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The \(x, y,\) and \(z\) components of a magnetic field are \(B_{x}=0.10 \mathrm{T}\) \(B_{y}=0.15 \mathrm{T},\) and \(B_{z}=0.17 \mathrm{T} .\) A \(25-\mathrm{cm}\) wire is oriented along the \(z\) axis and carries a current of \(4.3 \mathrm{A}\). What is the magnitude of the magnetic force that acts on this wire?

Short Answer

Expert verified
The magnetic force is approximately 0.1935 N.

Step by step solution

01

Understand the Problem

You're given a wire that carries a current along the z-axis in a magnetic field with components in three dimensions. We need to calculate the magnetic force on this wire using these components.
02

Recall the Formula for Magnetic Force

The magnetic force (\(F\)) on a current-carrying wire is given by the formula \(F = I \cdot L \cdot B \cdot \sin\theta\), where \(I\) is the current, \(L\) is the length of the wire, \(B\) is the magnetic field, and \(\theta\) is the angle between the current direction and the magnetic field.
03

Evaluate the Angle

Since the wire is oriented along the z-axis, we need to find the components of the magnetic field perpendicular to this axis. The components perpendicular are \(B_{x}\) and \(B_{y}\), as the z-component (\(B_{z}\)) will not contribute to the force (because \(\sin 0^\circ = 0\)).
04

Calculate Perpendicular Magnetic Field

Calculate the magnitude of the magnetic field perpendicular to the wire which is given by \(\sqrt{B_{x}^2 + B_{y}^2}\). Substitute the given values:\[B_{\perp} = \sqrt{(0.10)^2 + (0.15)^2} = \sqrt{0.01 + 0.0225} = \sqrt{0.0325} \approx 0.18\, T\]
05

Calculate the Force on the Wire

Substitute the known values into the formula \(F = I \cdot L \cdot B_{\perp}\):Let \(I = 4.3\, A\), \(L = 0.25\, m\) (convert cm to m), \(B_{\perp} = 0.18\, T\).Calculate the force:\[F = 4.3 \times 0.25 \times 0.18 = 0.1935\, N\]
06

Conclusion

The magnitude of the magnetic force that acts on the wire is approximately \(0.1935\, N\).

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Key Concepts

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

Magnetic Field Components
Understanding magnetic field components is critical to comprehending how magnetic forces operate. Magnetic fields are vector fields, meaning they have both magnitude and direction. In a three-dimensional space, these fields are typically described using their components along the x, y, and z axes.

In this context, the components were given as:
  • \( B_x = 0.10 \mathrm{T} \)
  • \( B_y = 0.15 \mathrm{T} \)
  • \( B_z = 0.17 \mathrm{T} \)
These components tell us how strong the magnetic field is along each direction. To understand the total effect on a current-carrying wire, it is often necessary to consider only the components that are perpendicular to the current. This is because the magnetic force is maximized when the field is perpendicular to the direction of the current. In this exercise, components \( B_x \) and \( B_y \) are considered since they are perpendicular to the wire which is aligned along the z-axis.

The combined effect of these perpendicular components creates a resultant magnetic field, which is crucial for calculating the magnetic force accurately.
Current-Carrying Wire
Wires that carry current can experience significant forces when placed in a magnetic field. When a current flows through a wire, it generates its own magnetic field. However, when this wire is subjected to an external magnetic field, the interaction between the two fields creates a magnetic force.

To analyze this in a practical example, consider a straight wire that is axial along the z-direction and carries a current of \(4.3 \mathrm{A}\). The length of the wire is \(25 \mathrm{cm}\), which needs to be converted to meters (\(0.25 \mathrm{m}\)) for calculations involving physical units in the SI system.

The magnetic force experienced by a current-carrying wire in a magnetic field can be computed using the formula:\[ F = I \cdot L \cdot B \cdot \sin \theta \]Where:
  • \( I \) is the current through the wire
  • \( L \) is the length of the wire
  • \( B \) is the magnetic field strength perpendicular to the wire
  • \( \theta \) is the angle between the current direction and the magnetic field
This fundamental interaction showcases the relevance of direction, specifically the alignment of the wire and the field, in determining the force magnitude.
Angle between Current and Magnetic Field
The angle between the direction of current and magnetic field is a crucial factor in calculating the magnetic force on a wire. The magnetic force is directly dependent on the sine of this angle \( \theta \).

In our example, since the wire is aligned along the z-axis, the z-component of the magnetic field, \( B_z \), will not contribute to the force. This is because the angle between \( B_z \) and the wire's direction is \(0^\circ\), and \( \sin 0^\circ = 0 \).

Thus, only \( B_x \) and \( B_y \) components contribute to the magnetic force, as these are perpendicular to the current direction. The resultant magnetic field perpendicular to the wire is given by the magnitude:\[ B_{\perp} = \sqrt{B_x^2 + B_y^2} \]The angle effectively becomes \(90^\circ\) between this resultant perpendicular magnetic field and the wire, ensuring maximum force since \( \sin 90^\circ = 1 \). This phenomenon underscores the significance of perpendicular alignment in electromagnetic interactions and helps in precise evaluations of force magnitudes when dealing with vector quantities.

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Most popular questions from this chapter

A particle that has an \(8.2 \mu \mathrm{C}\) charge moves with a velocity of magnitude \(5.0 \times 10^{5} \mathrm{m} / \mathrm{s}\) along the \(+x\) axis. It experiences no magnetic force, although there is a magnetic field present. The maximum possible magnetic force that the charge could experience has a magnitude of \(0.48 \mathrm{N}\). Find the magnitude and direction of the magnetic field. Note that there are two possible answers for the direction of the field.

A very long, hollow cylinder is formed by rolling up a thin sheet of copper. Electric charges flow along the copper sheet parallel to the axis of the cylinder. The arrangement is, in effect, a hollow tube of current \(I\). Use Ampere's law to show that the magnetic field (a) is \(\mu_{0} I /(2 \pi r)\) outside the cylinder at a distance \(r\) from the axis and (b) is zero at any point within the hollow interior of the cylinder. (Hint: For closed paths, use circles perpendicular to and centered on the axis of the cylinder.)

One component of a magnetic field has a magnitude of \(0.048 \mathrm{T}\) and points along the \(+x\) axis, while the other component has a magnitude of \(0.065 \mathrm{T}\) and points along the \(-y\) axis. A particle carrying a charge of \(+2.0 \times 10^{-5} \mathrm{C}\) is moving along the \(+z\) axis at a speed of \(4.2 \times 10^{3} \mathrm{m} / \mathrm{s} .\) (a) Find the magnitude of the net magnetic force that acts on the particle. (b) Determine the angle that the net force makes with respect to the \(+x\) axis.

A charge is moving perpendicular to a magnetic field and experiences a force whose magnitude is \(2.7 \times 10^{-3} \mathrm{N}\). If this same charge were to move at the same speed and the angle between its velocity and the same magnetic field were \(38^{\circ},\) what would be the magnitude of the magnetic force that the charge would experience?

Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The length of each rod is \(0.85 \mathrm{m},\) and the mass of each is \(0.073 \mathrm{kg}\). One rod is held in place above the ground, while the other floats beneath it at a distance of \(8.2 \times 10^{-3} \mathrm{m} .\) Determine the current in the rods.

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