A 3.0 cm segment of a wire, centred at the origin (0, 0, 0) lies along Y-axis. It carries a current of 6.0 A in positive Y-direction. The magnetic field due to this segment at a point (3.0 m, 0, 0) is [$$(\frac{μ _o}{4π})=10^{-7}$$​ Tm/A, and i, j and k are unit vectors along X-axis, Y-axis and Z-axis, respectively]

Correct option is D

​$$\text{The Biot-Savart law gives the magnetic field } \vec{B} \text{ due to a small current element:} \\\vec{B} = \frac{\mu_0 i \, d\vec{l} \times \hat{r}}{4 \pi r^2}$$

$$\text{Where:} \\\frac{\mu_0}{4\pi} = 10^{-7} \ \text{Tm/A (constant)}, \\i = 6.0 \ \text{A}, \\d\vec{l} = 0.03 \hat{j} \ \text{m (the current element along the Y-axis)}, \\\hat{r} = \hat{i} \text{ (unit vector along the X-axis)}, \\r = 3.0 \ \text{m (distance from the origin to the point on the X-axis)}.$$

$$\text{The vector cross product } d\vec{l} \times \hat{r} \text{ gives us:} \\d\vec{l} = 0.03 \hat{j}, \hat{r} = \hat{i} \\d\vec{l} \times \hat{r} = 0.03 \hat{j} \times \hat{i} = 0.03 \hat{k}$$

$$\text{Now substitute the values into the Biot-Savart law:} \\\vec{B} = \frac{10^{-7} \times 6.0 \times (0.03 \hat{k})}{(3)^2}$$

$$\text{Simplifying:} \\\vec{B} = 10^{-7} \times 6.0 \times 0.03 \hat{k} \div 9 \\\vec{B} = 2 \times 10^{-9} \hat{k} \ \text{T}$$

$$\text{The magnetic field at point } (3, 0, 0) \text{ is:} \\-2.0 \times 10^{-9} \hat{k} \ \text{T}$$​​​​​​​