Translated 9.1.6-9.1.15
en/9.1.9.md
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| + | <title>A triangular wire frame with current can rotate about a horizontal axis OO^\prime passing through the vertex of the triangle. The mass of a unit length of wire is \rho, the current in the frame is I. The frame is in a magnetic field of induction B directed along the gravitational field. Determine the angle of deviation of the plane of the triangle from the vertical.</title> | ||
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| + | <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | ||
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| + | Solutions of Savchenko Problems in Physics <br> | ||
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| + | <h3 id="back-link"><a href="../../#9.1">$\leftarrow$Back</a></h3> | ||
| + | |||
| + | <h3> Statement </h3> | ||
| + | <p> | ||
| + | $9.1.9.$ A triangular wire frame with current can rotate about a horizontal axis $OO^\prime$ passing through the vertex of the triangle. The mass of a unit length of wire is $\rho$, the current in the frame is $I$. The frame is in a magnetic field of induction $B$ directed along the gravitational field. Determine the angle of deviation of the plane of the triangle from the vertical. | ||
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| + | <figure> | ||
| + | <img src="https://savchenkosolutions.com/9/9.1.9/statement.png" | ||
| + | loading="lazy" width="200" /> | ||
| + | <figcaption> | ||
| + | For problem $9.1.9$ | ||
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| + | <p> | ||
| + | </p> | ||
| + | |||
| + | <h3>Solution</h3> | ||
| + | <p> | ||
| + | To solve this, we will use the fact that $\vec{N} = \vec{M} \times \vec{B}$ is valid for any contour $S$ with magnetic moment $M$ (for the proof, see <a href="../9.1.10">9.1.10</a>)</p><p> | ||
| + | The moment of magnetic field forces | ||
| + | $$ N_{B} = |\vec{M} \times \vec{B}| = ISB \cos\alpha\quad(1) $$ | ||
| + | The center of mass will be at the intersection point of the medians, the distance from this point to the suspension point: | ||
| + | $$ h_C=\frac{2h}{3} $$ | ||
| + | Moment of gravity | ||
| + | $$ M_{mg} = 3a\rho g\cdot \frac{2}{3}h \sin\alpha\quad(2) $$ | ||
| + | where $a$ is the side of the triangle, and its area $S = \frac{1}{2}ah$ | ||
| + | We substitute the result into $(2)$ | ||
| + | $$ M_{mg}= 4S\rho g\cdot \sin\alpha\quad(3) $$ | ||
| + | We equate the obtained moments from $(1)$ and $(3)$: | ||
| + | $$ ISB\cdot \cos\alpha = 4S\rho g \cdot \sin\alpha\Rightarrow \boxed{\tan\alpha = \frac{IB}{4\rho g}} $$ | ||
| + | |||
| + | </p> | ||
| + | |||
| + | <h4>Answer</h4> | ||
| + | <p> | ||
| + | $$\tan\alpha = \frac{IB}{4\rho g}$$ | ||
| + | </p> | ||
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| <meta name="author" content="Aliaksandr Melnichenka"> | |||
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| <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | |||
| </div> | |||
| </a> | |||
| <p class="author"> | |||
| Solutions of Savchenko Problems in Physics <br> | |||
| <i><b>knowledge must be free</b></i> | |||
| </p> | |||
| </header> | |||
| <h3 id="back-link"><a href="../../#9.1">$\leftarrow$Back</a></h3> | |||
| <h3> Statement </h3> | |||
| <p> | |||
| $9.1.9.$ A triangular wire frame with current can rotate about a horizontal axis $OO^\prime$ passing through the vertex of the triangle. The mass of a unit length of wire is $\rho$, the current in the frame is $I$. The frame is in a magnetic field of induction $B$ directed along the gravitational field. Determine the angle of deviation of the plane of the triangle from the vertical. | |||
| </p> | |||
| <center> | |||
| <figure> | |||
| <img src="https://savchenkosolutions.com/9/9.1.9/statement.png" | |||
| loading="lazy" width="200" /> | |||
| <figcaption> | |||
| For problem $9.1.9$ | |||
| </figcaption> | |||
| </figure> | |||
| </center> | |||
| <p> | |||
| </p> | |||
| <h3>Solution</h3> | |||
| <p> | |||
| To solve this, we will use the fact that $\vec{N} = \vec{M} \times \vec{B}$ is valid for any contour $S$ with magnetic moment $M$ (for the proof, see <a href="../9.1.10">9.1.10</a>)</p><p> | |||
| The moment of magnetic field forces | |||
| $$ N_{B} = |\vec{M} \times \vec{B}| = ISB \cos\alpha\quad(1) $$ | |||
| The center of mass will be at the intersection point of the medians, the distance from this point to the suspension point: | |||
| $$ h_C=\frac{2h}{3} $$ | |||
| Moment of gravity | |||
| $$ M_{mg} = 3a\rho g\cdot \frac{2}{3}h \sin\alpha\quad(2) $$ | |||
| where $a$ is the side of the triangle, and its area $S = \frac{1}{2}ah$ | |||
| We substitute the result into $(2)$ | |||
| $$ M_{mg}= 4S\rho g\cdot \sin\alpha\quad(3) $$ | |||
| We equate the obtained moments from $(1)$ and $(3)$: | |||
| $$ ISB\cdot \cos\alpha = 4S\rho g \cdot \sin\alpha\Rightarrow \boxed{\tan\alpha = \frac{IB}{4\rho g}} $$ | |||
| </p> | |||
| <h4>Answer</h4> | |||
| <p> | |||
| $$\tan\alpha = \frac{IB}{4\rho g}$$ | |||
| </p> | |||
| <p style="text-align: right; font-style: italic; font-size: 14;"> | |||
| Andrei Yersh<br> | |||
| </p> | |||
| <footer class="row container"> | |||
| <br> | |||
| <p> | |||
| <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small> | |||
| </p> | |||
| <p> | |||
| <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small> | |||
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