The position vector for the field point is:
r ⃗ = 0.5 j ^ \vec{r} = 0.5\hat{j} r = 0.5 j ^
The magnetic field produced by the current element Δ l \Delta l Δ l is given by the Biot-Savart law:
d B = μ 0 I ( Δ l ⃗ × r ⃗ ) 4 π r 3 dB = \frac{\mu_0 I (\Delta \vec{l} \times \vec{r})}{4\pi r^3} d B = 4 π r 3 μ 0 I ( Δ l × r )
Here,
Δ l ⃗ = Δ x i ^ = 1 100 i ^ m , r ⃗ = 0.5 j ^ m , r = ∣ r ⃗ ∣ = 0.5 m \Delta \vec{l} = \Delta x \hat{i} = \frac{1}{100}\hat{i} \, \text{m}, \quad \vec{r} = 0.5\hat{j} \, \text{m}, \quad r = |\vec{r}| = 0.5 \, \text{m} Δ l = Δ x i ^ = 100 1 i ^ m , r = 0.5 j ^ m , r = ∣ r ∣ = 0.5 m
Δ l ⃗ × r ⃗ = ∣ i ^ j ^ k ^ 1 100 0 0 0 0.5 0 ∣ = 1 100 × 0.5 k ^ = 1 200 k ^ \Delta \vec{l} \times \vec{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{1}{100} & 0 & 0 \\ 0 & 0.5 & 0 \end{vmatrix} = \frac{1}{100} \times 0.5 \hat{k} = \frac{1}{200} \hat{k} Δ l × r = i ^ 100 1 0 j ^ 0 0.5 k ^ 0 0 = 100 1 × 0.5 k ^ = 200 1 k ^
Substituting values into the Biot-Savart law:
d B = 1 0 − 7 × 10 × 1 200 ( 0.5 ) 3 T dB = \frac{10^{-7} \times 10 \times \frac{1}{200}}{(0.5)^3} \, \text{T} d B = ( 0.5 ) 3 1 0 − 7 × 10 × 200 1 T
d B = 1 0 − 7 × 10 × 1 200 1 8 = 4 × 1 0 − 8 T dB = \frac{10^{-7} \times 10 \times \frac{1}{200}}{\frac{1}{8}} = 4 \times 10^{-8} \, \text{T} d B = 8 1 1 0 − 7 × 10 × 200 1 = 4 × 1 0 − 8 T
Hence, the magnetic field is:
d B = 4 × 1 0 − 8 k ^ T dB = 4 \times 10^{-8} \hat{k} \, \text{T} d B = 4 × 1 0 − 8 k ^ T