Translated 9.2.10-9.2.17

astrosander правка от 17 окт. 2024 г., 19:43

правка #11455 GitHub ef227c0

en/9.2.10.md +106 −0

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		+	<meta name="date" content="2023-10" scheme="YYYY-MM">
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		+	<title>A current I flows through a ring of radius R. Determine the magnetic field induction at the center of the ring and on its axis at a distance h from the center 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>
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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="../../#9.2">$\leftarrow$Back</a></h3>
		+
		+	<h3> Statement </h3>
		+	<p>
		+	$9.2.10.$ A current $I$ flows through a ring of radius $R$. Determine the magnetic field induction at the center of the ring and on its axis at a distance $h$ from the center of the ring.
		+	</p>
		+
		+	<h3>Solution</h3>
		+	<p>
		+
		+	</p>
		+	<center>
		+	<figure>
		+	<img src="https://savchenkosolutions.com/9/9.2.10/9.2.10_1.png"
		+	loading="lazy" width="230" />
		+	<figcaption>
		+	The field induction at point $A$, located on the axis of the ring at a distance $z$ from its center
		+	</figcaption>
		+	</figure>
		+	</center>
		+	<p>
		+
		+	The magnetic field $B$ will be directed perpendicular to the plane of the ring. Let's consider a section of the ring of length $Rd\theta$ for it the formula from <a href="../9.2.8">9.2.8</a> will be valid then
		+	$$ dB = \frac{\mu_0 I R}{4\pi (R^2 + h^2)}\cdot d\theta $$
		+	but since the field will be directed along the axis, its projection onto this axis is needed there
		+	$$ dB\cdot \cos\alpha= \frac{\mu_0 I Rd\theta}{4\pi (R^2 + h^2)} \cdot \frac{R}{\sqrt{R^2 + h^2}} $$
		+
		+	$$ dB_z = \frac{\mu_0 I R^2d\theta}{4\pi (R^2 + h^2)^{3/2}} $$
		+	When integrating $dB_z$ from $0$ to $2\pi$ we get:
		+	$$ B(h)=\int_0^{2\pi}dB_z=\int_0^{2\pi}\frac{\mu_0 I R^2}{4\pi (R^2 + h^2)^{3/2}}~d\theta $$
		+
		+	$$ B(h) = \frac{\mu_0 I R^2}{2 (R^2 + h^2)^{3/2}} $$
		+	substituting $h = 0$ we get the formula you know for the inductance in the center of the ring:
		+	$$ B(0) = \frac{\mu_0 I}{2R} $$
		+
		+	</p>
		+
		+	<h4>Answer</h4>
		+	<p>
		+	$$ B=\frac{\mu_0 qv}{4\pi r^2} \sin\alpha; \quad B = \frac{\mu_0 Il}{4\pi r^2}\sin\alpha$$
		+	</p>
		+	<p style="text-align: right; font-style: italic; font-size: 14;">
		+	Andrei Yersh<br>
		+	</p>
		+
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			<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
			<meta name="description" content="A current I flows through a ring of radius R. Determine the magnetic field induction at the center of the ring and on its axis at a distance h from the center of the ring.">
			<meta name="author" content="Aliaksandr Melnichenka">
			<meta name="date" content="2023-10" scheme="YYYY-MM">
			<meta property="og:title" content="A current I flows through a ring of radius R. Determine the magnetic field induction at the center of the ring and on its axis at a distance h from the center of the ring.">
			<meta property="og:image" content="img/logo.png">
			<meta property="og:description" content="A current I flows through a ring of radius R. Determine the magnetic field induction at the center of the ring and on its axis at a distance h from the center of the ring.">
			<meta name="yandex-verification" content="6cfda41f74038368">
			<title>A current I flows through a ring of radius R. Determine the magnetic field induction at the center of the ring and on its axis at a distance h from the center of the ring.</title>
			<link rel="stylesheet" href="https://savchenkosolutions.com/css/css-latex/style.css">
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			<span><img src="../../img/book.png"></span><span>Savchenko Solutions</span>
			</div>
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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="../../#9.2">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>
			<p>
			$9.2.10.$ A current $I$ flows through a ring of radius $R$. Determine the magnetic field induction at the center of the ring and on its axis at a distance $h$ from the center of the ring.
			</p>

			<h3>Solution</h3>
			<p>

			</p>
			<center>
			<figure>
			<img src="https://savchenkosolutions.com/9/9.2.10/9.2.10_1.png"
			loading="lazy" width="230" />
			<figcaption>
			The field induction at point $A$, located on the axis of the ring at a distance $z$ from its center
			</figcaption>
			</figure>
			</center>
			<p>

			The magnetic field $B$ will be directed perpendicular to the plane of the ring. Let's consider a section of the ring of length $Rd\theta$ for it the formula from <a href="../9.2.8">9.2.8</a> will be valid then
			$$ dB = \frac{\mu_0 I R}{4\pi (R^2 + h^2)}\cdot d\theta $$
			but since the field will be directed along the axis, its projection onto this axis is needed there
			$$ dB\cdot \cos\alpha= \frac{\mu_0 I Rd\theta}{4\pi (R^2 + h^2)} \cdot \frac{R}{\sqrt{R^2 + h^2}} $$

			$$ dB_z = \frac{\mu_0 I R^2d\theta}{4\pi (R^2 + h^2)^{3/2}} $$
			When integrating $dB_z$ from $0$ to $2\pi$ we get:
			$$ B(h)=\int_0^{2\pi}dB_z=\int_0^{2\pi}\frac{\mu_0 I R^2}{4\pi (R^2 + h^2)^{3/2}}~d\theta $$

			$$ B(h) = \frac{\mu_0 I R^2}{2 (R^2 + h^2)^{3/2}} $$
			substituting $h = 0$ we get the formula you know for the inductance in the center of the ring:
			$$ B(0) = \frac{\mu_0 I}{2R} $$

			</p>

			<h4>Answer</h4>
			<p>
			$$ B=\frac{\mu_0 qv}{4\pi r^2} \sin\alpha; \quad B = \frac{\mu_0 Il}{4\pi r^2}\sin\alpha$$
			</p>
			<p style="text-align: right; font-style: italic; font-size: 14;">
			Andrei Yersh<br>
			</p>


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