A current element of length 0.8 cm carrying a current of 5 A towards +x- direction is placed symmetrically at the origin along the x-axis. The magnetic field at a point (0, 5 cm) is:

(i, j and k are unit vectors along the x-axis, y-axis and z-axis, respectively.)

Correct option is C

​$$\begin{aligned}&\textbf{ Given Data:} \\&dl = 0.8\,\text{cm} = 0.008\,\text{m} \quad (\text{along } +\hat{x}, \text{ so } \vec{dl} = 0.008\hat{i}\,\text{m}) \\&I = 5\,\text{A} \\&\text{Observation point: } (0, 5\,\text{cm}), \text{ so position vector } \vec{r} = 0.05\hat{j}\,\text{m} \\&\frac{\mu_0}{4\pi} = 10^{-7}\,\text{Tm/A} \\&\vec{B} = \frac{\mu_0}{4\pi} \cdot \frac{I \cdot \vec{dl} \times \vec{r}}{|\vec{r}|^3} \\\\&\vec{dl} \times \vec{r} = (0.008\hat{i}) \times (0.05\hat{j}) \\&\quad = 0.008 \times 0.05 (\hat{i} \times \hat{j}) \\&\quad = 0.0004\hat{k} \\&\vec{B} = 10^{-7} \times 5 \times \frac{0.0004\hat{k}}{(0.05)^3} \\&\text{Simplify denominator:} \\&(0.05)^3 = 1.25 \times 10^{-4}\end{aligned}$$​​