Slowly oscillating surface current on a solenoid

  • #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}##?
 

Answers and Replies

  • #2
Charles Link
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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
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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
Charles Link
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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
vela
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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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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
Charles Link
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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
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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
Charles Link
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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.
 

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