Calculating the Flux: Understanding Mean Radius & Circumference

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In summary: The little numbers at the top and bottom say 5mm. So from that, you can conclude that the outer radius is 5mm.
  • #1
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Question:

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?
 

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  • #2
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?
 
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  • #3
RUber said:
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?
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?
 
  • #4
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.
 
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  • #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
 

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  • #6
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.
 
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  • #7
RUber said:
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.
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.
 

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  • #8
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?
 
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  • #9
k, took me long enough, but I think I get it now. :) thanks a lot.
 
  • #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?
 
  • #11
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.
 

FAQ: Calculating the Flux: Understanding Mean Radius & Circumference

What is "flux" and why is it important in scientific calculations?

"Flux" refers to the flow or movement of a substance or energy through a surface or boundary. In scientific calculations, it is important because it helps us understand the rate at which a substance or energy is moving, and can provide insight into processes such as diffusion, heat transfer, and fluid dynamics.

How is mean radius calculated?

Mean radius is calculated by taking the average of all the radii of a circle or sphere. For a circle, this would involve adding all the radii and dividing by the number of radii. For a sphere, the mean radius is equal to the radius of the sphere.

What is the formula for calculating circumference?

The formula for calculating circumference depends on the shape of the object. For a circle, the formula is C = 2πr, where C is the circumference and r is the radius. For an ellipse, the formula is C = 2π√(a² + b²/2), where a and b are the lengths of the semi-major and semi-minor axes. For a regular polygon, the formula is C = ns, where n is the number of sides and s is the length of each side.

How can understanding mean radius and circumference help in real-world applications?

Understanding mean radius and circumference can be useful in a variety of real-world applications. For example, in engineering, these calculations can help determine the size and strength of materials needed for a structure. In physics, they can help predict the motion and behavior of objects. In astronomy, they can help measure the size and distance of celestial bodies. In medicine, they can aid in the design of medical devices and procedures.

Are there any common misconceptions about calculating flux, mean radius, and circumference?

One common misconception is that flux, mean radius, and circumference are only relevant to circles or spheres. In reality, these concepts can be applied to a wide range of shapes and can be used to understand various processes and phenomena. Additionally, it's important to note that these calculations are based on mathematical equations and may not always accurately reflect real-world situations due to factors such as imperfections in measurement or variations in shape.

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