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 \( (1,1,0) \): \[ \mathbf{B} \approx 3.5 \times 10^{-9} \, \hat{k} \, \text{T} \]

Answer 2

Concept: Instead of computing the vector cross product component by component, the Biot–Savart law can be applied by separately finding the magnitude of \( d\mathbf{B} \) using the scalar (sine) form of the law, and its direction using the right-hand rule.

Step 1: Given data.\[ dl = 1 \, \text{cm} = 10^{-2} \, \text{m}, \quad I = 10 \, \text{A} \] The current element points along \( +\hat{i} \), and the field point is \( (1, 1, 0) \), so the position vector from the element to the point is \( \mathbf{r} = \hat{i} + \hat{j} \), with magnitude \[ r = \sqrt{1^{2} + 1^{2}} = \sqrt{2} \, \text{m} \]

Step 2: Find the angle between the current element and \( \mathbf{r} \).
Since \( \mathbf{r} \) has equal \( x \) and \( y \) components, it makes an angle of \( 45^{\circ} \) with the \( x \)-axis, which is the direction of \( d\mathbf{l} \). So \[ \theta = 45^{\circ}, \qquad \sin\theta = \frac{1}{\sqrt{2}} \]

Step 3: Apply the scalar (magnitude) form of the Biot–Savart law.\[ dB = \frac{\mu_0}{4\pi} \frac{I \, dl \sin\theta}{r^{2}} = \frac{10^{-7} \times 10 \times 10^{-2} \times \frac{1}{\sqrt{2}}}{2} \]

Step 4: Simplify.\[ 10^{-7} \times 10 \times 10^{-2} = 10^{-8} \] \[ dB = \frac{10^{-8}}{2\sqrt{2}} \approx 3.5 \times 10^{-9} \, \text{T} \]

Step 5: Find the direction using the right-hand rule.
Point the fingers of the right hand along the current direction \( (+\hat{i}) \) and curl them toward the position vector \( \mathbf{r} \) (which lies in the first quadrant of the \( xy \)-plane). The thumb then points out of the plane, along \( +\hat{k} \), which fixes the direction of \( d\mathbf{B} \).

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