# Electromagnet with two cardboard pieces

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1. Aug 5, 2017

### peroAlex

Hello!

Recently I was going through some old exams and upon encountering this problem (which seemed pretty easy) I got stuck. Exams at my university are composed of individual tasks, each having three subquestions with four plausible answers respectively. Solution sheet gives results only, so there’s no way for me to check where my attempt at solution went wrong. I ask members of this forum for help, maybe someone will see where I made a mistake. Your help is very appreciated, so thank you in advance.

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1. The problem statement, all variables and given/known data

Given is a laminated electromagnet with magnetic field path $l = 0.4 m$, permittivity $\mu = 13500 \mu_0$ that has two gaps $\delta = 0.002$ in which we insert two cardboard pieces. Our electromagnet is wound with a coil , $N = 280$ plugged onto alternating voltage source ($U_{rms} = 230 V$, $\omega = 120 \pi s^{-1}$) and has surface area of cross section $S = 0.002 m^2$.

First question: compute coil inductance. Correct result should be $L = 0.0489 H$.
Second question: compute amplitude of electric current. Correct result should be $I_0 = 17.6 A$.
Third question: compute average value of magnetic force on one of the cardboard pieces. Correct result should be $F_avg = 945 N$.

2. Relevant equations

For the first question I used $L = \mu N^2 \frac{S}{l}$. For the second question I used $B = \mu H$, $NI = H_{core}l + H_{gaps}\delta = B (\frac{l}{\mu} + \frac{\delta}{\mu_0})$. For the third question I used $F = \frac{B^2 S}{2 \mu}$.

3. The attempt at a solution

Using $L = \mu N^2 \frac{S}{l}$ I obtained $L = 6.65H$, which is obviously incorrect. So I tried modifying the equation into $L = \mu N \frac{S}{l}$ (instead of $N^2$ I used $N$). This gave me $L = 0.02375H$, which – if multiplied by factor 2 – gives $L = 0.0475H$. Close enough, right? But why would I use the “modified” form?

OK, next up is the second question. From voltage amplitude $U_0 = U_{rms} \sqrt{2} = N B S \omega$ I derived $B = 1.54T$. If I stick with this value and plug it in $NI = B (\frac{l}{\mu} + \frac{\delta}{\mu_0})$, it produces: $I_0 = 8.8832 A$, again close enough because two times my first attempt yields $I_0 = 17.7664 A$. So I decided to find value of B-field that would return $I_0 = 17.6$. As it turns out, B-field should be $B = 3.05T$, again proving that something should be multiplied or divided by 2. At this point it becoming preposterous, since it all indicates that formulas I found on the Internet with my notes combined are inaccurate.

For third question, again, equation I noted above would return $F = 1887.26 N$, yet if I divide it by 2 it would give $F = 943.63N$.

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At this point I don’t know how to describe my situation. I’m either using completely wrong set of formulas, or I’m missing a point by not modifying them. Could someone be please kind enough to help me?

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2. Aug 5, 2017

3. Aug 5, 2017

### rude man

Re: 3rd question, why should there be a force on the cardboards?

4. Aug 5, 2017

### TSny

Good question!

Searching, I found a link that estimates the force of attraction between the poles at the air gap. See equation 3-71 here
http://www.vias.org/matsch_capmag/matsch_caps_magnetics_chap3_18.html

The formula appears to match the OP's formula except I guess you use $\mu$ for cardboard instead of air for this problem (equation 3-74). But I would think there would be negligible difference in $\mu$ for air and cardboard. (No permeability info was given for cardboard in the problem statement.)

945 N is a large force! This certainly cannot be the force on the pieces of cardboard.

5. Aug 5, 2017

### rude man

Yeah, so it seems, probably trick question.
Even if the cardboard had μ > μ0 there would be no force on the cardboard once it was wholly within the gap, since presumably no ∇B exists in the gap. There would of course be a force pulling in the cardboard as it was being inserted. Sound right?

6. Aug 5, 2017

### TSny

Yes. Sounds right to me. Even when inserting the cardboard, I would think the force on the cardboard would be very small.

7. Aug 6, 2017

### rude man

Yes, unless it was impregnated with hi-μ powder or whatever.

8. Aug 6, 2017

### peroAlex

To begin with, I would like to thank both for participating.
Third question, as explained above, has four plausible answers. Solution sheet gives $F=945N$, but you can chose from
• $F = 307 000N$
• $F=0.0038N$
• $F=0.0019N$
So if you deem that force of 945 newtons is inaccurate, could it be that solutions are wrong or that perhaps I'm dealing with unrealistic task?

9. Aug 6, 2017

### TSny

If I use the formula that you listed for the force, then I get about 940 N. Pretty close to 945. But this formula is for the force of attraction between the parallel faces of the core at the gap, not the force on some material placed in the gap. Asking for the force on the cardboard seems like a strange question to rude man and me. But, I'm certainly no expert in this area.

10. Aug 6, 2017

### peroAlex

Interesting to know that, because I know my translation is correct. It explicitly asks to compute force on cardboard sheets, so apparently it's a trick question. Thank you so much for helping me! Both you and @rude man, thank you!