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Chapter 7: Problem 3

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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Most popular questions from this chapter

A square filamentary differential current loop, \(d L\) on a side, is centered at the origin in the \(z=0\) plane in free space. The current \(I\) flows generally in the \(\mathbf{a}_{\phi}\) direction. ( \(a\) ) Assuming that \(r>>d L\), and following a method similar to that in Section \(4.7\), show that $$d \mathbf{A}=\frac{\mu_{0} I(d L)^{2} \sin \theta}{4 \pi r^{2}} \mathbf{a}_{\phi}$$ (b) Show that $$d \mathbf{H}=\frac{I(d L)^{2}}{4 \pi r^{3}}\left(2 \cos \theta \mathbf{a}_{r}+\sin \theta \mathbf{a}_{\theta}\right)$$ The square loop is one form of a magnetic dipole.

A toroid having a cross section of rectangular shape is defined by the following surfaces: the cylinders \(\rho=2\) and \(\rho=3 \mathrm{~cm}\), and the planes \(z=1\) and \(z=2.5 \mathrm{~cm}\). The toroid carries a surface current density of \(-50 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) on the surface \(\rho=3 \mathrm{~cm}\). Find \(\mathbf{H}\) at the point \(P(\rho, \phi, z):(a) P_{A}(1.5 \mathrm{~cm}, 0\), \(2 \mathrm{~cm}) ;\left(\right.\) b) \(P_{B}(2.1 \mathrm{~cm}, 0,2 \mathrm{~cm}) ;\) (c) \(P_{C}(2.7 \mathrm{~cm}, \pi / 2,2 \mathrm{~cm}) ;\) (d) \(P_{D}(3.5 \mathrm{~cm},\), \(\pi / 2,2 \mathrm{~cm})\)

(An inversion of Problem 7.20.) A solid, nonmagnetic conductor of circular cross section has a radius of \(2 \mathrm{~mm}\). The conductor is inhomogeneous, with \(\sigma=10^{6}\left(1+10^{6} \mathrm{\rho}^{2}\right) \mathrm{S} / \mathrm{m}\). If the conductor is \(1 \mathrm{~m}\) in length and has a voltage of \(1 \mathrm{mV}\) between its ends, find: \((a) \mathbf{H}\) inside; \((b)\) the total magnetic flux inside the conductor.

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).

A filamentary conductor on the \(z\) axis carries a current of \(16 \mathrm{~A}\) in the \(\mathbf{a}_{z}\) direction, a conducting shell at \(\rho=6\) carries a total current of \(12 \mathrm{~A}\) in the \(-\mathbf{a}_{z}\) direction, and another shell at \(\rho=10\) carries a total current of \(4 \mathrm{~A}\) in the \(-\mathbf{a}_{z}\) direction. \((a)\) Find \(\mathbf{H}\) for \(0<\rho<12 .\left(\right.\) b) Plot \(H_{\phi}\) versus \(\rho\). (c) Find the total flux \(\Phi\) crossing the surface \(1<\rho<7,0
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