$9.2.12.$ 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.
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 $(5)$ $$ B = \frac{\mu_0 I}{4R}\left(1+\frac{2}{\pi}\right) \Rightarrow \boxed{B = \frac{\mu_0 I}{4\pi R}(\pi + 2)} $$
$$B = \frac{\mu_0 I}{4\pi R}(\pi + 2) $$