Magnetism and Mechanics

Relevant wiki: Magnetic Effects of Current

Consider a particular position such that the wires are separated by a distance $2 x$ from each other as shown.

Now, consider a small portion of length $dy$ at a distance of $y$ from the midpoint of the wire of the right. Now, the force acting due to the left wire on this portion is given by $B I dy$ where $B$ is the magnetic field intensity produced by the left wire on that point. Applying the magnetic field intensity due to a finite wire = $\dfrac{\mu_{0} I}{4\pi d} \left(\sin{(A)} + \sin{(B)}\right)$

$\begin{aligned} dF & = \dfrac{\mu_{0}\times 4}{4 \pi (2x)} \left(\dfrac{(3-y)}{\sqrt{4x^2 + {(3-y)}^2}} + \dfrac{(3+y)}{\sqrt{4x^2 + {(3+y)}^2}}\right) \times 5 dy\\ \\ \int_{0}^{F} dF & = \int_{-3}^{3} \dfrac{\mu_{0}\times 4}{4 \pi (2x)} \left(\dfrac{(3-y)}{\sqrt{4x^2 + {(3-y)}^2}} + \dfrac{(3+y)}{\sqrt{4x^2 + {(3+y)}^2}} \right) \times 5 dy\\ \\ F & = \dfrac{5 \mu_{0}}{2 \pi x} \left[ \int_{0}^{6} \dfrac{t dt}{\sqrt{4x^2 + t^2}} \ + \ \int_{0}^{6} \dfrac{z dz}{\sqrt{4x^2 + z^2}} \right] \ {\color{#20A900}{\left(\text{Let 3 - y = t and 3 + y = z}\right)}} \\ \\ F & = \dfrac{5\mu_{0}}{2 \pi x} \left[ \int_{0}^{36} \dfrac{da}{2\sqrt{4x^2 + a}} + \int_{0}^{36} \dfrac{db}{2\sqrt{4x^2 + b}} \right] \ {\color{#20A900}{\left( \text{Let} \ t^2 \ \text{= a and} \ z^2 \ \text{= b} \right) }} \\ \\ F & = \dfrac{5 \mu_{0}}{\pi x} \left( \sqrt{4x^2 + 36} - 2x \right) \end{aligned}$

Now, this force provides the necessary acceleration for the motion of the wires. Since the complete setup is symmetric, the wires are always at equal distance from the mid-plane between them. Since $F = m v \dfrac{dv}{dx}$

$\begin{aligned} 2 v \dfrac{dv}{dx} & = \dfrac{5 \mu_{0}}{\pi x} \left( \sqrt{4x^2 + 36} - 2x \right)\\ \\ \int_{0}^{v} 2 v dv & = \int_{6}^{3} \left( \dfrac{5 \mu_{0}}{\pi x} \sqrt{4x^2 + 36} - \dfrac{10 \mu_{0}}{\pi} \right) dx \\ \\ v^2 & = \int_{6}^{3} \dfrac{5 \mu_{0}}{\pi x} \sqrt{4x^2 + 36} dx + \dfrac{30 \mu_{0}}{\pi}\\ \\ & = \int_{\arctan{(2)}}^{\frac{pi}{4}} \dfrac{5\mu_{0} \times 6{(\sec{\theta})}^{3} d\theta}{\pi \tan\theta} \ {\color{#20A900}{\text{Let x =} \ 3\tan\theta}} \\ \\ & = \dfrac{30 \mu_{0}}{\pi} \left[ \int_{\arctan{(2)}}^{\frac{\pi}{4}} \dfrac{d\theta}{\sin\theta {\cos}^2\theta} \ + \ 1 \right] \\ \\ & = \dfrac{30 \mu_{0}}{\pi} \left[ \int_{\arctan{(2)}}^{\frac{\pi}{4}} \dfrac{d\theta}{\sin\theta {\cos}^2\theta} \ + \ 1 \right] \\ \\ & = \dfrac{30 \mu_{0}}{\pi} \left[ \int_{\arctan{(2)}}^{\frac{\pi}{4}} \dfrac{\sin\theta d\theta}{(1 - {\cos}^2\theta){\cos}^2\theta} \ + \ 1 \right] \\ \\ & = \dfrac{30 \mu_{0}}{\pi} \left[1 - \int_{\frac{1}{\sqrt{5}}}^{\frac{1}{\sqrt{2}}} \dfrac{du}{u^2(1 - u^2)} \right] \ {\color{#20A900}{\left( \text{Let u =} \cos\theta \right)}} \\ \\ & = \dfrac{30 \mu_{0}}{\pi} \left[\dfrac{1}{u} - \dfrac{1}{2}\log\left(\dfrac{1+u}{1-u}\right) \right]_{\frac{1}{\sqrt{5}}}^{\frac{1}{\sqrt{2}}} + \dfrac{30 \mu_{0}}{\pi} \\ \\ \implies v & = \sqrt{12 \times {10}^{-6} \left[\sqrt{2} + 1 - \sqrt{5} + \log\left(\dfrac{\left(\sqrt{5} + 1\right) \left(\sqrt{2} - 1 \right)}{2}\right)\right]}\\ \\ \implies v & = {\color{#3D99F6}{\boxed{16.32}}} \times {10}^{-4} m/s \end{aligned}$