Quasi-stastic solenoid magnetic field

In summary, we are trying to find the fields far from an infinitely long, straight, right circular solenoid with radius a and N turns per unit length. The current, I(t), is the same throughout the solenoid and changes slowly in time. By using previous work on localized charge distributions and slowly changing currents, we can compute the dipole moment per unit length, which is then used to find the approximate fields far from the solenoid. To do this, we integrate along the z-axis and use the vector potential, A, to find the electric field, E. However, there is some uncertainty about the validity of integrating over field points as if they were source points.
  • #1
Pengwuino
Gold Member
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The construction of and excitation of an infinite, straight, right circular solenoid of radius a, with N turns per unit length, are such taht tis current I(t) is the same everywehre along its length and is changed very slowly in time. Show that the fields far from the solenoid are approximately:

[tex]E \approx \frac{{ - \mu _0 Na^2 }}{{2\rho }}\frac{{\partial I(t - \frac{\rho }{c})}}{{\partial t}}\hat \phi [/tex]

Attempt:

We know from previous work that for a localized charge distribution far from the origin for slowly changing currents with no charge density and only dipole terms is:

[tex]
E = \frac{{\mu _0 }}{{4\pi r^2 }}\hat r \times \frac{{\partial m(t - \frac{r}{c})}}{{\partial t}}[/tex]

So it should be a simple matter of computing the dipole moment per unit length. Now I've been second guessing myself continually on the charge density but I've kinda come to the conclusion that it the multipole moment should be

[tex]
\begin{array}{l}
m = \frac{1}{2}\int {I(t - \frac{\rho }{c})N\delta (\rho ' - a)\rho '^2 d\rho 'd\phi '(\hat \rho ' \times \hat \phi ')} \\
= 2\pi a^2 I(t - \frac{\rho }{c})N\hat z \\
\end{array}[/tex]

However, simply plugging this making the assumption that due to the symmetry, r becomes [tex]\rho[/tex], gives:

[tex]
E = \frac{{ - \mu _0 a^2 N}}{{2\rho ^2 }}\frac{{\partial I(t - \frac{\rho }{c})}}{{\partial t}}\hat \phi [/tex]

which is off by that [tex]\rho^2[/tex] instead of [tex]\rho[/tex]. Is there something obvious I am missing here?
 
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  • #2
It is an infinitely long solenoid. So you will need to integrate along the z-axis (since dm is proportional to dz), your A-vector will have the integral in it. Once you solve it, you will see the 1/r and not 1/r^2.
 
  • #3
Integrating along dz will blow the integral up. I can't find a way to do the z-integration without that happening.
 
  • #4
Hmm... So you have:

[tex]dA = \frac{\mu_0}{4\pi}\frac{dm\times r}{r^2}\hat{r}[/tex]

Next you know from a single current loop:

[tex]m_{loop} = IA \hat{z}[/tex]

So for an infinitesmal cross-section of our cylinder, we have:

[tex]dm = NIAdz \hat{z}[/tex]

Substituting this into our first equation gives:

[tex]dA =\frac{\mu_0 N I a^2}{4} \frac{dz}{r} \hat{z}\times\hat{r}[/tex]

Now you need to find an analytical form for [tex]\hat{z}\times\hat{r}[/tex] using only [tex]z[/tex] and [tex]r[/tex]. You know it will be [tex]sin(\theta)[/tex] where [tex]\theta[/tex] is the angle formed between [tex]\hat{z}[/tex] and [tex]\hat{r}[/tex].

So you will want to draw a picture of an infinite cylinder. Then draw a point far out on the x-axis, let's call [tex]\rho[/tex]. Then somewhere along the z-axis you will draw a vector out to this point and call it [tex]r[/tex]. So that:

[tex]\rho^2+z^2=r^2[/tex]

I set up the differential form for you in its most basic form. Now you will need to find this cross product and integrate it from -infinity to infinity. It won't blow up on you since it didn't blow up in Maple for me.
 
  • #5
I absolutely do not want to use the vector potential in this problem though. To find the B-field, i'd have to do [tex]\nabla \times \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} \over A[/tex] which would result in eventually having terms like [tex]\nabla I(t - \frac{\rho }{c})[/tex] eventually which I can't operate on. Then again using Jefimenko's equations, I might be able to get somewhere with using the moment.

Actually I don't even think your method is valid. Your dA is with respect to source points but you are doing your integration using field points (or maybe not, I'm still thinking about this...)
 
Last edited:
  • #6
You are not solving for the magnetic field. You want just the electric field, so you use:

[tex]E=-\frac{\partial A}{\partial t}[/tex]

I probably left out a few constants.

Also, it is very similar to integrating over charge points to get the full electric field at a point. Except now I am integrating over magnetic dipoles along an infinite wire to get the vector potential.
 
Last edited:
  • #7
Oh right, what's with me?

However, [tex]\[
\hat z \times \hat r = \frac{{ - \rho }}{r}\hat \phi
\]
[/tex]. Doing the integration along z, if valid, returns something funny.

[tex]\[
A = \frac{{\mu _0 NIa^2 }}{4}\int_{ - \infty }^\infty {\frac{{ - \rho }}{{(\rho ^2 + z^2 )}}dz = } \frac{{ - \pi \mu _0 NIa^2 }}{4}
\]
[/tex]

Did I miss something?
 
  • #8
I made a mistake, I lost a factor of 'r' in my original equation. The correct form is:

[tex]dA =\frac{\mu_0 N I a^2}{4} \frac{dz}{r^2} \hat{z}\times\hat{r}[/tex]

So you should have a power of 3/2 in your denominator of the integral.
 
  • #9
Yah I found that at the same time. The one thing I'm just hesitant about is integrating over field points as if they were source points. For example, if the solenoid was simply of length 2L, one would have to have a z-dependence on the electric field, but you would lose it running through the method we have done here I would think...
 
  • #10
This is similar to the magnetic field from an infinite wire. If it were a finite wire then the math would be more difficult and there would be a z-dependence on the magnetic field. Since it is infinite, then the magnetic field only depends on radial distance from the wire.
 

What is a quasi-static solenoid magnetic field?

A quasi-static solenoid magnetic field is a type of magnetic field that is produced by a solenoid, which is a cylindrical coil of wire. It is considered quasi-static because the magnetic field changes very slowly over time and can be approximated as a constant field.

How is a quasi-static solenoid magnetic field created?

A quasi-static solenoid magnetic field is created by passing an electric current through a coil of wire. The current creates a magnetic field around the coil, and the shape of the coil causes the magnetic field to be more concentrated inside the coil.

What are the characteristics of a quasi-static solenoid magnetic field?

A quasi-static solenoid magnetic field has several characteristics, including being relatively uniform within the coil, having a direction that is determined by the direction of the current, and having a strength that is directly proportional to the current and the number of turns in the coil.

What are some applications of a quasi-static solenoid magnetic field?

A quasi-static solenoid magnetic field has many practical applications, including in electromagnets used in industry, medical devices such as MRI machines, and electronic devices such as loudspeakers and electric motors.

How does a quasi-static solenoid magnetic field differ from a dynamic magnetic field?

A quasi-static solenoid magnetic field differs from a dynamic magnetic field in that it changes slowly over time and can be approximated as a constant field. In contrast, a dynamic magnetic field changes rapidly over time and cannot be approximated as a constant field. Additionally, a quasi-static solenoid magnetic field is typically produced by a direct current, while a dynamic magnetic field is often produced by an alternating current.

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