Question

Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

Hint:

For Biot–Savart problems:

Use vector form to avoid unit vector mistakes
Direction from cross product \( d\mathbf{l} \times \mathbf{r} \)
Always compute \( r^3 \) carefully

Answer 1

Concept: Magnetic field due to a current element is given by the Biot–Savart law: \[ d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I \, d\mathbf{l} \times \mathbf{\hat{r}}}{r^2} \] Vector form: \[ d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I \, (d\mathbf{l} \times \mathbf{r})}{r^3} \]
Step 1: Given data.

Length of segment: \( dl = 1 \, \text{cm} = 10^{-2} \, \text{m} \)
Current: \( I = 10 \, \text{A} \)
Segment along +x direction: \[ d\mathbf{l} = 10^{-2} \, \hat{i} \]
Field point: \( (1,1,0) \)
Position vector: \[ \mathbf{r} = \hat{i} + \hat{j} \] \[ r = \sqrt{1^2 + 1^2} = \sqrt{2} \]
Step 2: Compute cross product. \[ d\mathbf{l} \times \mathbf{r} = (10^{-2} \hat{i}) \times (\hat{i} + \hat{j}) \] Using cross products: \[ \hat{i} \times \hat{i} = 0, \quad \hat{i} \times \hat{j} = \hat{k} \] \[ d\mathbf{l} \times \mathbf{r} = 10^{-2} \hat{k} \]
Step 3: Apply Biot–Savart law. \[ d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I (10^{-2} \hat{k})}{(\sqrt{2})^3} \] \[ (\sqrt{2})^3 = 2\sqrt{2} \] \[ d\mathbf{B} = \frac{10^{-7} \times 10 \times 10^{-2}}{2\sqrt{2}} \hat{k} \]
Step 4: Simplify. \[ 10^{-7} \times 10 \times 10^{-2} = 10^{-8} \] \[ \mathbf{B} = \frac{10^{-8}}{2\sqrt{2}} \hat{k} = \frac{10^{-8}}{2.828} \hat{k} \] \[ \mathbf{B} \approx 3.5 \times 10^{-9} \, \hat{k} \, \text{T} \] Final Answers:

Vector expression (Biot–Savart law): \[ d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I (d\mathbf{l} \times \mathbf{r})}{r^3} \]
Magnetic field at \( (1,1,0) \): \[ \mathbf{B} \approx 3.5 \times 10^{-9} \, \hat{k} \, \text{T} \]