An element $\Delta l = \Delta x \hat{i}$ is placed at the origin and carries a large current $I = 10 \; A$ (Figure). What is the magnetic field on the $y$-axis at a distance of $0.5 \; m$? Given $\Delta x = 1 \; cm$.

(N/A) According to the Biot-Savart Law,the magnetic field $dB$ due to a current element $Idl$ is given by:
$|dB| = \frac{\mu_0}{4\pi} \frac{I dl \sin \theta}{r^2}$
Given values:
$dl = \Delta x = 1 \; cm = 10^{-2} \; m$
$I = 10 \; A$
$r = 0.5 \; m$
$\frac{\mu_0}{4\pi} = 10^{-7} \; T \cdot m/A$
Since the element is along the $x$-axis and the point is on the $y$-axis,the angle $\theta = 90^{\circ}$,so $\sin \theta = 1$.
Substituting the values:
$|dB| = \frac{10^{-7} \times 10 \times 10^{-2}}{(0.5)^2}$
$|dB| = \frac{10^{-8}}{0.25} = 4 \times 10^{-8} \; T$
The direction of the field is determined by the cross product $dl \times r$. Since $dl = \Delta x \hat{i}$ and the position vector $r = y \hat{j}$,the direction is $\hat{i} \times \hat{j} = \hat{k}$. Thus,the magnetic field is in the $+z$-direction.