Translated 9.2.10-9.2.17

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+ <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>
+ $$B = \frac{\mu_0 I}{4\pi R}(\pi + 2) $$
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