An element $d\vec l= dx\hat i$ (where $dx = 1$ cm) is placed at the origin and carries a large current $I = 10A$. What is the magnetic field on the y-axis at a distance of 0.5 m?

$4×10^{-8}\hat kT$

The correct answer is Option 2: $4×10^{-8}\hat kT$

The magnetic field $d\vec{B}$ at a point $\vec{r}$ due to a current element $Id\vec{l}$ is determined by the Biot-Savart Law:

$d\vec{B} = \frac{\mu_0}{4\pi} \frac{I(d\vec{l} \times \hat{r})}{r^2}$

For an element $d\vec{l} = dx\hat{i}$ at the origin and an observation point on the y-axis, the unit vector $\hat{r}$ is along $\hat{j}$. The cross product $\hat{i} \times \hat{j} = \hat{k}$ confirms the field points along the positive z-axis.

Given:

• $I = 10\text{ A}$

• $dx = 1\text{ cm} = 10^{-2}\text{ m}$

• $y = 0.5\text{ m}$

• $\frac{\mu_0}{4\pi} = 10^{-7}$

$B = \frac{10^{-7} \times 10 \times 10^{-2}}{(0.5)^2} = \frac{10^{-8}}{0.25} = 4 \times 10^{-8}\text{ T}$

The resulting magnetic field is: $\vec{B} = 4 \times 10^{-8}\hat{k}\text{ T}$