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Homework Help: Find the Current and Flux Density in a Feromagnetic Core

  1. Jan 21, 2014 #1
    1. The problem statement, all variables and given/known data
    A ferromagnetic core is shown the figure below. The depth of the core is 5 cm. The other
    dimensions of the core are as shown in the figure. Find the value of the current that will produce a flux of 0.003 Wb. With this current, what is the flux density at the top of the core? What is the flux density at the right side of the core? Assume that the relative permeability of the core is 1000.

    2. Relevant equations

    3. The attempt at a solution
    So I know to use this equation R=l/μA and I get what μ is and based on a similar problem in my book I also know what to do after I get Req. However what I am stuck on is how to get the mean path length l and the cross sectional areas A. I also know from the problem in the book I am supposed to seperate it into different regions but how exactly do I do this?

    Attached Files:

  2. jcsd
  3. Jan 22, 2014 #2
    This can help you






    And be careful with units
    Last edited by a moderator: Apr 17, 2017
  4. Jan 22, 2014 #3
    You can find the mean length of the flux path very easy.


    The red line represents the longest path which a flux line could take to go round the core. The blue line is the shortest path. Shown in green is the mean length of the flux path.
    le = (l_red + l_blue)/2

    l_red = 2*(10cm + 20cm + 5cm) +2*(15cm + 15cm + 15cm) = 160cm

    l_blue = 2*20cm + 2*15cm = 70cm


    le = (160cm + 70cm)/2 = 115cm

    Attached Files:

  5. Jan 22, 2014 #4
    Thanks for the help I will look at this later when I have more time and let you know if I have problems.
  6. Jan 22, 2014 #5

    rude man

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    You don't want to compute mean path lengths. You want to compute the reluctance around the core.

    You had the right formula for reluctance: R = l/μA.

    So compute the sum of this formula for the four sections of the core: R = ∑ Ri = (1/μ)∑ li/Ai, i = 1 to 4.

    Then apply mmf = Rø. ø = BA = flux and is constant around the core. mmf is magnetomotive force = NI. I is current.
  7. Jan 22, 2014 #6
    ok so I get the length stuff now but I still dont understand how the areas are being produced I know for a rectangular space its length * width but whenever I do it the answer is wrong. I keep thinking the area for the top is 7.5*27.5=206.25 this is obviously wrong what am I doing. Should I be taking the coil dimensions?
  8. Jan 22, 2014 #7
    I have l1= 55 cm and l2=l3=30 cm
  9. Jan 22, 2014 #8

    rude man

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    The area for the top is 15cm x 5cm = 75 cm2. It's the cross-sectional area you're looking for.
  10. Jan 22, 2014 #9
    Where are you taking the cross section from that looks like the right what am I not seeing?
  11. Jan 22, 2014 #10

    rude man

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    The cross-section on the right leg is 5 x 5 = 25 cm^2.
    The flux does not flow into or out of the page. It flows along the perimeter with depth = 5 cm everywhere.

    BTW you CAN do the problem by using the mean free path and doing some fancy area integration along the bends. That might even be more accurate than what I suggested. With either method the trick is to approximate the path area along the bends.

    Usually in a magnetic circuit like this there is an air gap which allows approximating R by d/μ0A where d = gap length and A = gap area. Reason is that μ is much greater in the paramagnetic (iron, steel etc.) core than in the air gap. So the length of the path in tghe core is immaterial. Unfortunately in this problem there is no such gap.
    Last edited: Apr 12, 2015
  12. Jan 22, 2014 #11
    ok so the left would be 10*5=50 because 10 cm is the diameter and 5 cm is the depth.
  13. Jan 22, 2014 #12
    If I read what you put right it is 5*5 because it has a 5cm radius and 5 cm depth.
  14. Jan 22, 2014 #13

    rude man

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    "Radius" and "diameter" refer to circular cross-sections. These cross-sections are rectangular.

    The left cross-section is 10 x 5. The top & bottom are 15 x 5. The right is 5 x 5.
  15. Jan 22, 2014 #14
    oh herp derp I got it now alright so back to the equation R1=l1/μA1=.55/((4pix10^-7)(1000)(.0075))=58.36, R2=l2/μA2=.30/((4pix10^-7)(1000)(.0025))=95.49, and R3=l3/μA3=.30/((4pix10^-7)(1000)(.0050))=47.74 the Req=R1+R2+R3= 58.36+95.49+47.74=201.59 then P=1/Req=1/201.59=.004960 then ø=FP so F=ø/P=.003/.004960=.604. F=Ni so .604=500i which gives I=1.2A then B=ø/A3=.003/.0075=0.4 T and B=ø/A1=.003/.0025=1.2 T
  16. Jan 22, 2014 #15
    Yah so its just basically rectangles 5 cm is on the right times its depth so that gives 5x5 on the left you have 10 and depth of 5 so 10*5 and top and bottom is 15*5 because 15 is at top and bottom
  17. Jan 22, 2014 #16

    rude man

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    Can't read your previous post.

    When you compute the path lengths don't count the corners twice!
    Total length is 25 + 25 + 15 +15.
  18. Nov 2, 2014 #17
    I got the same question.but how to calculate the length at the top??
  19. Apr 12, 2015 #18
    How l1=55
  20. Apr 12, 2015 #19
  21. Nov 8, 2015 #20
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