What is the effect of a 1 cm gap on the magnetic field in a solenoid?

[godiswatching_]

Hey, I was trying to figure out this problem. I got (a) using B = mu * NI/L
but I'm not sure how to start the part about the magnetic field in the gap after the solenoid is ripped in half with 1 cm gap.

Thanks for the help!

 

[ergospherical]

The basic idea is to consider a circular loop passing through the core and the gaps. If the field in the core is $H_{\mathrm{in}}$ and the field in the gaps is $H_{\mathrm{gap}}$, then Ampère's law tells you
\begin{align*}
\oint \mathbf{H} \cdot d\mathbf{r} &= NI \\
(l-2s)H_{\mathrm{in}} + 2sH_{\mathrm{gap}} &= NI
\end{align*}where $s = 1\, \mathrm{cm}$ is the gap width. Note that inside the core $\mu = \mu_0 \mu_r$, whilst inside the gap $\mu = \mu_0$. So
\begin{align*}
B_{\mathrm{in}}&= \mu_0 \mu_r H_{\mathrm{in}} \\
B_{\mathrm{gap}}&= \mu_0 H_{\mathrm{gap}}
\end{align*}Using these, re-write the equation in terms of the $\mathbf{B}$ fields. Finally, in order to write $B_{\mathrm{in}}$ in terms of $B_{\mathrm{gap}}$, what do you know about the boundary conditions on the $\mathbf{B}$ fields (specifically, the normal components)?

 

[godiswatching_]

The basic idea is to consider a circular loop passing through the core and the gaps. If the field in the core is $H_{\mathrm{in}}$ and the field in the gaps is $H_{\mathrm{gap}}$, then Ampère's law tells you
\begin{align*}
\oint \mathbf{H} \cdot d\mathbf{r} &= NI \\
(l-2s)H_{\mathrm{in}} + 2sH_{\mathrm{gap}} &= NI
\end{align*}where $s = 1\, \mathrm{cm}$ is the gap width. Note that inside the core $\mu = \mu_0 \mu_r$, whilst inside the gap $\mu = \mu_0$. So
\begin{align*}
B_{\mathrm{in}}&= \mu_0 \mu_r H_{\mathrm{in}} \\
B_{\mathrm{gap}}&= \mu_0 H_{\mathrm{gap}}
\end{align*}Using these, re-write the equation in terms of the $\mathbf{B}$ fields. Finally, in order to write $B_{\mathrm{in}}$ in terms of $B_{\mathrm{gap}}$, what do you know about the boundary conditions on the $\mathbf{B}$ fields (specifically, the normal components)?

why is it 2s? I am having a hard time visualizing it.

 

[ergospherical]

As far as I can tell there are two gaps; you cut the ring in half and then pull the two bits apart a little.

 

[hutchphd]

As far as I can tell there are two gaps; you cut the ring in half and then pull the two bits apart a little.

What ring? This is a long is straight solenoid yes ?

 

[ergospherical]

What ring? This is a long is straight solenoid yes ?

I had in mind a toroidal geometry (i.e. two horseshoes), but a straight solenoid also seems consistent with the question. In the latter case, one can perform an identical analysis except this time choosing a rectangular Ampérian loop with one edge through the solenoid.

 

[godiswatching_]

What ring? This is a long is straight solenoid yes ?

Yes it is a solid iron rod core.

 

[godiswatching_]

I had in mind a toroidal geometry (i.e. two horseshoes), but a straight solenoid also seems consistent with the question. In the latter case, one can perform an identical analysis except this time choosing a rectangular Ampérian loop with one edge through the solenoid.

Yeah so I was wondering what that would look like?

$$\oint \vec H \cdot d \vec l = NI$$
$$(l+s)H_{in} + sH_{gap} = NI$$

Is that valid? s is the gap length and l is the length of the original solenoid.

Then we use the same idea?

 

[ergospherical]

Nearly; it should rather be $(l-s) H_{\mathrm{in}} + s H_{\mathrm{gap}} = NI$.

Put that equation in terms of $\mathbf{B}$ fields, then think about what the relevant boundary condition is at the interface between the core and the gap.

 

[godiswatching_]

Nearly; it should rather be $(l-s) H_{\mathrm{in}} + s H_{\mathrm{gap}} = NI$.

Put that equation in terms of $\mathbf{B}$ fields, then think about what the relevant boundary condition is at the interface between the core and the gap.

So after that I just use
$$B_{{\bot}_{ab}} = B_{{\bot}_{bel}}$$

 

[ergospherical]

Pretty much, yeah. In this case it's just $B_{\mathrm{in}} = B_{\mathrm{gap}}$. So you have\begin{align*}
\dfrac{(l-s) B_{\mathrm{in}}}{\mu_0 \mu_r} + \dfrac{sB_{\mathrm{gap}}}{\mu_0} = NI
\end{align*}and finally you replace $B_{\mathrm{in}}$ with $B_{\mathrm{gap}}$ and solve for it.

 

[Charles Link]

If a really detailed answer was needed, I would use the pole method of magnetostatics, and compute the field for a surface magnetic charge density of $\sigma_m=\pm M$ on the two endfaces in the middle of the solenoid. It will not be completely uniform across the gap. Then the field from the windings must be added to that. They could have asked for the magnetic field in the center of the gap, and that could be fairly readily computed by this method.
For a much simpler answer, the magnetic field will still be approximately 2T. [Edit: See below. It turns out this simple answer is not correct].

Note for the above, the $M$, with $B=\mu_o H+M$ is nearly 2T, but a more refined calculation would subtract off the $\mu_o H$ from the windings. The assumption is made that the $M$ does not change as the gap is inserted into the problem, and that should be reasonably accurate.

Edit: I worked it by the pole method for the $B$ field in the center of the gap, using $M=2 T$. I need to double-check my work, but I got $B \approx (.29)(2T)$. It would appear that the simple answer of $B \approx 2T$ is not at all accurate. The permeability $\mu_r$ turns out to be large enough that the contribution of $B$ from the windings is minimal. The $B$ changes enough from 2T that the $M$ is likely to be non-uniform, but the computation accounting for any change in $M$ would be beyond what would be expected in an undergraduate E&M course.

 

[alan123hk]

I choose to use the following form to calculate.

\begin{align*}
\dfrac{l B_{\mathrm{in}}}{\mu_0 \mu_r} + \dfrac{sB_{\mathrm{gap}}}{\mu_0} = NI ~~~~~~~~l~\text{= length of the iron core,}~~~s~\text{ =width of the gap}
\end{align*}
By the way, I think this calculation method may be inaccurate, because the influence on the left end of the left iron core and the right end of the right iron core may not be ignored. Note that the magnetic lines of force at these two ends are radiating shapes, and the reluctance of the outer space is difficult to estimate, etc.

 

[Charles Link]

This approach uses a modified form of ampere's law $\nabla \times H=J$ with the result that $\oint H \cdot dl=NI$. It works for a toroidal geometry with a small gap because there are basically two regions of constant but differing $H$: in the material, and in the gap. For this long cylindrical case, the contribution from $H$ outside the cylinder will be large, and this $H$ is very non-uniform. I don't think this is a viable approach.

One additional input is that the approach I used above where I assumed $M$ to be constant also doesn't work. The problem is a good one for trying out a couple of things, but a toroidal geometry with a small gap is really what we are needing here.

 

[godiswatching_]

This approach uses a modified form of ampere's law $\nabla \times H=J$ with the result that $\oint H \cdot dl=NI$. It works for a toroidal geometry with a small gap because there are basically two regions of constant but differing $H$: in the material, and in the gap. For this long cylindrical case, the contribution from $H$ outside the cylinder will be large, and this $H$ is very non-uniform. I don't think this is a viable approach.

One additional input is that the approach I used above where I assumed $M$ to be constant also doesn't work. The problem is a good one for trying out a couple of things, but a toroidal geometry with a small gap is really what we are needing here.

So what do you suggest?

 

[Charles Link]

So what do you suggest?

IMO it is interesting to consider such a problem, but I don't see a good solution for it. It is more worthwhile to invest the time into problems like the toroid with a small gap, or to consider a hypothetical case of uniform magnetization $M$, and computing the magnetic field on-axis in the gap. This one doesn't seem to have a simple textbook-like solution.

 

[godiswatching_]

IMO it is interesting to consider such a problem, but I don't see a good solution for it. It is more worthwhile to invest the time into problems like the toroid with a small gap, or to consider a hypothetical case of uniform magnetization $M$, and computing the magnetic field on-axis in the gap. This one doesn't seem to have a simple textbook-like solution.

It’s from a practice midterm I was trying to solve. Sadly the professor does not provide solutions. I am assuming it is somewhat important if it is in a practice test.

 

[hutchphd]

The usual engineering approach is magnetic circuits and "reluctance" and magnetomotive force. This will provide average flux numbers trivially and well. The OP should understand them.
Anything more accurate is very much more complicated and usually not justified by the linearization assumptions already used. Real life can be a bear.

 

[Charles Link]

It’s from a practice midterm I was trying to solve. Sadly the professor does not provide solutions. I am assuming it is somewhat important if it is in a practice test.

I would have to believe that the professor does not have a good solution to it. Magnetostatics is a subject that takes a considerable amount of work to get proficient with solving some of the problems that arise in things like permanent magnets, transformers, and electromagnets. IMO the professor's problem is a rather poor choice for a problem for a midterm exam. It might be a good one for very advanced students to make a project out of it, and see what they might come up with.

 

[Charles Link]

The usual engineering approach is magnetic circuits and "reluctance" and magnetomotive force. This will provide average flux numbers trivially and well. The OP should understand them.
Anything more accurate is very much more complicated and usually not justified by the linearization assumptions already used. Real life can be a bear.

The magnetic circuit approach (as posted by others in this thread) seems to ignore the $\int H \cdot dl$ outside the cylinder. The MMF formula, which comes from a modified ampere's law, has $\oint H \cdot dl =NI$, with the integral over a closed loop. The contribution of $\int H \cdot dl$ outside the cylinder needs to be accounted for, and with the geometry at hand, that makes it very difficult.

 

[alan123hk]

It’s from a practice midterm I was trying to solve. Sadly the professor does not provide solutions. I am assuming it is somewhat important if it is in a practice test.

I think there is not much you can do. You can continue to choose to use a relatively simple approximation method to calculate. In any case, the problem itself has already been mentioned to ignore the end effects and hysteresis effects. As you can imagine, this question only require approximate numerical answer. Or if you think the professor will not mind, you can try to ask the professor some questions in your mind.

We can try to add one more term to the equation, as shown below.

$\dfrac{l_{in} B_{\mathrm{in}}}{\mu_0 \mu_r} + \dfrac{l_{gap}B_{\mathrm{gap}}}{\mu_0} + \dfrac{l_{out} B_{\mathrm{out}}} {\mu_0}= NI$

$l_{out}~\text{= Effective length of external space,}~~~~~B_{out}~\text{ =Effective B of external space}$

Please note that even in the case of air core solenoid inductor (ur=1), the influence of the last term $\frac {l_{out} B_{out}} {\mu_0}$ may not be ignored. What's more, when the relative permeability (ur) is large, the first term $\dfrac{l_{in} B_{\mathrm{in}}}{\mu_0 \mu_r}$ will become very small, and the influence of the last term will be even greater.

I think there is actually a way to evaluate the value of external reluctance. Because the reluctance and inductance are related to each other $\left ( \mathcal {R} = \frac {N^2}{L} \right )$, it is possible to use accurate inductance calculation formulas to evaluate the external reluctance of the cylindrical air-core coil. For example, the formula mentioned in the following links. Note that the Nagaoka coefficient (K) in the formula actually considers the influence of external magnetic reluctance.

https://en.wikipedia.org/wiki/Inductor
https://www.tdk.com/en/tech-mag/electronics_primer/1

Of course, what I said above is just a possible way to solve practical engineering problems. This process is very complicated and not suitable for ordinary students.