Translated 9.2.10-9.2.17

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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.">
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+ <meta name="date" content="2023-10" scheme="YYYY-MM">
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+ <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span>
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+ Solutions&nbsp;of&nbsp;Savchenko Problems&nbsp;in&nbsp;Physics <br>
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+ <h3 id="back-link"><a href="../../#9.2">$\leftarrow$Back</a></h3>
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+ <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>
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+ <figcaption>
+ For problem $9.2.14^*$
+ </figcaption>
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+<p>
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+
+ <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>
+$$B(0) = \frac{\mu_0I}{2\pi R}( \pi + 1)$$
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+ Andrei Yersh<br>
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