A wire lying in one plane consists of two long, straight, parallel sections connected by a semicircle. Current I flows through the wire. Determine the magnetic field induction at the center of the semicircle.

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

    <h3> Statement </h3>
    <p>
        $9.2.12.$ A wire lying in one plane consists of two long, straight, parallel sections connected by a semicircle. Current $I$ flows through the wire. Determine the magnetic field induction at the center of the semicircle.
    </p>

    <h3>Solution</h3>
    <p>
        The magnetic field at the desired point will be perpendicular to the plane of the wire, which means that to find the induction, we need to find the induction $\vec{B}_1$ of a semi-infinite wire and a semicircle $\vec{B}_2$

$$B_1 = \int\limits_{0}^{\infty}\varepsilon_0 \mu_0 \left[\frac{\vec{(dx\lambda )}}{4\pi\varepsilon_0 (x^2 + R^2)} \times \vec{v} \right]$$

$$B_1 = \int\limits_{0}^{\infty}\frac{Idx\mu_0}{4\pi (x^2+ R^2)}\cdot \sin\alpha\quad(1)$$
For geometric reasons
$$\sin\alpha= \frac{R}{\sqrt{x^2 + R^2}}\quad(2)$$
Next, we substitute $(2)$ into $(1)$
$$B_1 = \int\limits_0^\infty \frac{\mu_0 I dxR}{4\pi{(x^2 + R^2)}^{3/2}} = \frac{\mu_0 IR}{4\pi}\int\limits_0^\infty \frac{dx}{(x^2+R^2)^{3/2}}$$
We introduce the relative value $u=\frac{x}{R}$:
$$B_1 = \frac{\mu_0 I}{4\pi R}\int\limits_0^\infty \frac{du}{(u^2+1)^{3/2}}\quad(3)$$
We will return to this integral a little later, but for now we will denote its value as $(3)$
Now we will calculate the value of $B_2$
$$B_2 = \int\limits_0^\pi\frac{\mu_0 IR}{4\pi R^2}~d\alpha = \frac{\mu_0 I}{4\pi R}\int\limits_0^\pi d\alpha$$

$$\boxed{B_2= \frac{\mu_0 I}{4R}}\quad(4)$$
Then, given that there are two wires, the total magnetic induction
$$\boxed{B = 2B_1 + B_2}\quad(5)$$
Let's return to our $(3)$ integral 😋
$$A=\int\limits_0^\infty \frac{du}{(u^2+1)^{3/2}}$$
We make a trigonometric substitution
$$u = \tan \alpha;\quad du = \frac{d\alpha}{\cos^2 \alpha}$$

$$\int\limits_0^\infty \frac{du}{(u^2+1)^{3/2}} = \int\limits_0^\frac{\pi}{2} \frac{d\alpha}{\cos^2 \alpha \cdot (\tan^2\alpha +1)^{3/2}}$$

$$\boxed{A = \int\limits_0^\frac{\pi}{2} \cos\alpha ~d\alpha = 1}$$
Since the value of $(3)$ is $1$, then
$$\boxed{B_1 = \frac{\mu_0 I}{4\pi R}}\quad(6)$$
We substitute the expressions for components $B_1$ and $B_2$ from $(6)$ and $(5)$, respectively, into the expression for the total magnetic induction $(5)$
$$B = \frac{\mu_0 I}{4R}\left(1+\frac{2}{\pi}\right) \Rightarrow \boxed{B = \frac{\mu_0 I}{4\pi R}(\pi + 2)}$$

    </p>

    <h4>Answer</h4>
    <p>
        $$B = \frac{\mu_0 I}{4\pi R}(\pi + 2) $$
    </p>
	<p style="text-align: right; font-style: italic; font-size: 14;">
      Andrei Yersh<br>
    </p>


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