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Two semi-infinite filaments on the \(z\) axis lie in the regions \(-\infty

Short Answer

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Question: Given an arrangement of two semi-infinite filaments carrying current I, find the magnetic field \(\mathbf{H}\) at a point \((\rho, \phi, z=0)\). Calculate the value of \(a\) for which the magnitude of \(\mathbf{H}\) becomes half the value for an infinite filament. Solution: This problem is solved using the Biot-Savart law to find the individual contributions to the magnetic field, \(\mathbf{H}\), due to each filament. The total magnetic field is calculated by adding the contributions. Finally, the value of \(a\) that causes the magnitude to be half the value for an infinite filament is found. Steps for solving the problem include defining the magnetic field due to a single filament using Biot-Savart law, computing the magnetic field due to each filament, calculating the total magnetic field, and finding the value of \(a\) that reduces the magnitude by half.

Step by step solution

01

Define magnetic field due to a single filament using Biot-Savart law

The magnetic field \(\mathbf{dB}\) due to a small current element \(I \, d\mathbf{l}\) at point \(P\) with position vector \(\mathbf{r}\) is given by Biot-Savart law: $$ \mathbf{dB}=\frac{\mu_0}{4\pi}\frac{I \, d\mathbf{l} \times \mathbf{e}_r}{r^2} $$ In cylindrical coordinates \((\rho, \phi, z)\), we have \(d\mathbf{l}=dz'\mathbf{a}_z\), \(\mathbf{e}_r=-\mathbf{a}_\rho\) and \(r^2=(\rho^2+z'^2)\). Remember that \(d\mathbf{l}\) is pointed in the \(z\) direction.
02

Compute the magnetic field due to each filament

For each semi-infinite filament: the magnetic field contributions \(\mathbf{dB}\) will have \(\rho\) and \(\phi\) components and we need to compute these components separately. Thus, we have: $$ \mathbf{dB}_\rho=-\frac{\mu_0 I}{4\pi}\frac{dz'}{\rho^2+z'^2} \sin\phi $$ and $$ \mathbf{dB}_\phi=-\frac{\mu_0 I}{4\pi}\frac{dz'}{\rho^2+z'^2} \cos\phi $$ Now, compute the magnetic field due to each filament by integrating over their entire lengths, assuming the filament is located at \(z'=a\) and \(z'=-a\) respectively, i.e., integrate over \(-\infty<z'<-a\) and \(a<z'<\infty\).
03

Calculate the total magnetic field

The total magnetic field \(\mathbf{H}(\rho,\phi)\) is the sum of the magnetic fields due to both filaments. After adding the contributions, the components \(\mathbf{H_\rho}\) and \(\mathbf{H_\phi}\) can be expressed as a function of \(\rho\) and \(\phi\).
04

Find the value of \(a\) that reduces the magnitude by half

To find the value of \(a\) that causes the magnitude of \(\mathbf{H(\rho=1,z=0)}\) to be half of the value for an infinite filament, compare the total magnetic field magnitudes calculated in the previous step. Solve for \(a\) that makes their ratio equal to \(1/2\).

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Key Concepts

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

Biot-Savart Law
The Biot-Savart Law is a fundamental principle in electromagnetism that helps calculate the magnetic field generated by an electric current. It's particularly useful for determining the magnetic field around a current-carrying conductor and is expressed mathematically as:\[\mathbf{dB} = \frac{\mu_0}{4\pi} \frac{I \, d\mathbf{l} \times \mathbf{e}_r}{r^2}\]Here, \(\mu_0\) is the permeability of free space, \(I\) is the current, \(d\mathbf{l}\) is a differential element of the conductor, \(\mathbf{e}_r\) is the unit vector pointing from the conductor to the point of calculation, and \(r\) is the distance from the element to the point.
  • The law connects current and magnetic field, framing it as inversely proportional to the square of the distance.
  • The cross-product \(d\mathbf{l} \times \mathbf{e}_r\) gives direction, following the right-hand rule.
  • The magnetic field, \(\mathbf{dB}\), surrounds the filament in a looped manner.
Understanding Biot-Savart allows us to approximate the magnetic field around complex current geometries using integration, as was required in this problem for two semi-infinite current filaments.
Cylindrical Coordinates
Cylindrical coordinates are a three-dimensional coordinate system that extends two-dimensional polar coordinates by including a height dimension, \(z\). This system is perfect for problems dealing with symmetry around an axis, like a cylinder or a wire.### Cylindrical Coordinates Components- \(\rho\): The radial distance from the axis of symmetry (akin to radius in polar coordinates).- \(\phi\): The azimuthal angle, indicating position in the plane perpendicular to the axis.- \(z\): The height along the axis of symmetry, similar to the \(z\) coordinate in Cartesian coordinates.For this problem, we express vectors in cylindrical coordinates using these components. For example, the current element \(d\mathbf{l}\) in our problem involves only the \(z\) component, written as \(dz'\mathbf{a}_z\). Similarly, the distance \(r\) involved calculating the squared sum of \(\rho^2 + z'^2\), important for integrating along the filament. Utilizing cylindrical coordinates simplifies integration and calculation in systems with rotational or axial symmetry.
Current Filament
A current filament is a model representing a thin wire or path through which a current flows. In electromagnetism, when calculating fields, these are often assumed to have infinitesimal thickness but finite and defined length.### Characteristics of a Current Filament- **One-dimensional**: Models as having no thickness, only length.- **Carries current \(I\)**: The uniform current flows along its length, affecting the surrounding magnetic environment.- **Magnetic field interaction**: The magnetic field generated by a filament can be calculated using the Biot-Savart Law.In this exercise, two semi-infinite filaments carrying current are positioned on the \(z\)-axis. They represent sources of magnetic fields that interact with each other. By analyzing each individually and summing their effects, we understand the composite magnetic field experienced at a distance \(\rho\) from the axis at \(z=0\). Simplifying this with the concept of a filament is crucial for manageable computation.
Magnetic Field Integration
Integration is a mathematical tool that helps us combine small segments of a magnetic field into a complete picture. In this exercise, you needed to integrate the magnetic field contributions \(\mathbf{dB}\) from both current filaments.### Steps of Magnetic Field Integration1. **Calculate Segment Contribution**: Use Biot-Savart to find \(\mathbf{dB}\) for an infinitesimally small segment.2. **Set Limits of Integration**: For a semi-infinite filament, set limits depending on its \(z\)-position, \(-\infty < z' < -a\) and \(a < z' < \infty\).3. **Integrate Over Length**: Use these limits to find total contribution along the entire length for both filaments.This stepwise integration brings together individual contributions into a comprehensive magnetic field \(\mathbf{H}(\rho, \phi)\). Each filament's contribution is summed to calculate the total field at point \(z=0\). The integration accounts for cumulative effects over the entire path, providing an accurate result for complex current geometries.

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Most popular questions from this chapter

Show that \(\nabla_{2}\left(1 / R_{12}\right)=-\nabla_{1}\left(1 / R_{12}\right)=\mathbf{R}_{21} / R_{12}^{3}\).

A filamentary conductor carrying current \(I\) in the \(\mathbf{a}_{z}\) direction extends along the entire negative \(z\) axis. At \(z=0\) it connects to a copper sheet that fills the \(x>0, y>0\) quadrant of the \(x y\) plane. \((a)\) Set up the Biot-Savart law and find \(\mathrm{H}\) everywhere on the \(z\) axis; \((b)\) repeat part \((a)\), but with the copper sheet occupying the entire \(x y\) plane (Hint: express \(\mathbf{a}_{\phi}\) in terms of \(\mathbf{a}_{x}\) and \(\mathbf{a}_{y}\) and angle \(\phi\) in the integral).

The free space region defined by \(1

The magnetic field intensity is given in a certain region of space as \(\mathbf{H}=\) \(\left[(x+2 y) / z^{2}\right] \mathbf{a}_{y}+(2 / z) \mathbf{a}_{z} \mathrm{~A} / \mathrm{m} .(a)\) Find \(\nabla \times \mathbf{H} .(b)\) Find \(\mathbf{J} .(c)\) Use \(\mathbf{J}\) to find the total current passing through the surface \(z=4,1 \leq x \leq 2,3 \leq z \leq 5\), in the \(\mathbf{a}_{z}\) direction. ( \(d\) ) Show that the same result is obtained using the other side of Stokes' theorem.

A long, straight, nonmagnetic conductor of \(0.2 \mathrm{~mm}\) radius carries a uniformly distributed current of 2 A dc. \((a)\) Find \(J\) within the conductor. (b) Use Ampère's circuital law to find \(\mathbf{H}\) and \(\mathbf{B}\) within the conductor. (c) Show that \(\nabla \times \mathbf{H}=\mathbf{J}\) within the conductor. \((d)\) Find \(\mathbf{H}\) and \(\mathbf{B}\) outside the conductor. \((e)\) Show that \(\nabla \times \mathbf{H}=\mathbf{J}\) outside the conductor.

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