Slowly oscillating surface current on a solenoid

In summary, the magnetic field in a solenoid is generated by the same field as would be generated by a cylinder of the same dimensions of uniform magnetization.
  • #1
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Homework Statement


From an original surface current ##\vec{K}=K\hat{\phi}## on a finite solenoid, I got ##\vec{B}=\mu_{0}Kf(z)\hat{k}##, for ##r<R##. Assuming that ##\vec{K}## now slowly oscillates in time such as: ##\vec{K(t)}=K_{0}\cos\left(\omega t\right)\hat{\phi}##, so that I still can use the original expression for ##\vec{B}##, how can I find ##\vec{E}##?

Homework Equations


Maxwell's Equations

The Attempt at a Solution


Since ##\vec{B}=B_{k}(z)\hat{k}##, then: ##\nabla\times\vec{B}=0##, thus ##\frac {\partial{\vec E}} {\partial t}=\frac{-\vec{J}}{\epsilon_{0}}##.
Also, ##\nabla\times\vec{E}=\mu_{0}K_{0}f(z)\omega \sin\left(\omega t\right)\hat{k}##
How can I find the x and y components of ##\vec{E}##?
 
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  • #2
In an introductory course, the assumption is usually made that the solenoid is very long, so that ## f(z) \approx 1 ##. (Otherwise the problem becomes very difficult). The magnetic field ## B ## is then uniform inside the solenoid. Stokes theorem can be used and ## \oint E \cdot dl=-A\frac{\partial{B}}{\partial{t}} ##. By symmetry ## \oint E \cdot dl=E(2 \pi r) ##. ## \\ ## Your statement that ## \nabla \times B =0 ## is incorrect. In addition, there is no ## \mu_o ## on the right side of the ## \nabla \times E ## expression. It correctly reads ## \nabla \times E=-\frac{\partial{B}}{\partial{t}} ##. This equation is integrated over ## dA ## over the circular area ## A ## to get the integral result of Stokes theorem: ## \int \nabla \times E \, dA=\oint E \cdot dl=-A \frac{\partial{B}}{\partial{t}} ##.
 
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  • #3
Charles Link said:
In an introductory course, the assumption is usually made that the solenoid is very long, so that ## f(z) \approx 1 ##. (Otherwise the ptoblem becomes very difficult). The magnetic field ## B ## is then uniform inside the solenoid. Stokes theorem can be used and ## \oint E \cdot dl=-A\frac{\partial{B}}{\partial{t}} ##. By symmetry ## \oint E \cdot dl=E(2 \pi r) ##. ## \\ ## Your statement that ## \nabla \times B =0 ## is incorrect. In addition, there is no ## \mu_o ## on the right side of the ## \nabla \times E ## expression. It correctly reads ## \nabla \times E=-\frac{\partial{B}}{\partial{t}} ##.
The problem is about a finite solenoid, so the approximation won't hold. ## \nabla \times E=-\frac{\partial{B}}{\partial{t}} ##, but there is a ## \mu_o ## on the expression for ##\vec{B(t)}##. Why ##\nabla \times \vec{B} =0## is not correct? The vector only has ##\hat{k}## component and it is a function of ##z## only.
 
  • #4
On the first part, my mistake=you have the correct ## \mu_o ## in the expression for ## B ##. ## \\ ## For the second question, ## \nabla \times B=\mu_o J + \mu_o \epsilon_o \frac{\partial{E}}{\partial{t}} ##. In general, ## \nabla \times B \neq 0 ##. In a static case with ## J=0 ##, then ## \nabla \times B=0 ##, but here that is not the case. ## \\ ## The finite solenoid makes the problem more complicated with ## f(z) \neq 1 ##, and, in addition, the magnetic field ## B ## will then include x and y components. It can get very complicated...
 
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  • #5
digogalvao said:
From an original surface current ##\vec{K}=K\hat{\phi}## on a finite solenoid, I got ##\vec{B}=\mu_{0}Kf(z)\hat{k}##, for ##r<R##.
Can you show us how you got this expression for the magnetic field? I know you can derive an expression like this for the field along the axis of the solenoid, but it doesn't seem like the field would only have a z-component off the axis, especially near the ends of the solenoid.
 
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  • #6
vela said:
Can you show us how you got this expression for the magnetic field? I know you can derive an expression like this for the field along the axis of the solenoid, but it doesn't seem like the field would only have a z-component off the axis, especially near the ends of the solenoid.
Ops, I forgot to mention that it was done along the z-axis. The electrical field is also to be found near the z-axis.
 
  • #7
vela said:
Can you show us how you got this expression for the magnetic field? I know you can derive an expression like this for the field along the axis of the solenoid, but it doesn't seem like the field would only have a z-component off the axis, especially near the ends of the solenoid.
I will answer this one for the OP, because I don't think it is likely that he has previously seen the exact solution: A solenoid of finite length generates the same magnetic field as a cylinder of the same dimensions of uniform magnetization ##M ##, that has corresponding surface current per unit length ## K_m=\frac{M \times \hat{n}}{\mu_o } ##. (Using the equation ## B=\mu_o H+M ## for the definition of ## M ##). The following is a result that follows from the "pole" model of E&M theory. There is an ## H ## field found by using the inverse square law from the end faces, where magnetic surface charge density ## \sigma_m=M \cdot \hat{n} ##. The magnetic field is then found everywhere by the equation ## B=\mu_o H+M ##, and the ## \mu_o H ## from the end faces turns out to be the exact correction to ## B ## inside the cylinder. Where ## M=0 ## outside the cylinder, ## B=\mu_o H ## is precisely the magnetic field ## B ## from the solenoid, where ## H ## is the ## H ## that results from the end faces above. For the case of the cylinder of infinite length ## B=M=\mu_o K_m =\mu_o n I ## inside the cylinder.
 
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  • #9
Charles Link said:
I will answer this one for the OP, because I don't think it is likely that he has previously seen the exact solution: A solenoid of finite length generates the same magnetic field as a cylinder of the same dimensions of uniform magnetization ##M ##, that has corresponding surface current per unit length ## K_m=\frac{M \times \hat{n}}{\mu_o } ##. (Using the equation ## B=\mu_o H+M ## for the definition of ## M ##). There is an ## H ## field found by using the inverse square law from the end faces, where magnetic surface charge density ## \sigma_m=M \cdot \hat{n} ##. The magnetic field is then found everywhere by the equation ## B=\mu_o H+M ##, and the ## \mu_o H ## from the end faces turns out to be the exact correction to ## B ##. For the case of the cylinder of infinite length ## B=M=\mu_o K_m =\mu_o n I ##.
I did find the exact solution both by Biot-Savart law and by magnetic vector potential. So it is very likely that I know the exact solution ;)
 
  • #10
digogalvao said:
I did find the exact solution both by Biot-Savart law and by magnetic vector potential. So it is very likely that I know the exact solution ;)
The exact solution for a solenoid of finite length off-axis using Biot-Savart is non-trivial. (Perhaps you are referring to the on-axis solution. If so, then yes, you could compute the result with Biot-Savart). You may have an integral expression for it off-axis, but I don't believe you have successfully evaluated it. Even the ## H ## correction from the poles that is described above is non-trivial, but can be evaluated exactly on axis. The off-axis evaluation using ## H ## from the poles is much simpler than using Biot-Savart, but it is still non-trivial, and can only be done numerically. I don't believe a closed form expression exists for the off-axis result.
 

1. What is a solenoid?

A solenoid is a coil of wire that produces a magnetic field when an electric current is passed through it. It is often used in electronic devices such as motors, speakers, and electromagnets.

2. How does a solenoid create a slowly oscillating surface current?

When an electric current is passed through a solenoid, it creates a magnetic field that can induce currents in nearby conductive materials. This induced current is known as a surface current, and it can oscillate at a slow rate due to the changing magnetic field of the solenoid.

3. What factors affect the amplitude and frequency of the oscillating surface current on a solenoid?

The amplitude and frequency of the oscillating surface current on a solenoid can be affected by the strength of the electric current passing through the solenoid, the number of turns in the coil, and the material and distance of the conductive surface from the solenoid.

4. How is the slowly oscillating surface current on a solenoid used in practical applications?

The slowly oscillating surface current on a solenoid has many practical applications, such as in wireless charging systems, inductive heating, and electromagnetic sensors. It is also used in electronic devices to produce sound and movement.

5. What are the potential drawbacks of using a slowly oscillating surface current on a solenoid?

One potential drawback of using a slowly oscillating surface current on a solenoid is that it can cause interference with other electronic devices, leading to signal distortion or malfunction. Additionally, the oscillating current can cause heat buildup in the solenoid, which can affect its performance and longevity.

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