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.

Savchenko Solutions

Solutions of Savchenko Problems in Physics
knowledge must be free

    <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.

    <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 9.2.10:
$$B_2 = \frac{\mu_0 I R^2}{2 (R^2 + h^2)^{3/2}}$$
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

When $h = 0$ we substitute $(1)$ into the expression

    <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>

$$B(0) = \frac{\mu_0I}{2\pi R}( \pi + 1)$$

Andrei Yersh

<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>