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

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 from element to point: \[ \mathbf{r} = \hat{i} + \hat{j}, \quad r = \sqrt{1^2 + 1^2} = \sqrt{2} \]
Step 2: Compute the cross product \( d\mathbf{l} \times \mathbf{r} \). \[ d\mathbf{l} \times \mathbf{r} = (10^{-2} \hat{i}) \times (\hat{i} + \hat{j}) = 10^{-2} (\hat{i} \times \hat{i} + \hat{i} \times \hat{j}) = 10^{-2} \hat{k} \]
Step 3: Apply the Biot–Savart law. \[ d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I (d\mathbf{l} \times \mathbf{r})}{r^3} = \frac{10^{-7} \cdot 10 \cdot 10^{-2}}{(\sqrt{2})^3} \hat{k} \] \[ (\sqrt{2})^3 = 2\sqrt{2}, \quad 10^{-7} \cdot 10 \cdot 10^{-2} = 10^{-8} \] \[ d\mathbf{B} = \frac{10^{-8}}{2\sqrt{2}} \hat{k} = \frac{10^{-8}}{2.828} \hat{k} \approx 3.5 \times 10^{-9} \, \hat{k} \, \text{T} \]
Final Answers:
Vector form (Biot–Savart law): \[ d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I (d\mathbf{l} \times \mathbf{r})}{r^3} \] Magnetic field at \( (1,1,0) \): \[ \mathbf{B} \approx 3.5 \times 10^{-9} \, \hat{k} \, \text{T} \]

Answer 2

Step 1: Use the unit-vector form of the Biot–Savart law.
An equivalent way to write the law uses the unit vector along \( \mathbf{r} \) instead of \( \mathbf{r} \) itself: \[ d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I \left(d\mathbf{l} \times \hat{\mathbf{r}}\right)}{r^{2}} \]

Step 2: Compute \( \hat{\mathbf{r}} \).
With \( \mathbf{r} = \hat{i} + \hat{j} \) and \( r = \sqrt{2} \): \[ \hat{\mathbf{r}} = \frac{\hat{i} + \hat{j}}{\sqrt{2}} \]

Step 3: Evaluate the cross product.\[ d\mathbf{l} \times \hat{\mathbf{r}} = \left(10^{-2}\hat{i}\right) \times \left(\frac{\hat{i} + \hat{j}}{\sqrt{2}}\right) = \frac{10^{-2}}{\sqrt{2}} \left(\hat{i} \times \hat{i} + \hat{i} \times \hat{j}\right) = \frac{10^{-2}}{\sqrt{2}} \hat{k} \]

Step 4: Substitute into the law.\[ d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I}{r^{2}} \cdot \frac{10^{-2}}{\sqrt{2}} \hat{k} = 10^{-7} \times \frac{10}{2} \times \frac{10^{-2}}{\sqrt{2}} \hat{k} \]

Step 5: Simplify.\[ d\mathbf{B} = \frac{5 \times 10^{-7} \times 10^{-2}}{\sqrt{2}} \hat{k} = \frac{5 \times 10^{-9}}{1.414} \hat{k} \approx 3.5 \times 10^{-9} \, \hat{k} \, \text{T} \]

Final Answer:\[ \mathbf{B} \approx 3.5 \times 10^{-9} \, \hat{k} \, \text{T} \]