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Calculating the flux

  1. Sep 10, 2015 #1

    The magnetic circuit shown (in the pictures) is constructed with a semi-circular piece of metal and a metal bar. The c.s.a of the metal parts is 10mmx10mm. The coil has 500 turns and a resistance of 10Ω. μr of metal parts = 1500.
    μo = 4π x 10-7

    My attempt:

    I completely understand the concept of how to work out the flux, but there's just one part that I can't seem to understand.

    1. The first picture shows the diagram.

    2. The second picture shows all the values I obtained throughout the entire solution.

    3. The third picture shows the working out of the reluctance of the semi circle.

    4. The fourth picture shows the working out of the reluctance of the other piece of metal and the 2 air gaps.

    5. The last picture shows me adding the reluctance of all the parts together and working out the answer, which is the flux.

    The working in the third picture is what I don't understand. I got this from a friend and I don't understand how he works out the "mean radius" and "mean circumference." Where does the 3cm - 0.5cm come from? Can someone please explain?

    Attached Files:

  2. jcsd
  3. Sep 10, 2015 #2


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    The mean radius (assuming the semi-circle is 1cm thick) is just that, the (inner radius + outer radius)/2, which gives 2.5cm.
    The mean circumference similar, the circumference around the middle of the semi-circle.
    Does it make sense that you would be looking for values in reference to the center of the semi-circle?
  4. Sep 10, 2015 #3
    Yea, I'm sorry, I don't understand.

    How by looking at that picture, which is all that is given in the question, do I work out the mean radius. I understand the mean circumference now, but I still don't understand where the 3cm - 0.5cm comes from. Can you explain step by step the simplest form you can?
  5. Sep 10, 2015 #4


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    I am not seeing anywhere in your diagram or work that states that the semi-circle is 1 cm thick, or that the internal radius of the semi-circle in 2cm. However, the only way that subtracting .5cm to get the mean radius makes sense is if that is that case.

    The outer radius is 3cm, this should be apparent from the measure of 6cm from end to end. If the thickness is 1cm, then the central (a.k.a. mean) radius is 3cm - 1/2*1cm, subtracting off half the thickness.
  6. Sep 10, 2015 #5
    I'm sorry for going on but I'm trying to completely understand this.

    That picture is a picture of the marking scheme written up. I don't know if you can make it out, but if you look at the top right of the semi circle, the arrows inside it say the measurement is 10mm, which is obviously 1cm as you said. The little numbers at the top and bottom say 5mm.

    I just don't understand where the 1cm is coming into this. Can I just randomly assume it's 1cm?

    Also, what is the equation to work out the mean of this if there is any because I'm struggling to understand the 3-1/2*1

    Attached Files:

  7. Sep 10, 2015 #6


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    I see now. You can see that from the center of the semi-circle it is 5mm to the outside edge and 5mm to the inside edge. Then the radius of the center of the semi-circle is just 3cm - 5mm. That is the formula.
    Alternate ways:
    You know that the outer edge is a constant 3cm radius, and the thickness is a constant 1cm, so the central radius is 3 - 1/2 * 1cm.
    Or, you know that the outer radius is 3cm, the inner radius is 3cm - 1cm = 2cm. The mean of 3cm and 2cm is (3cm+2cm)/2.
  8. Sep 10, 2015 #7
    I've never did this stuff before so sorry if I'm asking dumb questions... just a few more.

    That is the exact question right there in the picture. Which part shows that from end to end it measures 10mm?

    Also, how is the outer radius 3cm? Why would I just half 6, I don't get it.

    Attached Files:

  9. Sep 10, 2015 #8


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    It says the cross-sectional area of metal parts is 10mm x 10mm. From that, you know that the thickness is 10mm = 1cm.
    If you have a circle with diameter of 6cm, how do you find the radius?
    If you have a semi-circle with width of 6cm, would that not be just half of a circle with diameter 6cm?
  10. Sep 10, 2015 #9
    k, took me long enough, but I think I get it now. :) thanks a lot.
  11. Sep 10, 2015 #10
    I have another question. Why to work this out do I need to half the semi-circle? What is the reason for doing it?
  12. Sep 10, 2015 #11


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    You are finding the center of the semi-circle. I am not 100% clear what assumptions are required to make this simplification possible, but it makes sense that the principle field produced would be generally described through the center of the metal piece.
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