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Given the field \(\mathbf{H}=20 \rho^{2} \mathbf{a}_{\phi} \mathrm{A} / \mathrm{m}:(a)\) Determine the current density \(\mathbf{J}\). (b) Integrate \(\mathbf{J}\) over the circular surface \(\rho \leq 1,0<\phi<2 \pi, z=0\), to determine the total current passing through that surface in the \(\mathbf{a}_{z}\) direction. (c) Find the total current once more, this time by a line integral around the circular path \(\rho=1,0<\phi<2 \pi, z=0 .\)

Short Answer

Expert verified
Question: Determine the current density and the total current passing through a circular surface with radius 1 and located at z=0, given the magnetic field in cylindrical coordinates as \(\mathbf{H}=20 \rho^{2} \mathbf{a}_{\phi} \mathrm{A} /\mathrm{m}\). Verify the total current using a line integral.

Step by step solution

01

Calculate Current Density from Given Magnetic Field

We have the magnetic field \(\mathbf{H}=20 \rho^{2} \mathbf{a}_{\phi} \mathrm{A} /\mathrm{m}\), and we need to find the current density \(\mathbf{J}\). From Ampere's law, we know that \(\nabla \times \mathbf{H} = \mathbf{J}\). Using the formula for curl in cylindrical coordinates, we have: $$ \mathbf{J} = \nabla \times \mathbf{H} = \left( \frac{1}{\rho} \frac{\partial H_z}{\partial \phi} - \frac{\partial H_\phi}{\partial z} \right) \mathbf{a}_{\rho} + \left( \frac{\partial H_\rho}{\partial z} - \frac{\partial H_z}{\partial \rho} \right) \mathbf{a}_{\phi} + \left( \frac{1}{\rho} \frac{\partial (\rho H_\phi)}{\partial \rho} - \frac{1}{\rho} \frac{\partial H_\rho}{\partial \phi} \right) \mathbf{a}_{z} $$ Given \(\mathbf{H}=20 \rho^{2} \mathbf{a}_{\phi}\), we can compute the partial derivatives in the above expression to find \(\mathbf{J}\).
02

Calculate Total Current using Surface Integral

Now, we need to compute the total current passing through the circular surface \(\rho \leq 1, 0 < \phi < 2\pi, z=0\) in the \(\mathbf{a}_{z}\) direction which requires the integration: $$ I = \int\int_S \mathbf{J} \cdot d\mathbf{S} $$ Where \(S\) is the given circular surface and \(d\mathbf{S}\) denotes the infinitesimal surface vector.
03

Calculate Total Current using Line Integral

Finally, we compute the total current once again, this time by a line integral around the circular path \(\rho = 1, 0 < \phi < 2 \pi, z = 0\). Using Ampere's law in integral form: $$ \oint_C \mathbf{H} \cdot d\mathbf{l} = I_{enc} $$ Where \(C\) is the given circular path, \(d\mathbf{l}\) denotes the infinitesimal arc length vector, and \(I_{enc}\) is the total enclosed current. By evaluating Steps 1, 2 and 3, we can find the current density \(\mathbf{J}\), the total current passing through the given surface, and verify the total current using line integral.

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Key Concepts

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

Magnetic Field
Magnetic fields are invisible forces that influence charged particles. In electromagnetics, the magnetic field is often represented by the vector \( \mathbf{H} \). It can be thought of as a map that shows the direction and strength of the magnetic force at different locations. In our exercise, \( \mathbf{H} = 20 \rho^{2} \mathbf{a}_{\phi} \mathrm{A} / \mathrm{m} \) describes how the magnetic field varies with the radial distance \( \rho \).

When dealing with problems involving magnetic fields, it's crucial to understand how these fields interact with currents and create forces. The direction of the field is indicated by \( \mathbf{a}_{\phi} \), which points along the azimuthal direction in cylindrical coordinates. The factor of \( \rho^2 \) tells us that the field strength increases with the square of the radial distance.
Current Density
Current density, denoted as \( \mathbf{J} \), is a measure of how much electric current flows through a particular area. In our problem, we need to find \( \mathbf{J} \) from the magnetic field \( \mathbf{H} \). According to Ampere’s Law, \( abla \times \mathbf{H} = \mathbf{J} \), meaning the curl of the magnetic field gives us the current density.

This operation essentially measures how the magnetic field 'twists' around a point, and in our context, helps determine the distribution of currents in space. By using the curl formula in cylindrical coordinates, we translate changes in the magnetic field into information about current flow.

The result is a vector that shows where currents are strong or weak, or even zero, providing us insight into how electric circuits or materials respond to the magnetic fields.
Ampere's Law
Ampere’s Law is a fundamental law of electromagnetism that relates magnetic fields to the currents that produce them. There are both differential and integral forms of Ampere's Law. In our exercise, both forms help us verify the total current.

  • The differential form \( abla \times \mathbf{H} = \mathbf{J} \) helps calculate the current density based on how the magnetic field changes in space.
  • The integral form \( \oint_C \mathbf{H} \cdot d\mathbf{l} = I_{enc} \) allows us to determine the total current enclosed by a path by integrating the magnetic field along that path.

This law is useful when analyzing circuits or geometry like the circular surfaces in our problem. It makes linking magnetic fields and currents clearer, offering insight into their interdependency.
Cylindrical Coordinates
Cylindrical coordinates are a system of coordinates that lend themselves to problems with rotational symmetry around an axis, typical in electromagnetics. Instead of using \(x\), \(y\), and \(z\), cylindrical coordinates use \( \rho \), \( \phi \), and \( z \).

  • \( \rho \) is the radial distance from the origin to the projection of the point in the \(xy\)-plane.
  • \( \phi \) is the angle of rotation around the \(z\)-axis.
  • \( z \) represents the height from the \(xy\)-plane.

This system is advantageous when dealing with problems like ours, where symmetry greatly simplifies calculations. Converting the vector field equations to cylindrical coordinates enables us to apply Ampere's Law more easily and integrate vectors over surfaces and paths that match the problem's symmetry.

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Most popular questions from this chapter

A current filament on the \(z\) axis carries a current of \(7 \mathrm{~mA}\) in the \(\mathbf{a}_{z}\) direction, and current sheets of \(0.5 \mathrm{a}_{z} \mathrm{~A} / \mathrm{m}\) and \(-0.2 \mathrm{a}_{z} \mathrm{~A} / \mathrm{m}\) are located at \(\rho=1 \mathrm{~cm}\) and \(\rho=0.5 \mathrm{~cm}\), respectively. Calculate \(\mathbf{H}\) at: \((\) a \() \rho=0.5 \mathrm{~cm} ;(b) \rho=\) \(1.5 \mathrm{~cm} ;(c) \rho=4 \mathrm{~cm} .(d)\) What current sheet should be located at \(\rho=4 \mathrm{~cm}\) so that \(\mathbf{H}=0\) for all \(\rho>4 \mathrm{~cm}\) ?

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).

Use an expansion in rectangular coordinates to show that the curl of the gradient of any scalar field \(G\) is identically equal to zero.

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 \(\mathbf{H}\).

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