# Determine flux in air gap (Magnetic Core)

1. Dec 22, 2016

### ElectroViruz

1. The problem statement, all variables and given/known data
As shown above

2. Relevant equations
*Not too sure if they are relevant. Pretty sure there's more. We have textbooks from the library as references but we still can't quite figure it out :/
Φ = BcAc = BgAg

B = Φ/Ac = (Ni/R)/Ac

R = lc/(Ac.μ)

3. The attempt at a solution
Attempt has been very little. Me and my classmates aren't a smart bunch. So we haven't gotten really far. There were several arguments on how we should handle the flux directions as well.

Our attempt:

To find Φ (which is flux in core first),
Hence using Φ = BA, we find A first in the question Where A = 40cm2 = 0.004m2

To find B, we used:
B = Φ/Ac = (Ni/R)/Ac

So to find R, we used:
R = lc/(Ac.μ) where we used μr=2500 and μ0 to get μ for the R equation.

So we used lc which is 1.2m (use the entire core length right ?). We arrived at
R= 95480.6 At/Wb

To get B we need Ni which is the At, but we don't know how to deal with the two coils and the flux.

We've resorted to look for answers online and we hope someone could help us with this. At this point we don't even know if our attempts are appropriate.

We appreciate any help given. Thank you.

Last edited: Dec 22, 2016
2. Dec 22, 2016

### cnh1995

Use the right hand rule to find the direction of flux from each coil and compute the total mmf accordingly. Replace the two coils with a single coil supplying the resultant mmf.

3. Dec 22, 2016

### ElectroViruz

But isn't the flux going against each other at the corner ? Left coil flux goes up, and the top coil goes to the left.
How do we deal with that ?

4. Dec 22, 2016

### cnh1995

Subtract the smaller mmf from the larger mmf. Isn't it like connecting two batteries in series opposition?

5. Dec 22, 2016

### ElectroViruz

So we'd get the net flux of 2000AT which is only going in one direction clockwise right ?
Some of my buds were saying it's supposed to be parallel because both ends of the coil connects to each other.

6. Dec 22, 2016

### cnh1995

I don't think so. The coils are wound on the same flux path, i.e. they are in series.

Last edited: Dec 22, 2016
7. Dec 22, 2016

### ElectroViruz

When the question says neglect fringing and flux leakage, it's safe to say that Φ = BcAc = BgAg right ? So can we also say that Ac = Ag ?

The problem now is, to get the MMF, I require the Rc and Rg, and Rg requires Ag that the question does not provide.

Otherwise, how do you suggest that I find Ag (or Rg for this matter) ? :)

8. Dec 22, 2016

### cnh1995

Yes.
You can first find reluctances of all the sections and compute the equivalent reluctance using series-parallel reduction (just like in electrical circuits). Then you can find the total flux using mmf and net reluctance. Using flux division (just like current division in electrical circuits), you can determine the flux through the air gap.

9. Dec 22, 2016

### ElectroViruz

I really appreciate your help, cn1995. I apologise if this question seem effortless to solve, but I am having difficulties with it.
I'm doing my mechanical degree first year but I have electric subjects as well.

Can you help me verify if
RCORE= 95480.6 1/H or 39789 1/H
?

I noticed that difference between me and my friends' answers were that we used different length.
Following what you said: to replace the two coils with a single coil supplying the resultant mmf.
Hence when calculating the RCORE, I immediately used the length 1.2m which gave me 95480.6 1/H

However my friends did a very complex way, such as splitting the resistance (they did it the electrical circuit way where the core reluctance was the resistance) into several parts and summing them up.
Hence the length they used were 0.5m instead (their workings were too messy for me to understand how they got there).
They got 39789 1/H.

They based their calculations on someone else's anyway so they didn't know how to explain.

10. Dec 22, 2016

### cnh1995

Well, you should go with the electrical circuits way. You can't use the length of the core directly because the flux is not same throughout the length. It is getting divided in the middle portion. I also got the air gap reluctance as 397888 At/Wb.

It's really late here in India now, so I'd better be off to bed. Will look at this tomorrow. Good luck!

11. Dec 22, 2016

### ElectroViruz

Alright. Thank you so much once again

12. Dec 22, 2016

### cnh1995

First, identify all possible flux paths.
There are three paths:
1)The left C section (length=1.202m)
2)The middle limb (again has two iron paths and one air gap)
3The right C (reverse C) section (length=1.202m).
Calculate reluctances of these paths.

Think on how they are interconnected in terms of series or parallel fashion. Form an electrical equivalent circuit for simplicity. Compute the equivalent reluctance.

13. Dec 22, 2016

### ElectroViruz

Thank you !
I was able to get the answer :)
Here's what I did

*After finding the RC1, RC2, RC3 and Rgap
Φ3 is the flux that goes through the air gap

Using KVL:
Loop 1: -2000 + (RC2 + Rgap) Φ3 + RC1Φ1 = 0
15915.49 Φ3 + 397887.36 Φ3 + 39788.74 Φ1 = 2000
39788.74 Φ1 + 413802.85 Φ3 = 2000 — ①

Loop 2: (- Rgap - RC2) Φ3 + RC3Φ2 = 0
- 397887.36 Φ3 – 15915.49 Φ3 + 39788.74 (Φ1 – Φ3) = 0
- 397887.36 Φ3 – 15915.49 Φ3 – 39788.74 Φ3 + 39788.74 Φ1
- 453591.59 Φ3 + 39788.74 Φ1 = 0 — ②

Solve ① and ② simultaneously:
39788.74 Φ1 + 413802.85 Φ3 = 2000
39788.74 Φ1 - 453591.59 Φ3 = 0

Therefore, Φ1= 0.0262 Wb
Φ3 = 0.0231 Wb

However, is it wrong to use mesh analysis for this ? Using mesh analysis, I got 3mWb. So my friends suggested I used KVL instead.
Also is there a simpler way to calculate this question ?

14. Dec 22, 2016

### cnh1995

Do you mean nodal analysis? Because mesh analysis involves use of KVL.
This way, you need not keep track of signs and solve for multiple unknowns. So, this method may be simpler.

15. Dec 22, 2016

### ElectroViruz

Hmm I must've miscalculated when doing mesh then. I did mesh but my answer was 3mWb. I'm supposed to get the same answer as the way I did KVL right ?

I see. Thanks again man :)

16. Dec 23, 2016

### cnh1995

You're welcome!