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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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Most popular questions from this chapter

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

A current filament carrying \(I\) in the \(-\mathbf{a}_{z}\) direction lies along the entire positive \(z\) axis. At the origin, it connects to a conducting sheet that forms the \(x y\) plane. (a) Find \(\mathbf{K}\) in the conducting sheet. \((b)\) Use Ampere's circuital law to find \(\mathbf{H}\) everywhere for \(z>0 ;(c)\) find \(\mathbf{H}\) for \(z<0\).

Assume that there is a region with cylindrical symmetry in which the conductivity is given by \(\sigma=1.5 e^{-150 \rho} \mathrm{kS} / \mathrm{m}\). An electric field of \(30 \mathbf{a}_{z} \mathrm{~V} / \mathrm{m}\) is present. ( \(a\) ) Find \(\mathbf{J}\). \((b)\) Find the total current crossing the surface \(\rho<\rho_{0}\), \(z=0\), all \(\phi\). ( \(c\) ) Make use of Ampère's circuital law to find \(\mathbf{H}\).

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.

An infinite filament on the \(z\) axis carries \(20 \pi \mathrm{mA}\) in the \(\mathbf{a}_{z}\) direction. Three \(\mathbf{a}_{z}\) -directed uniform cylindrical current sheets are also present: \(400 \mathrm{~mA} / \mathrm{m}\) at \(\rho=1 \mathrm{~cm},-250 \mathrm{~mA} / \mathrm{m}\) at \(\rho=2 \mathrm{~cm}\), and \(-300 \mathrm{~mA} / \mathrm{m}\) at \(\rho=3 \mathrm{~cm}\). Calculate \(H_{\phi}\) at \(\rho=0.5,1.5,2.5\), and \(3.5 \mathrm{~cm}\).

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