An element $\vec{Δl} = 0.5\hat j\, cm$ is placed at the origin and carries a large current $I = 10 A$. The magnetic filed at point (50 cm, 0) is

$2 × 10^{-8} (-\hat k) T$

The correct answer is Option (4) → $2 × 10^{-8} (-\hat k) T$

Given:

Current element: $\vec{dl} = 0.5 \hat{j} \, \text{cm} = 0.005 \hat{j} \, \text{m}$

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

Point P: $\vec{r} = 50 \, \text{cm} \, \hat{i} = 0.5 \, \hat{i} \, \text{m}$

Using Biot–Savart law:

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

Distance: $r = |\vec{r}| = 0.5 \, \text{m}$

Unit vector: $\hat{r} = \frac{\vec{r}}{r} = \hat{i}$

Cross product: $\vec{dl} \times \hat{r} = (0.005 \hat{j}) \times \hat{i} = 0.005 (\hat{j} \times \hat{i}) = -0.005 \hat{k}$

Magnitude of magnetic field:

$dB = \frac{\mu_0}{4 \pi} \frac{I |\vec{dl} \times \hat{r}|}{r^2} = 10^{-7} \frac{10 \cdot 0.005}{0.5^2}$

$dB = 10^{-7} \frac{0.05}{0.25} = 10^{-7} \cdot 0.2 = 2 \times 10^{-8} \, \text{T}$

Direction: $-\hat{k}$ (into the page)

Answer: $\vec{B} = 2 \times 10^{-8} \, -\hat{k} \, \text{T}$ (into the page)