Slowly oscillating surface current on a solenoid

[digogalvao]

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 $\vec{E}$?

 

[Charles Link]

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: $\int \nabla \times E \, dA=\oint E \cdot dl=-A \frac{\partial{B}}{\partial{t}}$.

 

[digogalvao]

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.

 

[Charles Link]

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...

 

[vela]

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.

 

[digogalvao]

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.

 

[Charles Link]

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.

 

[Charles Link]

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.

There is a simple expression for the on-axis field of a solenoid of finite length. I will try to find a "link". See: https://notes.tyrocity.com/magnetic-field-along-axis-of-solenoid/

 

[digogalvao]

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 ;)

 

[Charles Link]

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.