Translated 9.2.10-9.2.17

astrosander edited Oct 17, 2024, 7:43 PM

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		+	<header style="text-align:center;">
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		+	<div id="logo">
		+	<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>
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		+	<h3 id="back-link"><a href="../../#9.2">$\leftarrow$Back</a></h3>
		+
		+	<h3> Statement </h3>
		+	<p>
		+	$9.2.14^*.$ A straight wire has a turn of radius $R$. Current $I$ flows through the wire. Determine the magnetic field induction at the center of the turn and on its axis at a distance $h$ from its center.
		+	</p>
		+	<center>
		+	<figure>
		+	<img src="https://savchenkosolutions.com/9/9.2.14/statement.png"
		+	loading="lazy" width="230" />
		+	<figcaption>
		+	For problem $9.2.14^*$
		+	</figcaption>
		+	</figure>
		+	</center>
		+	<p>
		+	</p>
		+
		+	<h3>Solution</h3>
		+	<p>
		+	Let's split the system into a separate wire and a ring with current with the corresponding fields $B_1 $ and $B_2$</p><p>
		+	From the school course it is known that the field of an infinite wire
		+	$$ B_1 = \frac{\mu_0I}{2\pi l} = \frac{\mu_0 I}{2\sqrt{R^2 + h^2}} $$
		+	We found the ring field in the problem <a href="../9.2.10">9.2.10</a>:
		+	$$ B_2 = \frac{\mu_0 I R^2}{2 (R^2 + h^2)^{3/2}} $$
		+	It is worth noting that $\vec{B_1}$ and $\vec{B_2}$ have different directions, the vector $\vec{B_1}$ is perpendicular to the plane of the point and the wire, and the vector $\vec{B_2}$ is directed along the axis. It is easy to show that the cosine of the angle between these vectors
		+	$$ \cos\gamma = \frac{R}{\sqrt{R^2+h^2}} $$
		+	тогда
		+	<div class="scroll-wrapper">
		+	$$ \|\vec B\| = \|\vec B_1 + \vec B_2\| = \sqrt{{B_1}^2+{B_2}^2+2B_1B_2\cos\gamma} $$
		+	</div>
		+	<div class="scroll-wrapper">
		+	$$ \boxed{B=\frac{\mu_0I}{2}\left[\frac{1}{\pi^2(R^2+h^2)} + \frac{R^4}{(R^2+h^2)^3} + \frac{2R^3}{\pi (R^2+h^2)^{5/2}}\right]^{1/2}}\quad(1) $$
		+	</div>
		+	When $h = 0$ we substitute $(1)$ into the expression
		+	<div class="scroll-wrapper">
		+	$$ B(0) = \frac{\mu_0I}{2R}\sqrt{\frac{1}{\pi^2} + 1 + \frac{2}{\pi}} \Rightarrow \boxed{B(0) = \frac{\mu_0I}{2\pi R}(1 + \pi)} $$
		+	</div>
		+	</p>
		+
		+	<h4>Answer</h4>
		+	<p>
		+	<div class="scroll-wrapper">
		+	$$B(h)=\frac{\mu_0I}{2}\sqrt{\frac{1}{\pi^2(R^2+h^2)} + \frac{R^4}{(R^2+h^2)^3} + \frac{2R^3}{\pi (R^2+h^2)^{5/2}}}$$
		+	</div>
		+	$$B(0) = \frac{\mu_0I}{2\pi R}( \pi + 1)$$
		+	</p>
		+	<p style="text-align: right; font-style: italic; font-size: 14;">
		+	Andrei Yersh<br>
		+	</p>
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			<meta name="description" content="A straight wire has a turn of radius R. Current I flows through the wire. Determine the magnetic field induction at the center of the turn and on its axis at a distance h from its center.">
			<meta name="author" content="Aliaksandr Melnichenka">
			<meta name="date" content="2023-10" scheme="YYYY-MM">
			<meta property="og:title" content="A straight wire has a turn of radius R. Current I flows through the wire. Determine the magnetic field induction at the center of the turn and on its axis at a distance h from its center.">
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			</p>
			</header>

			<h3 id="back-link"><a href="../../#9.2">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>
			<p>
			$9.2.14^*.$ A straight wire has a turn of radius $R$. Current $I$ flows through the wire. Determine the magnetic field induction at the center of the turn and on its axis at a distance $h$ from its center.
			</p>
			<center>
			<figure>
			<img src="https://savchenkosolutions.com/9/9.2.14/statement.png"
			loading="lazy" width="230" />
			<figcaption>
			For problem $9.2.14^*$
			</figcaption>
			</figure>
			</center>
			<p>
			</p>

			<h3>Solution</h3>
			<p>
			Let's split the system into a separate wire and a ring with current with the corresponding fields $B_1 $ and $B_2$</p><p>
			From the school course it is known that the field of an infinite wire
			$$ B_1 = \frac{\mu_0I}{2\pi l} = \frac{\mu_0 I}{2\sqrt{R^2 + h^2}} $$
			We found the ring field in the problem <a href="../9.2.10">9.2.10</a>:
			$$ B_2 = \frac{\mu_0 I R^2}{2 (R^2 + h^2)^{3/2}} $$
			It is worth noting that $\vec{B_1}$ and $\vec{B_2}$ have different directions, the vector $\vec{B_1}$ is perpendicular to the plane of the point and the wire, and the vector $\vec{B_2}$ is directed along the axis. It is easy to show that the cosine of the angle between these vectors
			$$ \cos\gamma = \frac{R}{\sqrt{R^2+h^2}} $$
			тогда
			<div class="scroll-wrapper">
			$$ \|\vec B\| = \|\vec B_1 + \vec B_2\| = \sqrt{{B_1}^2+{B_2}^2+2B_1B_2\cos\gamma} $$
			</div>
			<div class="scroll-wrapper">
			$$ \boxed{B=\frac{\mu_0I}{2}\left[\frac{1}{\pi^2(R^2+h^2)} + \frac{R^4}{(R^2+h^2)^3} + \frac{2R^3}{\pi (R^2+h^2)^{5/2}}\right]^{1/2}}\quad(1) $$
			</div>
			When $h = 0$ we substitute $(1)$ into the expression
			<div class="scroll-wrapper">
			$$ B(0) = \frac{\mu_0I}{2R}\sqrt{\frac{1}{\pi^2} + 1 + \frac{2}{\pi}} \Rightarrow \boxed{B(0) = \frac{\mu_0I}{2\pi R}(1 + \pi)} $$
			</div>
			</p>

			<h4>Answer</h4>
			<p>
			<div class="scroll-wrapper">
			$$B(h)=\frac{\mu_0I}{2}\sqrt{\frac{1}{\pi^2(R^2+h^2)} + \frac{R^4}{(R^2+h^2)^3} + \frac{2R^3}{\pi (R^2+h^2)^{5/2}}}$$
			</div>
			$$B(0) = \frac{\mu_0I}{2\pi R}( \pi + 1)$$
			</p>
			<p style="text-align: right; font-style: italic; font-size: 14;">
			Andrei Yersh<br>
			</p>


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