Translated 9.1.6-9.1.15
en/9.1.12.md
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| + | <meta property="og:description" content="A coil of radius R was bent along its diameter at a right angle and placed in a uniform magnetic field of induction B so that one of the planes of the coil was located at an angle \alpha, the other at an angle \frac{\pi}{2}-\alpha to the direction of induction B. The current in the coil is I. Determine the moment of forces acting on the coil."> | ||
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| + | <title>A coil of radius R was bent along its diameter at a right angle and placed in a uniform magnetic field of induction B so that one of the planes of the coil was located at an angle \alpha, the other at an angle \frac{\pi}{2}-\alpha to the direction of induction B. The current in the coil is I. Determine the moment of forces acting on the coil.</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> | ||
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| + | <h3 id="back-link"><a href="../../#9.1">$\leftarrow$Back</a></h3> | ||
| + | |||
| + | <h3> Statement </h3> | ||
| + | <p> | ||
| + | $9.1.12.$ A coil of radius $R$ was bent along its diameter at a right angle and placed in a uniform magnetic field of induction B so that one of the planes of the coil was located at an angle $\alpha$, the other at an angle $\frac{\pi}{2}-\alpha$ to the direction of induction $B$. The current in the coil is $I$. Determine the moment of forces acting on the coil. | ||
| + | </p> | ||
| + | <center> | ||
| + | <figure> | ||
| + | <img src="https://savchenkosolutions.com/9/9.1.12/statement.png" | ||
| + | loading="lazy" width="180" /> | ||
| + | <figcaption> | ||
| + | For problem $9.1.12$ | ||
| + | </figcaption> | ||
| + | </figure> | ||
| + | </center> | ||
| + | <p> | ||
| + | </p> | ||
| + | |||
| + | <h3>Solution</h3> | ||
| + | <p> | ||
| + | Let's split the circuit into two circuits with the same current, but in different planes.</p><p> | ||
| + | Let's find the moment of forces for each of them: | ||
| + | $$ M_1 = \frac{1}{2}\pi R^2IB\cdot \sin\alpha $$ | ||
| + | $$ M_2 = \frac{1}{2}\pi R^2IB\cdot \sin\left(\frac{\pi}{2} -\alpha\right) = \frac{1}{2}\pi R^2IB\cdot \cos\alpha $$ | ||
| + | Total moment of forces | ||
| + | $$ \boxed{M = M_1 + M_2 = \frac{1}{2}\pi R^2IB\cdot (\sin\alpha + \cos\alpha )} $$ | ||
| + | |||
| + | </p> | ||
| + | |||
| + | <h4>Answer</h4> | ||
| + | <p> | ||
| + | $$M = \frac{1}{2}\pi R^2IB\cdot (\sin\alpha + \cos\alpha )$$ | ||
| + | </p> | ||
| + | <p style="text-align: right; font-style: italic; font-size: 14;"> | ||
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| <meta name="description" content="A coil of radius R was bent along its diameter at a right angle and placed in a uniform magnetic field of induction B so that one of the planes of the coil was located at an angle \alpha, the other at an angle \frac{\pi}{2}-\alpha to the direction of induction B. The current in the coil is I. Determine the moment of forces acting on the coil."> | |||
| <meta name="author" content="Aliaksandr Melnichenka"> | |||
| <meta name="date" content="2023-10" scheme="YYYY-MM"> | |||
| <meta property="og:title" content="A coil of radius R was bent along its diameter at a right angle and placed in a uniform magnetic field of induction B so that one of the planes of the coil was located at an angle \alpha, the other at an angle \frac{\pi}{2}-\alpha to the direction of induction B. The current in the coil is I. Determine the moment of forces acting on the coil."> | |||
| <meta property="og:image" content="img/logo.png"> | |||
| <meta property="og:description" content="A coil of radius R was bent along its diameter at a right angle and placed in a uniform magnetic field of induction B so that one of the planes of the coil was located at an angle \alpha, the other at an angle \frac{\pi}{2}-\alpha to the direction of induction B. The current in the coil is I. Determine the moment of forces acting on the coil."> | |||
| <meta name="yandex-verification" content="6cfda41f74038368"> | |||
| <title>A coil of radius R was bent along its diameter at a right angle and placed in a uniform magnetic field of induction B so that one of the planes of the coil was located at an angle \alpha, the other at an angle \frac{\pi}{2}-\alpha to the direction of induction B. The current in the coil is I. Determine the moment of forces acting on the coil.</title> | |||
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| </head> | |||
| <body style=""> | |||
| <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> | |||
| $9.1.12.$ A coil of radius $R$ was bent along its diameter at a right angle and placed in a uniform magnetic field of induction B so that one of the planes of the coil was located at an angle $\alpha$, the other at an angle $\frac{\pi}{2}-\alpha$ to the direction of induction $B$. The current in the coil is $I$. Determine the moment of forces acting on the coil. | |||
| </p> | |||
| <center> | |||
| <figure> | |||
| <img src="https://savchenkosolutions.com/9/9.1.12/statement.png" | |||
| loading="lazy" width="180" /> | |||
| <figcaption> | |||
| For problem $9.1.12$ | |||
| </figcaption> | |||
| </figure> | |||
| </center> | |||
| <p> | |||
| </p> | |||
| <h3>Solution</h3> | |||
| <p> | |||
| Let's split the circuit into two circuits with the same current, but in different planes.</p><p> | |||
| Let's find the moment of forces for each of them: | |||
| $$ M_1 = \frac{1}{2}\pi R^2IB\cdot \sin\alpha $$ | |||
| $$ M_2 = \frac{1}{2}\pi R^2IB\cdot \sin\left(\frac{\pi}{2} -\alpha\right) = \frac{1}{2}\pi R^2IB\cdot \cos\alpha $$ | |||
| Total moment of forces | |||
| $$ \boxed{M = M_1 + M_2 = \frac{1}{2}\pi R^2IB\cdot (\sin\alpha + \cos\alpha )} $$ | |||
| </p> | |||
| <h4>Answer</h4> | |||
| <p> | |||
| $$M = \frac{1}{2}\pi R^2IB\cdot (\sin\alpha + \cos\alpha )$$ | |||
| </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> | |||
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| </html> | |||