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I've been given the following problem:

The magnetic flux density (B) of the air space should be 0.5T. I=5A and the coil has N=200 turns. Calculate the length of the air space e.

electromagnet problem

What I've done so far:
I've calculated $$\Phi=B\cdot A=7.5\times 10^{-3}Vs$$ $$=\frac{\Theta}{R_{m,Fe}+R_{m,\delta}}=\frac{N\cdot I}{R_{m,Fe}+R_{m,\delta}}$$ $$\Rightarrow R_{m,Fe}+R_{m,\delta}=133,33\times 10^3 \frac{1}{H}$$

My plan is to calculate the magnetic resistance of the iron and use that to calculate the magnetic resistance of the air and from there the length of the air space.

My questions:
Am I going about this the right way?
Given that we have been told to calculate the length of the iron via a middle line, from where should the length of the larger section be calculated? iron length
In other words, does the length run from the dot or the dashed line? Or put another way, is the length of the larger section \$d-e\$ or \$d-e+a\$? So far, we have only seen magnets which are circular. Does this one need to be treated as a parallel circuit (see below), and if so are my calculations completely wrong? magnetic circuit diagram

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  • \$\begingroup\$ "In other words, does the length run from the dot or the dashed line?" Eh? \$\endgroup\$
    – Andy aka
    Commented Jan 5, 2017 at 13:08
  • \$\begingroup\$ In the second image I've drawn in the midlines and added dots and a dashed line. I need the length of the larger section to calculate the resistance. I'm assuming that it's d-e (shown in the first diagram) \$\endgroup\$
    – Andy Grey
    Commented Jan 5, 2017 at 13:11
  • \$\begingroup\$ Just use the mid-lines for both parallel sections is my advice. \$\endgroup\$
    – Andy aka
    Commented Jan 5, 2017 at 13:20
  • \$\begingroup\$ Which two parallel sections? I'm afraid i dont follow. \$\endgroup\$
    – Andy Grey
    Commented Jan 5, 2017 at 13:25
  • \$\begingroup\$ With μr=5000 core contribution to total magnetic resistance is going to negligible by far. 1cm air gap is going to be same as 5000cm=50m of core. \$\endgroup\$
    – carloc
    Commented Jan 5, 2017 at 13:28

2 Answers 2

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Am I going about this the right way?

This is my method: -

Aim to find the effective permeability of the core with the air gap

Using B/H = \$\mu_e\$

  • B is 0.5 T as stated
  • H is ampere-turns per metre (5 x 200 / length)
  • The "per metre" part is the effective length of the flux around the core

This is really the only contentious part - how do you estimate the effective length of the core - if you ignore the longer path (to the right) you can argue that the left side route has an effective length of 44 cm. This is based on mid-point positions through the entire core including the air gap length (unknown).

So B/H = 0.00022 = \$\mu_0\mu_e\$ where \$\mu_0\$ = \$4\pi\times 10^{-7}\$

Therefore \$\mu_e\$ = 175

Then relate effective permeability to ungapped permeability

\$\dfrac{l_g}{l_e} = \dfrac{1}{\mu_e}-\dfrac{1}{\mu_i}\$

Where \$\mu_i\$ is the initial relative permeability of the core (5000)

So, the gap to length ratio = 1/181.35 and given that the length is 44 cm, this makes the gap 2.43 mm.

Of course, if I did a more detailed analysis of the core I would probably conclude that the core effective length is more like 50 cm.

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  • \$\begingroup\$ That is very different and much more elegant from my method or that of my friend. I arrived at 2,21mm so I'll check for rounding errors and maybe post my own answer to see if its feasible,. \$\endgroup\$
    – Andy Grey
    Commented Jan 5, 2017 at 14:29
  • \$\begingroup\$ @AndyGrey If I use 50 cm for the effective length my number changes slightly to 2.41 mm. I thought you said your friend calculated 2.42 cm? \$\endgroup\$
    – Andy aka
    Commented Jan 5, 2017 at 14:56
  • \$\begingroup\$ Yes he did, but with a different method again, although closer to mine \$\endgroup\$
    – Andy Grey
    Commented Jan 5, 2017 at 15:02
  • \$\begingroup\$ Using reluctance is fine but it still all rests on doing a sound mechanical analysis and then there is a further consideration; the mean path is actually smaller than the mechanical mean path because the flux tends to slightly "hog" the inner path. Sometimes this results in excessive saturation on the inside of cores because the mean length reduces. \$\endgroup\$
    – Andy aka
    Commented Jan 5, 2017 at 15:07
  • \$\begingroup\$ I'm not sure I understand, but thanks anyway. I'm doing an introduction course for Engineering and we didn't get an explanation or a reason for using the midline (the mechanical mean path?) \$\endgroup\$
    – Andy Grey
    Commented Jan 5, 2017 at 15:20
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Sorry about the image. Would take to long for me to type out

my solution attempt

Is this feasible?

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  • \$\begingroup\$ I'm sure it's another valid approach but it's a bit difficult to read the scrawl in places LOL. \$\endgroup\$
    – Andy aka
    Commented Jan 5, 2017 at 14:59
  • \$\begingroup\$ Some is in German :) I'm studying in Berlin. \$\endgroup\$
    – Andy Grey
    Commented Jan 5, 2017 at 15:01

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