- #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.
--------------------------------------------------------
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 ##.
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}##.
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##.
--------------------------------------------------------
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?
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.
--------------------------------------------------------
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##.
--------------------------------------------------------
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?