Правка разделов «Statement», «Solution», «Answer»
en/11.5.23.md
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| @@ -1,6 +1,6 @@ | |||
| ### Statement | |||
| − | $11.5.23.$ [Insert the problem statement] | ||
| + | $11.5.23.$ A copper ring of radius $r$ and mass $m$ hangs on a thread, making small torsional vibrations with a period of $T$. The inductance of the ring is $L$. How will the oscillation period of the ring change if it is placed in a horizontal homogeneous magnetic field of induction $B$ parallel to the plane of the ring in the equilibrium position? The moment of inertia of the ring relative to the axis passing through the diameter is equal to $J$. Ignore the resistance of the ring. | ||
| ### Solution | |||
| The oscillation period of a conventional torsional pendulum with a small amplitude is: | |||
| $$T = 2\pi \sqrt{\frac{J}{k}}$$ | |||
| Where $k$ is the torsional stiffness of the thread, we express it: | |||
| $$k = \frac{4 \pi^2 J}{T^2}$$ | |||
| The moment acting on the frame from the side of the magnetic field is equal to: | |||
| @@ -16,7 +16,7 @@Solution | |||
| $$M = \mu \times B \approx \mu B$$ | |||
| − | where $\mu = \pi r^2 | ||
| + | where $\mu = \pi r^2I$ is the magnetic moment of the frame. | ||
| The magnetic flux through the frame is: | |||
| $$\Phi = L I = B \pi r^2 \varphi$$ | |||
| $$I = \frac{B \pi r^2}{L} \varphi$$ | |||
| Let's make up the equation of oscillations in 2ZN for rotational motion: | |||
| $$J \beta = k = \frac{4 \pi^2 J}{T^2} \varphi + \frac{B^2 \pi^2 r^4}{L} \varphi$$ | |||
| The period is similar to the usual fluctuations, after the transformations: | |||
| $$T' = \frac{T}{\sqrt{1+\frac{B^2 r^4 T^2}{4LJ}}}$$ | |||
| @@ -34,4 +34,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $$T' = \frac{T}{\sqrt{1+\frac{B^2 r^4 T^2}{4LJ}}}$$ | ||
| @@ -1,6 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $11.5.23.$ [Insert the problem statement] | $11.5.23.$ A copper ring of radius $r$ and mass $m$ hangs on a thread, making small torsional vibrations with a period of $T$. The inductance of the ring is $L$. How will the oscillation period of the ring change if it is placed in a horizontal homogeneous magnetic field of induction $B$ parallel to the plane of the ring in the equilibrium position? The moment of inertia of the ring relative to the axis passing through the diameter is equal to $J$. Ignore the resistance of the ring. | ||
| ### Solution | ### Solution | ||
| The oscillation period of a conventional torsional pendulum with a small amplitude is: | The oscillation period of a conventional torsional pendulum with a small amplitude is: | ||
| $$T = 2\pi \sqrt{\frac{J}{k}}$$ | $$T = 2\pi \sqrt{\frac{J}{k}}$$ | ||
| Where $k$ is the torsional stiffness of the thread, we express it: | Where $k$ is the torsional stiffness of the thread, we express it: | ||
| $$k = \frac{4 \pi^2 J}{T^2}$$ | $$k = \frac{4 \pi^2 J}{T^2}$$ | ||
| The moment acting on the frame from the side of the magnetic field is equal to: | The moment acting on the frame from the side of the magnetic field is equal to: | ||
| @@ -16,7 +16,7 @@Solution | |||
| $$M = \mu \times B \approx \mu B$$ | $$M = \mu \times B \approx \mu B$$ | ||
| where $\mu = \pi r^2 |
where $\mu = \pi r^2I$ is the magnetic moment of the frame. | ||
| The magnetic flux through the frame is: | The magnetic flux through the frame is: | ||
| $$\Phi = L I = B \pi r^2 \varphi$$ | $$\Phi = L I = B \pi r^2 \varphi$$ | ||
| $$I = \frac{B \pi r^2}{L} \varphi$$ | $$I = \frac{B \pi r^2}{L} \varphi$$ | ||
| Let's make up the equation of oscillations in 2ZN for rotational motion: | Let's make up the equation of oscillations in 2ZN for rotational motion: | ||
| $$J \beta = k = \frac{4 \pi^2 J}{T^2} \varphi + \frac{B^2 \pi^2 r^4}{L} \varphi$$ | $$J \beta = k = \frac{4 \pi^2 J}{T^2} \varphi + \frac{B^2 \pi^2 r^4}{L} \varphi$$ | ||
| The period is similar to the usual fluctuations, after the transformations: | The period is similar to the usual fluctuations, after the transformations: | ||
| $$T' = \frac{T}{\sqrt{1+\frac{B^2 r^4 T^2}{4LJ}}}$$ | $$T' = \frac{T}{\sqrt{1+\frac{B^2 r^4 T^2}{4LJ}}}$$ | ||
| @@ -34,4 +34,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $$T' = \frac{T}{\sqrt{1+\frac{B^2 r^4 T^2}{4LJ}}}$$ | ||