A current element Δl = Δx î (Δx = 1 cm) at the origin carries a current I = 10 A. The magnetic field at a point 0.5 m away on the y-axis has magnitude:

1. Biot–Savart magnitude: $|d\mathbf{B}| = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}$. 2. The element is along x and the point is on the y-axis, so $\theta = 90^\circ$, $\sin\theta = 1$. 3. Substitute $\dfrac{\mu_0}{4\pi}=10^{-7}$, $I=10$, $dl=10^{-2}$, $r=0.5$, $r^2=0.25$. 4. $|d\mathbf{B}| = \dfrac{10^{-7}\times10\times10^{-2}}{0.25} = \dfrac{10^{-8}}{0.25}$. 5. $|d\mathbf{B}| = 4\times10^{-8}$ T. _Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.144_