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en/11.3.2.md
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| + | ### Statement | ||
| + | |||
| + | $11.3.2.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | The mutual inductance between the solenoid and the coil is obtained by calculating the magnetic flux of the solenoid through the coil. | ||
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| + | 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. | ||
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| + | The bent coil is formed by two half-turns of radius r, each with area$ S = \pi r^2/2$. | ||
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| + | 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: | ||
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| + | $\Phi = B\frac{\pi r^2}{2}\left[\cos\!\left(\frac{\pi}{2}-\alpha\right) + \cos\alpha\right] = \mu_0 n I\frac{\pi r^2}{2}(\sin\alpha + \cos\alpha)$ | ||
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| + | The mutual inductance is $M = \Phi/I$ So | ||
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| + | $\boxed{M = \frac{1}{2}\,\mu_0 n \pi r^2 (\sin\alpha + \cos\alpha)}$ | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $11.3.2.$ [Insert the problem statement] | |||
| ### 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 | |||
| $\boxed{M = \frac{1}{2}\,\mu_0 n \pi r^2 (\sin\alpha + \cos\alpha)}$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||