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Uniform current sheets are located in free space as follows: \(8 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) at \(y=0,-4 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) at \(y=1\), and \(-4 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) at \(y=-1\). Find the vector force per meter length exerted on a current filament carrying \(7 \mathrm{~mA}\) in the \(\mathbf{a}_{L}\) direction if the filament is located at \((a) x=0, y=0.5\), and \(\mathbf{a}_{L}=\mathbf{a}_{z} ;\) (b) \(y=0.5, z=0\), and \(\mathbf{a}_{L}=\mathbf{a}_{x} ;(c) x=0, y=1.5\), and \(\mathbf{a}_{L}=\mathbf{a}_{z}\)

Short Answer

Expert verified
After analyzing the given problem and finding the magnetic fields produced by the current sheets, we can conclude that the vector force exerted on the current filament in all the three cases is zero. This is because the force on a current filament due to a magnetic field is perpendicular both to the magnetic field and to the direction of the current flow, and in all the cases, the magnetic field is either parallel or anti-parallel to the current flow direction.

Step by step solution

01

Calculate the magnetic field produced by each current sheet

First, we need to find the magnetic field produced by each current sheet. Since the current is flowing in the \(\mathbf{a}_z\) direction and the current sheets are on the \(y\) plane, the magnetic field will be in the \(\mathbf{a}_x\) direction. Using Amperè's law, we can find that: - Magnetic field due to the first current sheet at \(y=0\): \(\vec{B_1} = 4 \pi \times 8 \times 10^{-7} \mathbf{a}_x \mathrm{~T}\). For \(y<0\), \(\vec{B_1}\) is positive. For \(y>0\), \(\vec{B_1}\) is negative. - Magnetic field due to the second current sheet at \(y=1\): \(\vec{B_2} = 2 \pi \times 4 \times 10^{-7} \mathbf{a}_x \mathrm{~T}\), which is always positive. - Magnetic field due to the third current sheet at \(y=-1\): \(\vec{B_3} = 2 \pi \times 4 \times 10^{-7} \mathbf{a}_x \mathrm{~T}\), which is always negative.
02

Compute the total magnetic field at the location of the current filament

For each case, we will now compute the total magnetic field at the location of the current filament by summing the magnetic fields produced by each current sheet: 1. Located at \((x, y, z) = (0, 0.5, 0)\): \(\vec{B} = -\vec{B_1} + \vec{B_2} = -2\pi \times 4 \times 10^{-7} \mathbf{a}_x \mathrm{~T}\). 2. Located at \((x, y, z) = (0, 0.5, 0)\): same as in the first case: \(\vec{B} = -2\pi \times 4 \times 10^{-7} \mathbf{a}_x \mathrm{~T}\). 3. Located at \((x, y, z) = (0, 1.5, 0)\): \(\vec{B} = -\vec{B_1} - \vec{B_2} = -6\pi \times 4 \times 10^{-7} \mathbf{a}_x \mathrm{~T}\).
03

Calculate the vector force exerted on the current filament

Now, we need to find the force exerted on the current filament using the formula \(\vec{F} = \vec{I} \times \vec{B}\), where \(\vec{I}\) is the current in the filament and \(\vec{B}\) is the total magnetic field at the location of the filament: 1. Located in the \(\mathbf{a}_z\) direction: \(\vec{F} = (7 \times 10^{-3} \mathbf{a}_z) \times (-2\pi \times 4 \times 10^{-7} \mathbf{a}_x) = 0\). 2. Located in the \(\mathbf{a}_x\) direction: \(\vec{F} = (7 \times 10^{-3} \mathbf{a}_x) \times (-2\pi \times 4 \times 10^{-7} \mathbf{a}_x) = 0\). 3. Located in the \(\mathbf{a}_z\) direction: \(\vec{F} = (7 \times 10^{-3} \mathbf{a}_z) \times (-6\pi \times 4 \times 10^{-7} \mathbf{a}_x) = 0\). In all three cases, the vector force exerted on the current filament is zero. The reason is that the force on a current filament due to a magnetic field is perpendicular both to the magnetic field and to the direction of the current flow. In all cases, the magnetic field is either parallel or anti-parallel to the current flow direction, so the force is zero.

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Key Concepts

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

Magnetic Field
A magnetic field is a vector field surrounding magnets and electric currents. It affects the movement of charged particles, producing observable phenomena like magnetism.
A magnetic field is represented by field lines that indicate the direction and strength of the field. The denser the lines, the stronger the field in that region.
In this exercise, magnetic fields are produced by current sheets lying in the yz-plane. These fields are oriented in the x-direction, perpendicular to the direction of the current flow. Understanding the 3D orientation is crucial to grasp how the field impact is calculated.
Current Density
Current density is a measure of electrical current flowing per unit area, typically expressed in amperes per square meter (A/m²). It differs from total current by considering the distribution over a specific region.
In the given exercise, uniform current sheets with specified current densities are the sources of the magnetic fields. For instance, the sheet at y=0 has a current density of 8 A/m in the z-direction. This creates a specific distribution and direction for the generated field.
Visualizing current density helps to comprehend how different regions contribute uniquely to the magnetic fields they produce.
Electromagnetic Forces
Electromagnetic forces are the forces between charged particles due to electric and magnetic fields. They are one of the four fundamental forces in nature.
The force on a current-carrying conductor in a magnetic field is given by the Lorentz force law: \( \vec{F} = \vec{I} \times \vec{B} \). This cross-product indicates that the force is perpendicular to both the current and magnetic field directions.
In the scenario provided, the alignment of the magnetic field and current direction results in no perpendicular component, leading to zero force on the filament.
Amperè's Law
Amperè's Law is a fundamental principle used to determine magnetic fields produced by current-carrying conductors. It states that the line integral of the magnetic field \( \vec{B} \) around a closed path is proportional to the total current \( I \) passing through the area enclosed by the path: \[\oint \vec{B} \cdot d\vec{l} = \mu_0 I\] Here, \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{T}\cdot \text{m/A} \).
In the exercise, Amperè's Law is applied to calculate the magnetic field vectors due to each current sheet. This approach provides an effective way to understand the cumulative and directional effects of the fields on nearby objects. The method showcases how streamline calculations are possible for symmetrical scenarios.

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Most popular questions from this chapter

A current of 6 A flows from \(M(2,0,5)\) to \(N(5,0,5)\) in a straight, solid conductor in free space. An infinite current filament lies along the \(z\) axis and carries \(50 \mathrm{~A}\) in the \(\mathbf{a}_{z}\) direction. Compute the vector torque on the wire segment using an origin at: \((a)(0,0,5) ;(b)(0,0,0) ;(c)(3,0,0)\).

Assume that an electron is describing a circular orbit of radius \(a\) about a positively charged nucleus. (a) By selecting an appropriate current and area, show that the equivalent orbital dipole moment is \(e a^{2} \omega / 2\), where \(\omega\) is the electron's angular velocity. \((b)\) Show that the torque produced by a magnetic field parallel to the plane of the orbit is \(e a^{2} \omega B / 2 .(c)\) By equating the Coulomb and centrifugal forces, show that \(\omega\) is \(\left(4 \pi \epsilon_{0} m_{e} a^{3} / e^{2}\right)^{-1 / 2}\), where \(m_{e}\) is the electron mass. \((d)\) Find values for the angular velocity, torque, and the orbital magnetic moment for a hydrogen atom, where \(a\) is about \(6 \times 10^{-11} \mathrm{~m} ;\) let \(B=0.5 \mathrm{~T}\).

(a) Find an expression for the magnetic energy stored per unit length in a coaxial transmission line consisting of conducting sleeves of negligible thickness, having radii \(a\) and \(b\). A medium of relative permeability \(\mu_{r}\) fills the region between conductors. Assume current \(I\) flows in both conductors in opposite directions. (b) Obtain the inductance, \(L\), per unit length of line by equating the energy to \((1 / 2) L I^{2}\).

Show that the external inductance per unit length of a two-wire transmission line carrying equal and opposite currents is approximately \((\mu / \pi) \ln (d / a)\) \(\mathrm{H} / \mathrm{m}\), where \(a\) is the radius of each wire and \(d\) is the center-to-center wire spacing. On what basis is the approximation valid?

The dimensions of the outer conductor of a coaxial cable are \(b\) and \(c\), where \(c>b\). Assuming \(\mu=\mu_{0}\), find the magnetic energy stored per unit length in the region \(b<\rho

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