal and the field. Thus, the total flux is: | ||
| $\Phi = B\frac{\pi r^2}{2}\left[\cos\!\left(\frac{\pi}{2}-\alpha\right) + \cos\alpha\right] = \mu_0 n I\frac{\pi r^2}{2}(\sin\alpha + \cos\alpha)$ | $\Phi = B\frac{\pi r^2}{2}\left[\cos\!\left(\frac{\pi}{2}-\alpha\right) + \cos\alpha\right] = \mu_0 n I\frac{\pi r^2}{2}(\sin\alpha + \cos\alpha)$ | ||
| The mutual inductance is $M = \Phi/I$ So | The mutual inductance is $M = \Phi/I$ So | ||
| $\boxed{M = \frac{1}{2}\,\mu_0 n \pi r^2 (\sin\alpha + \cos\alpha)}$ | $\boxed{M = \frac{1}{2}\,\mu_0 n \pi r^2 (\sin\alpha + \cos\alpha)}$ | ||
| @@ -20,4 +26,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $\boxed{M = \frac{1}{2}\,\mu_0 n \pi r^2 (\sin\alpha + \cos\alpha)}$ | ||