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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.

Short Answer

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

Step by step solution

01

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 \(-0.730 \hat{\imath} + 0.390 \hat{k} \, \text{m}\).
02

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.
03

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 \(0.500 \times 10^{-3} \times 0.390 \hat{\jmath} = 0.195 \times 10^{-3} \hat{\jmath}\).
04

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 \((0.830)^3 \approx 0.571 \, \text{m}^3\).
05

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:\[d\overrightarrow{B} \approx 1.843 \times 10^{-9} \hat{\jmath} \, \text{T}\]
06

Express the result

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A 15.0-cm-long solenoid with radius 0.750 cm is closely wound with 600 turns of wire. The current in the windings is 8.00 A. Compute the magnetic field at a point near the center of the solenoid.

(a) How large a current would a very long, straight wire have to carry so that the magnetic field 2.00 cm from the wire is equal to 1.00 G (comparable to the earth's northward-pointing magnetic field)? (b) If the wire is horizontal with the current running from east to west, at what locations would the magnetic field of the wire point in the same direction as the horizontal component of the earth's magnetic field? (c) Repeat part (b) except the wire is vertical with the current going upward.

A long, straight, solid cylinder, oriented with its axis in the \(z\)-direction, carries a current whose current density is \(\overrightarrow{J}\). The current density, although symmetric about the cylinder axis, is not constant and varies according to the relationship $$\overrightarrow{J} = (\frac{b}{r})e^{(r a)/\delta}\hat{k} \space for \space r \leq a$$ $$=0 \space for \space r \geq a$$ where the radius of the cylinder is a = 5.00 cm, \(r\) is the radial distance from the cylinder axis, \(b\) is a constant equal to 600 A/m, and \(\delta\) is a constant equal to 2.50 cm. (a) Let \(I_0\) be the total current passing through the entire cross section of the wire. Obtain an expression for \(I_0\) in terms of \(b\), \(\delta\), and a. Evaluate your expression to obtain a numerical value for I0. (b) Using Ampere's law, derive an expression for the magnetic field \(\overrightarrow{B}\) in the region \(r \leq a\). Express your answer in terms of \(I_0\) rather than b. (c) Obtain an expression for the current \(I\) contained in a circular cross section of radius \(r \leq a\) and centered at the cylinder axis. Express your answer in terms of \(I_0\) rather than b. (d) Using Ampere's law, derive an expression for the magnetic field \(\overrightarrow{B}\) in the region \(r \leq a\). (e) Evaluate the magnitude of the magnetic field at \(r = \delta\), \(r = a\), and \(r = 2a\).

Long, straight conductors with square cross sections and each carrying current \(I\) are laid side by side to form an infinite current sheet (Fig. P28.73). The conductors lie in the \(xy\)-plane, are parallel to the \(y\)-axis, and carry current in the +\(y\)-direction. There are \(n\) conductors per unit length measured along the \(x\)-axis. (a) What are the magnitude and direction of the magnetic field a distance \(a\) below the current sheet? (b) What are the magnitude and direction of the magnetic field a distance a above the current sheet?

Two concentric circular loops of wire lie on a tabletop, one inside the other. The inner wire has a diameter of 20.0 cm and carries a clockwise current of 12.0 A, as viewed from above, and the outer wire has a diameter of 30.0 cm. What must be the magnitude and direction (as viewed from above) of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires?

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