Translated 9.1.6-9.1.15
en/9.1.14.md
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| + | <meta property="og:title" content="A ring of radius R, through which a current I circulates, is placed in a non-uniform axially symmetric field. The axis of the ring coincides with the axis of symmetry of the magnetic field. The induction of the magnetic field B acting on the current is directed at an angle \alpha to the axis of symmetry of the field. The mass of the ring is m. Determine the acceleration of the ring."> | ||
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| + | <title>A ring of radius R, through which a current I circulates, is placed in a non-uniform axially symmetric field. The axis of the ring coincides with the axis of symmetry of the magnetic field. The induction of the magnetic field B acting on the current is directed at an angle \alpha to the axis of symmetry of the field. The mass of the ring is m. Determine the acceleration of the ring.</title> | ||
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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> | ||
| + | $9.1.14.$ A ring of radius $R$, through which a current $I$ circulates, is placed in a non-uniform axially symmetric field. The axis of the ring coincides with the axis of symmetry of the magnetic field. The induction of the magnetic field $B$ acting on the current is directed at an angle $\alpha$ to the axis of symmetry of the field. The mass of the ring is $m$. Determine the acceleration of the ring. | ||
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| + | For problem $9.1.14$ | ||
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| + | <p> | ||
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| + | <h3>Solution</h3> | ||
| + | <p> | ||
| + | The ring will be accelerated by the tangential component of magnetic induction $B_{\|}$ since its perpendicular component $B_\bot$ is directed at the deformation of the ring. </p><p> | ||
| + | At each point of the ring this is $B\sin\alpha$ since the acceleration does not change direction when turning $dl$ then: | ||
| + | $$ ma = 2\pi RIB\sin\alpha\Rightarrow \boxed{a = 2\pi\frac{RIB\sin\alpha}{m}} $$ | ||
| + | |||
| + | </p> | ||
| + | |||
| + | <h4>Answer</h4> | ||
| + | <p> | ||
| + | $$a = 2\pi\frac{RIB\sin\alpha}{m} $$ | ||
| + | </p> | ||
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| <meta name="author" content="Aliaksandr Melnichenka"> | |||
| <meta name="date" content="2023-10" scheme="YYYY-MM"> | |||
| <meta property="og:title" content="A ring of radius R, through which a current I circulates, is placed in a non-uniform axially symmetric field. The axis of the ring coincides with the axis of symmetry of the magnetic field. The induction of the magnetic field B acting on the current is directed at an angle \alpha to the axis of symmetry of the field. The mass of the ring is m. Determine the acceleration of the ring."> | |||
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| <meta name="yandex-verification" content="6cfda41f74038368"> | |||
| <title>A ring of radius R, through which a current I circulates, is placed in a non-uniform axially symmetric field. The axis of the ring coincides with the axis of symmetry of the magnetic field. The induction of the magnetic field B acting on the current is directed at an angle \alpha to the axis of symmetry of the field. The mass of the ring is m. Determine the acceleration of the ring.</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> | |||
| $9.1.14.$ A ring of radius $R$, through which a current $I$ circulates, is placed in a non-uniform axially symmetric field. The axis of the ring coincides with the axis of symmetry of the magnetic field. The induction of the magnetic field $B$ acting on the current is directed at an angle $\alpha$ to the axis of symmetry of the field. The mass of the ring is $m$. Determine the acceleration of the ring. | |||
| </p> | |||
| <center> | |||
| <figure> | |||
| <img src="https://savchenkosolutions.com/9/9.1.14/statement.png" | |||
| loading="lazy" width="230" /> | |||
| <figcaption> | |||
| For problem $9.1.14$ | |||
| </figcaption> | |||
| </figure> | |||
| </center> | |||
| <p> | |||
| </p> | |||
| <h3>Solution</h3> | |||
| <p> | |||
| The ring will be accelerated by the tangential component of magnetic induction $B_{\|}$ since its perpendicular component $B_\bot$ is directed at the deformation of the ring. </p><p> | |||
| At each point of the ring this is $B\sin\alpha$ since the acceleration does not change direction when turning $dl$ then: | |||
| $$ ma = 2\pi RIB\sin\alpha\Rightarrow \boxed{a = 2\pi\frac{RIB\sin\alpha}{m}} $$ | |||
| </p> | |||
| <h4>Answer</h4> | |||
| <p> | |||
| $$a = 2\pi\frac{RIB\sin\alpha}{m} $$ | |||
| </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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