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A wire of \(3 \mathrm{~mm}\) radius is made up of an inner material \((0<\rho<2 \mathrm{~mm})\) for which \(\sigma=10^{7} \mathrm{~S} / \mathrm{m}\), and an outer material ( \(2 \mathrm{~mm}<\rho<3 \mathrm{~mm}\) ) for which \(\sigma=4 \times 10^{7} \mathrm{~S} / \mathrm{m}\). If the wire carries a total current of \(100 \mathrm{~mA}\) dc, determine \(\mathbf{H}\) everywhere as a function of \(\rho\).

Short Answer

Expert verified
Question: Determine the magnetic field intensity (H) for every point within a cylindrical wire with two concentric regions, each with different electrical conductivity, as a function of radial distance (蟻) from its axis. Answer: H(蟻) = { I鈧 / (2蟺蟻) if 0 < 蟻 < 2 mm , I鈧 / (2蟺蟻) if 2 mm < 蟻 < 3 mm }

Step by step solution

01

Determine current distribution

First, we need to determine the distribution of the total current in both the inner and outer materials of the wire. We know the total current is 100 mA and the conductivities of the materials are 蟽鈧 = 10^7 S/m and 蟽鈧 = 4脳10^7 S/m. The current density J is the current per unit area and can be related to the conductivity and the electric field as follows: J = 蟽E Since the electric field E is constant within each cylindrical material, their current densities J鈧 and J鈧 inside each material are also constant. Let the current in the inner material be I鈧 and the current in the outer material be I鈧. We can express the total current I (100 mA) in terms of I鈧 and I鈧: I = I鈧 + I鈧 Then, we can write the current densities J鈧 and J鈧 in terms of I鈧 and I鈧: J鈧 = I鈧 / A鈧 J鈧 = I鈧 / A鈧 Where A鈧 and A鈧 are the cross-sectional areas of the inner and outer materials. A鈧 = 蟺(2 mm)虏 A鈧 = 蟺[(3 mm)虏 - (2 mm)虏] Now, we can use the proportions of conductivities to find the relation between I鈧 and I鈧: J鈧 / J鈧 = 蟽鈧 / 蟽鈧 (I鈧 / A鈧) / (I鈧 / A鈧) = 蟽鈧 / 蟽鈧
02

Solve for I鈧 and I鈧

Now we have two equations with two unknowns (I鈧 and I鈧), which we can express as follows: I鈧 + I鈧 = 100 mA (I鈧 / A鈧) / (I鈧 / A鈧) = 蟽鈧 / 蟽鈧 Solve these linear equations simultaneously to find I鈧 and I鈧.
03

Apply Ampere's circuital law

For an Azimuthal magnetic field \(\textbf{H}\) in a cylindrical coordinate system, Ampere's circuital law is given as: 鈭瓾 dL =NI For the inner region (0 < 蟻 < 2 mm), we have I = I鈧, and for the outer region (2 mm < 蟻 < 3 mm), I = I鈧.
04

Calculate H for each region

Now we can calculate the magnetic field intensity H as a function of 蟻 for both the inner and outer regions individually. For the inner region (0 < 蟻 < 2 mm): H(蟻) = I鈧 / (2蟺蟻) For the outer region (2 mm < 蟻 < 3 mm): H(蟻) = I鈧 / (2蟺蟻)
05

Write the final expression for H(蟻)

Combine the expressions for the magnetic field intensity H(蟻) in both regions into a single expression: H(蟻) = { I鈧 / (2蟺蟻) if 0 < 蟻 < 2 mm , I鈧 / (2蟺蟻) if 2 mm < 蟻 < 3 mm } This is the magnetic field intensity H(蟻) everywhere as a function of 蟻 for the given wire.

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Key Concepts

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

Current Distribution
Understanding current distribution is crucial for evaluating electromagnetic fields within complex wire systems. In this exercise, we have a wire divided into two regions, each made of different materials with distinct electrical conductivities. Here, the inner region (\(0<\rho<2\mathrm{~mm}\)) has a conductivity (\(\sigma_1 = 10^7 \mathrm{~S} / \mathrm{m}\)) and the outer region (\(2\mathrm{~mm}<\rho<3\mathrm{~mm}\)) has a higher conductivity (\(\sigma_2 = 4 \times 10^7 \mathrm{~S} / \mathrm{m}\)).
To determine how the total current of 100 mA distributes between these two regions, we look at current densities, \(J_1\) and \(J_2\), which are proportional to their respective conductivities. Using \(J = \sigma E\), where \(\sigma\) is conductivity and \(E\) is electric field, we derive the relationship between the current in each region and match it with their respective areas:
  • The area of the inner region is \(A_1 = \pi (2 \mathrm{~mm})^2\).
  • The area of the outer region is \(A_2 = \pi [(3 \mathrm{~mm})^2 - (2 \mathrm{~mm})^2]\).
By setting up the equation in terms of current \(I_1\) in the inner region and \(I_2\) in the outer region, we can solve for how much current each section carries. This is an essential step because it influences the resultant magnetic fields formed around the wire.
Ampere's Circuital Law
Ampere鈥檚 Circuital Law is a cornerstone for determining magnetic fields created by current-carrying conductors. It states that the magnetic field \(\textbf{H}\) around a closed loop is proportional to the current passing through the loop:
\[\oint \textbf{H} \cdot dL = NI\]where \(N\) is the number of turns (which is 1 in this scenario) and \(I\) is the current enclosed by the loop. The integration path \(dL\) essentially forms a closed loop around the wire.
For this exercise, we apply Ampere鈥檚 Law differently in two regions:
  • For the inner region (\(0 < \rho < 2 \mathrm{~mm}\)), only the current \(I_1\) is enclosed by the loop.
  • For the outer region (\(2 \mathrm{~mm} < \rho < 3 \mathrm{~mm}\)), \(I_2\) is the relevant current within the loop.
This application helps in understanding how different sections of the wire cumulatively contribute to magnetic field formation, thus giving deeper insights into how current affects its electromagnetic environment.
Magnetic Field Intensity
Magnetic field intensity \(H\), also expressed as \(\textbf{H}\), signifies how strong a magnetic field is at a certain point in space around current-carrying conductors. It is influenced by the distribution and magnitude of the current flowing through the wire's sections.
To compute \(H\) as a function of radius \(\rho\), we consider:
  • In the inner region (\(0 < \rho < 2 \mathrm{~mm}\)), \(H(\rho) = \frac{I_1}{2\pi\rho}\).
  • In the outer region (\(2 \mathrm{~mm} < \rho < 3 \mathrm{~mm}\)), \(H(\rho) = \frac{I_2}{2\pi\rho}\).
These expressions arise from applying Ampere鈥檚 Circuital Law to concentric hypothetical circles centered around the axis of the wire. They reveal how the magnetic field diminishes with increasing distance from the wire, a phenomenon inherent to electromagnetic fields. Additionally, they highlight how the two regions individually affect \(H\) based on their specific currents, helping us foresee the magnetic influence the wire exerts within each material section.

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Most popular questions from this chapter

A cylindrical wire of radius \(a\) is oriented with the \(z\) axis down its center line. The wire carries a nonuniform current down its length of density \(\mathbf{J}=b \rho \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}^{2}\) where \(b\) is a constant. ( \(a\) ) What total current flows in the wire? \((b)\) Find \(\mathbf{H}_{i n}(0<\rhoa)\), as a function of \(\rho ;(d)\) verify your results of parts \((b)\) and \((c)\) by using \(\nabla \times \mathbf{H}=\mathbf{J}\).

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 solid cylinder of radius \(a\) and length \(L\), where \(L \gg a\), contains volume charge of uniform density \(\rho_{0} \mathrm{C} / \mathrm{m}^{3}\). The cylinder rotates about its axis (the \(z\) axis) at angular velocity \(\Omega \mathrm{rad} / \mathrm{s}\). (a) Determine the current density \(\mathbf{J}\) as a function of position within the rotating cylinder. (b) Determine \(\mathbf{H}\) on-axis by applying the results of Problem 7.6. ( \(c\) ) Determine the magnetic field intensity \(\mathbf{H}\) inside and outside. \((d)\) Check your result of part ( \(c\) ) by taking the curl of \(\mathbf{H}\).

Consider a sphere of radius \(r=4\) centered at \((0,0,3)\). Let \(S_{1}\) be that portion of the spherical surface that lies above the \(x y\) plane. Find \(\int_{S_{1}}(\nabla \times \mathbf{H}) \cdot d \mathbf{S}\) if \(\mathbf{H}=3 \rho \mathbf{a}_{\phi}\) in cylindrical coordinates.

Given \(\mathbf{H}=\left(3 r^{2} / \sin \theta\right) \mathbf{a}_{\theta}+54 r \cos \theta \mathbf{a}_{\phi} \mathrm{A} / \mathrm{m}\) in free space: \((a)\) Find the total current in the \(\mathbf{a}_{\theta}\) direction through the conical surface \(\theta=20^{\circ}, 0 \leq \phi \leq 2 \pi\), \(0 \leq r \leq 5\), by whatever side of Stokes' theorem you like the best. \((b)\) Check the result by using the other side of Stokes' theorem.

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