Translated 9.1.6-9.1.15
en/9.1.6.md
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| + | <title>A square frame with current is fixed so that it can rotate freely around its horizontal side. The frame is in a vertical uniform magnetic field of induction B. The angle of the frame to the horizon is \alpha, its mass is m, and the length of the side is a. Find the current in the frame.</title> | ||
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| + | <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | ||
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| + | <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> | ||
| + | $9.1.6.$ A square frame with current is fixed so that it can rotate freely around its horizontal side. The frame is in a vertical uniform magnetic field of induction $B$. The angle of the frame to the horizon is $\alpha$, its mass is $m$, and the length of the side is $a$. Find the current in the frame. | ||
| + | </p> | ||
| + | <center> | ||
| + | <figure> | ||
| + | <img src="https://savchenkosolutions.com/9/9.1.6/statement.png" | ||
| + | loading="lazy" width="200" /> | ||
| + | <figcaption> | ||
| + | For problem $9.1.6$ | ||
| + | </figcaption> | ||
| + | </figure> | ||
| + | </center> | ||
| + | <p> | ||
| + | </p> | ||
| + | |||
| + | <h3>Solution</h3> | ||
| + | <p> | ||
| + | The moment of forces acting on the frame from the magnetic field caused by the Ampere force | ||
| + | $$ M_B = Fa = Ba^2I\sin\alpha $$ | ||
| + | Moment of gravity: | ||
| + | $$ M_{mg} = mg\cdot \frac{a}{2}\cos\alpha $$ | ||
| + | In the equilibrium condition, the sum of the moments of external forces, taking into account the direction, is equal to zero $(M_B = M_{mg})$: | ||
| + | $$ B \sin\alpha~\cdot aI = \frac{1}{2}mg\cdot \cos\alpha $$ | ||
| + | Where do we find the current in the frame? | ||
| + | $$ \boxed{I = \frac{mg\cdot \cot\alpha}{2Ba}} $$ | ||
| + | </p> | ||
| + | |||
| + | <h4>Answer</h4> | ||
| + | <p> | ||
| + | $$I = \frac{mg\cdot \cot\alpha}{2Ba} $$ | ||
| + | </p> | ||
| + | <p style="text-align: right; font-style: italic; font-size: 14;"> | ||
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| <meta name="description" content="A square frame with current is fixed so that it can rotate freely around its horizontal side. The frame is in a vertical uniform magnetic field of induction B. The angle of the frame to the horizon is \alpha, its mass is m, and the length of the side is a. Find the current in the frame."> | |||
| <meta name="author" content="Aliaksandr Melnichenka"> | |||
| <meta name="date" content="2023-10" scheme="YYYY-MM"> | |||
| <meta property="og:title" content="A square frame with current is fixed so that it can rotate freely around its horizontal side. The frame is in a vertical uniform magnetic field of induction B. The angle of the frame to the horizon is \alpha, its mass is m, and the length of the side is a. Find the current in the frame."> | |||
| <meta property="og:image" content="img/logo.png"> | |||
| <meta property="og:description" content="A square frame with current is fixed so that it can rotate freely around its horizontal side. The frame is in a vertical uniform magnetic field of induction B. The angle of the frame to the horizon is \alpha, its mass is m, and the length of the side is a. Find the current in the frame."> | |||
| <meta name="yandex-verification" content="6cfda41f74038368"> | |||
| <title>A square frame with current is fixed so that it can rotate freely around its horizontal side. The frame is in a vertical uniform magnetic field of induction B. The angle of the frame to the horizon is \alpha, its mass is m, and the length of the side is a. Find the current in the frame.</title> | |||
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| <header style="text-align:center;"> | |||
| <a href="../../" style="text-decoration: none;"> | |||
| <div id="logo"> | |||
| <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> | |||
| $9.1.6.$ A square frame with current is fixed so that it can rotate freely around its horizontal side. The frame is in a vertical uniform magnetic field of induction $B$. The angle of the frame to the horizon is $\alpha$, its mass is $m$, and the length of the side is $a$. Find the current in the frame. | |||
| </p> | |||
| <center> | |||
| <figure> | |||
| <img src="https://savchenkosolutions.com/9/9.1.6/statement.png" | |||
| loading="lazy" width="200" /> | |||
| <figcaption> | |||
| For problem $9.1.6$ | |||
| </figcaption> | |||
| </figure> | |||
| </center> | |||
| <p> | |||
| </p> | |||
| <h3>Solution</h3> | |||
| <p> | |||
| The moment of forces acting on the frame from the magnetic field caused by the Ampere force | |||
| $$ M_B = Fa = Ba^2I\sin\alpha $$ | |||
| Moment of gravity: | |||
| $$ M_{mg} = mg\cdot \frac{a}{2}\cos\alpha $$ | |||
| In the equilibrium condition, the sum of the moments of external forces, taking into account the direction, is equal to zero $(M_B = M_{mg})$: | |||
| $$ B \sin\alpha~\cdot aI = \frac{1}{2}mg\cdot \cos\alpha $$ | |||
| Where do we find the current in the frame? | |||
| $$ \boxed{I = \frac{mg\cdot \cot\alpha}{2Ba}} $$ | |||
| </p> | |||
| <h4>Answer</h4> | |||
| <p> | |||
| $$I = \frac{mg\cdot \cot\alpha}{2Ba} $$ | |||
| </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> | |||
| </p> | |||
| </footer> | |||
| </body> | |||
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