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Reluctance in a cast steel ring

  1. Mar 27, 2016 #1
    1. The problem statement, all variables and given/known data
    A coil of 450 turns is wound uniformly round a cast steel ring. The ring has a crosssectional area of 750mm2 and a mean circumference of 600mm. A flux of 0.825mWb is produced in the ring. (Given that H = 800At/m at that flux).
    a.
    Calculate the reluctance of the ring
    b.
    Calculate the current required to produce the given flux.

    2. Relevant equations
    Φ=BA
    μ=B/H
    S=l/μoμrA
    Φ=F/S

    3. The attempt at a solution
    a.
    Φ=BA
    B=Φ/A
    Flux density in ring: 0.825x10-3/750x10-6
    = 1.1T

    μ=B/H
    =1.1/800
    =1.375x10-3

    S=l/μoμrA
    =600x10-3/1.375x10-3x4πx10-7x750x10-6
    =4.63x1011A/Wb

    b. Φ=F/S
    =0.825x10-3=mmf/4.63x1011
    mmf=0.825x10-3x4.63x1011
    =381975000A

    These answers look very wrong to me but I am unsure where I have gone wrong?
     
  2. jcsd
  3. Mar 27, 2016 #2
    If you are familiar with Ohm's law, then reluctance is analogous to resistance in a circuit, with electro-motive force (voltage) being analogous to magneto-motive force (ampere-turns) and resulting current flow (amperes) being analogous to magnetic flux (webers). The B-field that you may be more familiar with is flux density (webers/square meter), and the H-field (ampere-turns/meter) is analogous to the electric field (volts/metre).

    a) So from ohms law R=V/I which for a magnetic loop becomes Reluctance = [Ampere.turns/meter * meters] / [resultant flux].

    In your case this comes to (800At/m * 600mm / 0.825mWb) => 776*103 At/Wb (allowing for the millimeters and milliwebers).

    b) You have been given the magnetising field intensity (800 ampere-turns/meter) required to produce the flux and you know the mean circumference of the core (600mm) so that multiplying these together gives the ampere-turns (as we already did in (a)). Dividing by the turns then gives you the amperes:

    In your case this comes to (800At/m * 600mm / 450t) => 1.07 Amps

    Rather easy when you understand the quantities that you are dealing with.
     
  4. Mar 28, 2016 #3
    Thank you jwinter.
    There are so many different quantities that it becomes confusing, thank you for your thorough explanation.

    I am not getting to the same answer as you have provided in a).
    S=800x600x10^-3/0.825x10^-3
    =581818.1818

    What am I doing wrong?
     
  5. Mar 28, 2016 #4
    Nothing - my calculator must have made a mistake :smile:
     
  6. Mar 28, 2016 #5
    :nb)
    My answer of 581818.18A/Wb is correct?
     
  7. Mar 28, 2016 #6
    Yes it is correct. But don't keep so many significant figures - it looks stupid. You were given at most a 3 significant figure input so only 3 significant figures are useful in the output.
     
  8. Mar 28, 2016 #7
    I.E 0.582μA/Wb?
     
  9. Mar 28, 2016 #8
    Gosh you need to think for yourself at least a tiny bit if you are ever going to pass an exam and actually become useful in a subject! If you are not then why bother studying at all?

    Do you know the difference between micro (μ) and mega?
     
  10. Mar 29, 2016 #9
    micro = 10^-6
    mega=10^6
     
  11. Mar 29, 2016 #10
    So you noticed that you got it wrong?
     
  12. Mar 29, 2016 #11
    Yeah I went the wrong way my bad 0.582MA/Wb
     
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