Problem 10 - University Physics with Modern Physics

Question

A short current element \(\overrightarrow{dl}\) S = (0.500 mm)\(\hat{\imath}\) carries a current of 5.40 A in the same direction as \(\overrightarrow{dl}\). Point P is located at \(\overrightarrow{r} =\) (-0.730 m)\(\hat{\imath}\) + (0.390 m)\(\hat{k}\) . Use unit vectors to express the magnetic field at \(P\) produced by this current element.

Step-by-Step Solution

Answer

\(\overrightarrow{B} = 1.843 \times 10^{-9} \hat{\jmath} \, \text{T}\)

Step 1: Understand the problem

We need to calculate the magnetic field at point P due to a current-carrying wire segment using the Biot-Savart Law. The current element is given by \(\overrightarrow{dl} = 0.500 \times 10^{-3} \hat{\imath} \, \text{m}\) with a current \(I = 5.40 \, \text{A}\), and we find \(\overrightarrow{r}\) at point P which is \(-0.730 \hat{\imath} + 0.390 \hat{k} \, \text{m}\).

Step 2: Applying the Biot-Savart Law

The Biot-Savart Law states that the magnetic field \(d\overrightarrow{B}\) due to a current element \(\overrightarrow{dl}\) at point P is given by:\[d\overrightarrow{B} = \frac{\mu_0}{4\pi} \frac{I \overrightarrow{dl} \times \overrightarrow{r}}{r^3}\]where \(\mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}\) is the permeability of free space, and \(\overrightarrow{r}\) is the position vector from the current element to the point P.

Step 3: Calculate the cross product \(\overrightarrow{dl} \times \overrightarrow{r}\)

Perform the cross product \(\overrightarrow{dl} \times \overrightarrow{r}\) with \(\overrightarrow{dl} = 0.500 \times 10^{-3} \hat{\imath}\) and \(\overrightarrow{r} = -0.730 \hat{\imath} + 0.390 \hat{k}\):\[\overrightarrow{dl} \times \overrightarrow{r} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \ 0.500 \times 10^{-3} & 0 & 0 \ -0.730 & 0 & 0.390 \end{vmatrix}\]This yields \(0.500 \times 10^{-3} \times 0.390 \hat{\jmath} = 0.195 \times 10^{-3} \hat{\jmath}\).

Step 4: Calculate \(r^3\)

Calculate the magnitude \(r\) of the position vector \(\overrightarrow{r}\):\[r = \sqrt{(-0.730)^2 + 0.390^2} = \sqrt{0.533\ \approx 0.730} \approximately\, 0.830\, \text{m}\]\(r^3\) is then \((0.830)^3 \approx 0.571 \, \text{m}^3\).

Step 5: Calculate the magnetic field dB

Substitute \(\overrightarrow{dl} \times \overrightarrow{r}\), \(r^3\), current \(I\), and \(\mu_0\) into the Biot-Savart law to find \(d\overrightarrow{B}\):\[d\overrightarrow{B} = \frac{4\pi \times 10^{-7}}{4\pi} \times \frac{5.40 \times 0.195 \times 10^{-3} \hat{\jmath}}{0.571}\]Simplifying gives:\[d\overrightarrow{B} \approx 1.843 \times 10^{-9} \hat{\jmath} \, \text{T}\]

Step 6: Express the result

The magnetic field at point P due to the current element is:\[\overrightarrow{B} = 1.843 \times 10^{-9} \hat{\jmath} \, \text{T}\]

Key Concepts

Magnetic Field Calculation

The Biot-Savart Law is a fundamental principle used to calculate the magnetic field produced by a current element in space. This law provides a way to determine the magnetic field

• at a point P due to a current-carrying wire element, taking into consideration the direction and magnitude of the current.
• The formula indicates that the magnetic field is dependent on factors such as the element's length, the magnitude of the current, and the spatial relationships between the current element and the point P.
• The Biot-Savart law is expressed as: \[d\overrightarrow{B} = \frac{\mu_0}{4\pi} \frac{I \overrightarrow{dl} \times \overrightarrow{r}}{r^3}\]where \(\mu_0\) is the magnetic permeability of free space, ensuring that the units are consistent with the magnetic field interpretation of magnetic flux density.

Understanding this formula requires knowledge of vectors and operations like the cross product, as the direction of the magnetic field is orthogonally related to both the current direction and the point of observation.

Current Element

A current element is a small segment of a wire through which electric current flows. In our exercise, the current element is given as.

• \(\overrightarrow{dl} = 0.500 \times 10^{-3} \hat{\imath} \ \text{m}\)
• This represents a segment of the wire with a length of 0.500 mm along the \(\hat{\imath}\) or x-direction.
• The current is measured as \(5.40 \, \text{A}\), flowing along the same direction as \(\overrightarrow{dl}\).

The current element's significance lies in its influence on the surrounding magnetic field. As current flows through the segment, it generates a magnetic field around itself, the strength and direction of which are determined by the Biot-Savart Law. Studying current elements helps understand how current distribution affects magnetic field behavior.

Cross Product

The cross product \( \overrightarrow{dl} \times \overrightarrow{r} \) is a key vector operation used to determine the direction and magnitude of the magnetic field. In this context, it involves:

• Taking two vectors: the current element \(\overrightarrow{dl}\) and the position vector \(\overrightarrow{r}\) from the current element to the point P.
• Calculating the determinant to find the resulting vector perpendicular to both:

The cross product for our specific example can be visualized as a determinant involving unit vectors \( \hat{\imath}, \hat{\jmath}, \hat{k} \), leading to:\[\overrightarrow{dl} \times \overrightarrow{r} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \0.500 \times 10^{-3} & 0 & 0 \-0.730 & 0 & 0.390 \end{vmatrix} = 0.195 \times 10^{-3} \hat{\jmath}\]This vector is perpendicular to both \(\overrightarrow{dl}\) and \(\overrightarrow{r}\), reflecting how vector operations dictate the orientation of resulting physical phenomena like the magnetic field.

Unit Vectors

Unit vectors are essential in describing direction within three-dimensional space. They have a magnitude of one and serve as a standard basis for expressing other vectors.

• \(\hat{\imath}\), \(\hat{\jmath}\), and \(\hat{k}\) represent unit vectors along the x, y, and z axes, respectively.
• These elements simplify the expression of vectors by indicating direction while scaling for magnitude with other components.

For the given problem, the current element \(\overrightarrow{dl}\) is along the \(\hat{\imath}\) direction, whereas the position vector \(\overrightarrow{r}\) is defined in terms of \(\hat{\imath}\) and \(\hat{k}\), showing its cross-sectional positioning in space.In the solution, the result of the cross product uses \(\hat{\jmath}\), demonstrating how the magnetic field's direction is perpendicular to both original vectors due to established orthogonal vectors.