Magnetic Field from Current Strip

Let $\vec{l}=x\hat{i}+z\hat{k}$ be the parametrization of the metal strip, where $z$ goes from $-\infty$ to $\infty$ and $x$ goes from $0$ to $1$ . Let's find the magnetic field in a point $\vec{P}=d\hat{i}$ . By Biot-Savart Law we have: $\begin{aligned} \vec{B} &= \dfrac{\mu_0 I}{4\pi} \int_0^1 \int_{-\infty}^{\infty} \dfrac{\mathrm{d}\vec{l} \times \left(\vec{P}-\vec{l}\right)}{\left\lVert \vec{P}-\vec{l}\right\rVert^3} \mathrm{d}x\\ &= \dfrac{\mu_0 I}{4\pi} \int_0^1 \int_{-\infty}^{\infty} \dfrac{(d-x)\hat{j}}{\left\lVert (d-x)\hat{i}-z\hat{k} \right\rVert^3} \mathrm{d}z\mathrm{d}x\\ &= \dfrac{\mu_0 I}{4\pi} \int_0^1 \int_{-\infty}^{\infty} \dfrac{(d-x)\hat{j}}{((d-x)^2+z^2)^{3/2}}\mathrm{d}z\mathrm{d}x\\ &= \dfrac{\mu_0 I}{4\pi} \int_0^1 \dfrac{1}{d-x} \int_{-\pi/2}^{\pi/2} \cos \theta \hat{j}\mathrm{d}\theta\mathrm{d}x\\ &= \dfrac{\mu_0 I}{2\pi} \int_0^1 \dfrac{1}{d-x} \mathrm{d}x\\ &= \dfrac{\mu_0 I}{2\pi} (\ln(d)-\ln(d-1)) \end{aligned}$ Finally, substituting $d=2\text{ m}$ and $I=-1\text{ A}$ we have $\vec{B} = -2 \times 10^{-7} \ln(2)\hat{j}\text{ T}$ , and $\left\lVert \vec{B} \right\rVert \approx \boxed{138.629}\text{ nT}$ .