Current, $I = 10\,\text{A}$

Position for the magnetic field: $(0,\,0.5,\,0)$ (i.e., on Y-axis, $r = 0.5\,\text{m}$)

$\Delta x = 1\,\text{cm} = 0.01\,\text{m}$

$\dfrac{\mu_0}{4\pi} = 10^{-7}\,\text{SI units}$

$\sin 90^\circ = 1$

Step 1: Biot-Savart Law for a Small Element

The magnetic field at point $P$ due to a small current element is:

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

Step 2: Set Up the Cross Product

$d\vec{l} = \Delta x\,\hat{i}$
$\vec{r} = 0\,\hat{i} + 0.5\,\hat{j} + 0\,\hat{k}$

So,
$\hat{r} = \frac{\vec{r}}{|\vec{r}|} = \hat{j}$
Angle between $d\vec{l}$ and $\vec{r}$ is $90^\circ$ (since $d\vec{l}$ along $x$, $\vec{r}$ along $y$).

So,

$ d\vec{l} \times \hat{r} = \Delta x\,\hat{i} \times \hat{j} = \Delta x\,\hat{k} $

Step 3: Calculate the Magnitude of $dB$

$ dB = \frac{\mu_0}{4\pi} \frac{I\, \Delta x \sin 90^\circ}{r^2} $

$ dB = \frac{\mu_0}{4\pi}\frac{I\,\Delta x}{r^2} $

Substitute values:

$ dB = 10^{-7} \times \frac{10 \times 0.01}{(0.5)^2} $

$ dB = 10^{-7} \times \frac{0.1}{0.25} $

$ dB = 10^{-7} \times 0.4 $

$ dB = 4 \times 10^{-8}~\mathrm{T} $

Correct option:

$ \boxed{4 \times 10^{-8}~\mathrm{T}} $

So the answer is Option C.

Moving Charges And Magnetism

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