Electromagnet with two cardboard pieces

In summary, the student attempted to solve a problem involving a magnetic field, but incorrectly used a modified version of the equation. He then attempted to solve a second problem involving voltage amplitude, but incorrectly used the incorrect equation. He was able to solve the third problem using the correct equation.
  • #1
peroAlex
35
4
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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Homework Statement



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 ##.

Homework 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}##.

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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  • #3
Re: 3rd question, why should there be a force on the cardboards?
 
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  • #4
rude man said:
Re: 3rd question, why should there be a force on the cardboards?
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.
 
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  • #5
TSny said:
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.
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?
 
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  • #6
rude man said:
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?
Yes. Sounds right to me. Even when inserting the cardboard, I would think the force on the cardboard would be very small.
 
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  • #7
TSny said:
Yes. Sounds right to me. Even when inserting the cardboard, I would think the force on the cardboard would be very small.
Yes, unless it was impregnated with hi-μ powder or whatever.
 
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  • #8
TSny said:
945 N is a large force! This certainly cannot be the force on the pieces of cardboard.

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
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.
 
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  • #10
TSny said:
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.

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!
 

1. How does an electromagnet work?

An electromagnet works by using electricity to create a magnetic field. When an electric current flows through a wire, it creates a magnetic field around the wire. By coiling the wire and passing a current through it, the magnetic field becomes stronger, creating an electromagnet.

2. What materials are needed to make an electromagnet with two cardboard pieces?

The materials needed to make an electromagnet with two cardboard pieces include a battery, copper wire, an iron nail, two small pieces of cardboard, and a switch. The cardboard pieces will serve as the base for the electromagnet.

3. How can the strength of an electromagnet be increased?

The strength of an electromagnet can be increased by increasing the number of coils in the wire, increasing the current flowing through the wire, and using a magnetic core made of a ferromagnetic material such as iron. The closer the wire is to the core, the stronger the magnetic field will be.

4. What happens if the current is reversed in an electromagnet?

If the current is reversed in an electromagnet, the north and south poles of the magnet will also be reversed. This means that the end that was previously attracting another magnet will now repel it, and vice versa.

5. How is an electromagnet different from a permanent magnet?

An electromagnet is different from a permanent magnet because it only produces a magnetic field when an electric current is flowing through it. When the current is turned off, the magnetic field disappears. A permanent magnet, on the other hand, always has a magnetic field without the need for an electric current.

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