Step 1 (Formula / Governing Law & Given Data): [1 Mark]

The magnetic field $\vec{dB}$ due to a current element $I\vec{dl}$ at a position vector $\vec{r}$ is given by the Biot-Savart Law:

$$d\vec{B} = \frac{\mu_0}{4\pi} \frac{I(d\vec{l} \times \vec{r})}{r^3}$$

Given Data:

• Current $I = 10\text{ A}$
• Length element $dl = 1\text{ cm} = 0.01\text{ m}$
• Direction of current: $\hat{i}$ (positive x-direction), so $d\vec{l} = 0.01\hat{i}\text{ m}$
• Position vector $\vec{r} = (1\hat{i} + 1\hat{j})\text{ m}$
• Magnitude $r = \sqrt{1^2 + 1^2} = \sqrt{2}\text{ m}$
• Constant $\frac{\mu_0}{4\pi} = 10^{-7}\text{ T}\cdot\text{m/A}$

Step 2 (Step-by-Step Mathematical Working): [1.5 Marks]

First, calculate the cross product $d\vec{l} \times \vec{r}$:

$$d\vec{l} \times \vec{r} = (0.01\hat{i}) \times (1\hat{i} + 1\hat{j})$$ $$d\vec{l} \times \vec{r} = 0.01(\hat{i} \times \hat{i}) + 0.01(\hat{i} \times \hat{j})$$

Since $\hat{i} \times \hat{i} = 0$ and $\hat{i} \times \hat{j} = \hat{k}$:

$$d\vec{l} \times \vec{r} = 0.01\hat{k}\text{ m}^2$$

Now, substitute the values into the Biot-Savart formula:

$$\vec{B} = 10^{-7} \times \frac{10 \times 0.01\hat{k}}{(\sqrt{2})^3}$$ $$\vec{B} = 10^{-7} \times \frac{0.1\hat{k}}{2\sqrt{2}}$$ $$\vec{B} = \frac{10^{-8}}{2.828}\hat{k}$$

Step 3 (Final Concluded Answer): [0.5 Mark]

Calculating the final value:

$$\vec{B} \approx 3.536 \times 10^{-9}\hat{k}\text{ T}$$

The magnetic field at the point $(1, 1, 0)$ is $3.54 \times 10^{-9}\hat{k}\text{ Tesla}$.

💡 EL10 Concept Insight:

Think of the Biot-Savart Law like a "magnetic compass" rule: the current creates a swirling magnetic field around it, and the cross-product tells us that the field always points in a direction perpendicular to both the wire and the path to the point we are measuring.