# Flux and Current of Magnetic Core With Air Gap

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1. May 8, 2017

### peroAlex

I am a student of electrical engineering. This task appears in our textbook. However, there were no solutions provided. I tried understanding the task but got lost in the process. This is why I seek help here on this site. I would like to ask you for guidance and any sort of advice on how to solve this task.

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

Given is a magnetic structure (check the attached png file). Primary winding has $N_1 = 210$
and secondary winding has $N_2 = 40$. Voltage source which drives the current $i_1$ is alternating (sketch is obviously wrongfully displaying DC voltage source) $u(t) = 500 cos(300/s t)$. Surface area of cross section is constant and equals $S=0.004 m^2$. Air gap is $\delta = 0.002 m$ wide.
Compute magnetic flux $\Phi_2$ and both currents $i_1$ and $i_2$.

2. Relevant equations
I assume that I need to implement Faraday's Law $$EMF = -N \frac{d \Phi}{dt}$$ probably inductance $$L = \frac{N \Phi}{I}$$ I sincerely hope I have mention edenough equations.

3. The attempt at a solution
I began trying to compute flux and magnetic flux density from given information. If $EMF = -N \frac{d \Phi}{dt} = NBS\omega sin(\omega t)$ then $B \cong 2.6455 T$. Now that evidently leads to $\Phi_1 = 0.00794 Wb$. No data for permittivity was given so I assume I need to continue with my computation using magnetic flux density.
From now on I'm lost. I would really appreciate at least some help or at least guidance. I will be happy with any sort of tips or pieces of advice given. Thank you in advance!!!

#### Attached Files:

• ###### Fakulteta za elektrotehniko01.png
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2. May 8, 2017

I think you need permittivity data. Knowing the current in the primary winding will tell you what $H$ is inside the coil by ampere's law, but to get the response of the material, and thereby the magnetic field $B$, you need permittivity data. The part of solving for the magnetic field in the air gap is also somewhat of a puzzle. If it were a permanent magnet, with a saturated value of magnetization $M$ everywhere, it would be a straightforward calculation. In this ac case, the value of $M$ will depend upon the ac permittivity, and even then, it doesn't appear a simple matter to determine how the magnetization will respond in regions outside of the primary coil, and in particular, exactly what it will do when it encounters the split where the path goes two directions. Perhaps the assumption is made that the magnetization is uniform everywhere=probably a reasonably good assumption, (an alternative assumption would be that $B$ spits in two because $\nabla \cdot B=0$), but my engineering expertise in this sort of ac magnetic problem is somewhat limited.

3. May 9, 2017

### peroAlex

OK, I will try to ask the professor whether or not there is any sort of permittivity data given. I was lost exactly because I could not find the relation between $B$ and $H$ field. I assume that magnetization is uniform since this is an introductory course. Thank you so much for letting me know I need extra information!

4. May 12, 2017

### rude man

Usually, permittivity is assumed infinite which I'll bet your prof will tell you/has told you. Tha's because permittivity of iron, steel etc is >> permittivity of air.

You don't want or need to calculate B. You want Φ which you have done indirectly. Assume Φ1 = Φ2 (no flux leakage). So that takes care of Φ. Notice you haven't even needed to compute air gap reluctance or anything else.

To compute i1 you need to invoke Ampere's law relating mmf = Ni to path reluctance and Φ.
The value of i2 should be VERY obvious since no load is specified for the secondary winding.

5. May 12, 2017

IMO the problem is somewhat poorly specified. By having the output coil be short-circuited, it makes it a much more complex problem than the open-circuited secondary, in which case $\frac{\mathcal{E}_2}{\mathcal{E}_1}=\frac{N_2}{N_1}$. This simple equation does not apply for a short-circuited secondary for which the secondary current will affect the value of $\Phi_2$. $\\$ For an introductory problem, I think the instructor did not need to include an air-gap and an alternative flux path, and he should also have open-circuited the secondary coil. In addition, if he wants you to compute $i_1$ (assuming the geometry of a long solenoid), I do think that you require magnetic permittivity data. The flux $\Phi_1$ can be computed from the voltage and the number of turns $N_1$ as follows: $u_o cos(\omega t)=-N_1 \frac{d \Phi_1}{dt}$ , from which $\Phi_1=-\int \frac{u_o}{N_1} cos(\omega t)\, dt$. The current $i_1$ depends on the magnetic permittivity, and will be smaller as the permittivity increases. (For a long solenoid (by Ampere's law) $H=\frac{N}{L}i$ and $B=\mu H$, thereby the permittivity $\mu$ is needed along with the length $L$). $\\$ IMO, it would make a good problem for a more advanced class, but the permittivity data needs to be supplied as well as information on the length of the part that is wound. For an introductory problem, it would be much more practical to ask the student to compute $\Phi_1$ (along with $\Phi_2$) and $\mathcal{E_2}$ , with an open-circuited secondary and no alternative flux path through the middle.

Last edited: May 12, 2017
6. May 12, 2017

### rude man

1. there is no mention of a shorted secondary. EDIT: OK, the illustration indicates it is shorted. However, in that case, primary and/or secondary leakages and resistances must be finite or the problem is unbounded.

2. Regardlress, the flux does not change with secondary current, so it makes no difference to the flux whether the secondary is shorted, open, or any finite load of any kind. emf = -N dΦ/dt and there is only once source of emf.

Last edited: May 12, 2017
7. May 12, 2017

The picture is drawn with a shorted secondary. Meanwhile, any current in the secondary produces a magnetic flux in the opposite direction. IMO, the instructor needs to be more clear in presenting a homework assignment. In addition, the current $i_2$ will depend on what the load resistance is in the secondary circuit.

8. May 15, 2017

### marcusl

1) The assumption that $\Phi_1=\Phi_2$ is incorrect because some of the flux from the driven coil is shunted through the center leg. This is what keeps the problem from being elementary.
2) Permittivity is an electric quantity. It is permeability that describes a magnetic material.
3) So long as the permeability is large, $\mu>>1$, its exact value is unneeded. The effective reluctance of a leg with a small air gap is determined by the gap. That's why ferrite transformer cores ("pot cores") are manufactured with a gap--the reluctance of the entire path is determined by precise machining (the gap width) rather than by poorly controlled chemical engineering (mu).
EDIT:
4) The permeability is needed, in general, to calculate how the flux divides between the center leg and leg #2. There's something special about this problem, however, that means we don't need its exact value.
5) In an idealized case where mu is truly infinite, as assumed by rude man, then it's true that $\Phi_1=\Phi_2$ and the center leg becomes irrelevant. Since the problem specifies the leg and gap, I assume the opposite: that mu must instead be finite (why include the leg if it's irrelevant to the solution?).

Last edited: May 15, 2017
9. May 20, 2017

### rude man

The gap dimension is in m, not m^2, and is as stated gap width, not area. The area is obvioulsy also needed to compute overall reluctance. The core reluctance is obviously intended to be zero.

10. May 20, 2017

### marcusl

Yes, that's what I said--gap width. The core permeability doesn't need to be specified exactly so long as it's large compared to one. I think the solution is this: The shorted secondary prevents any flux from passing through leg 2, so $\Phi_2=0$, by Lenz's Law. Since $\Phi_1=\Phi_2+\Phi_3$, where 3 refers to the central leg, then $\Phi_1=\Phi_3$. The path that conducts flux now contains an air gap whose width (in m) dominates and determines the effective reluctance of the path. The flux is now calculated from of the magnetomotive force mmf (as you suggest), together with the effective reluctance of the portion of the core containing legs 1 and 3.

11. May 20, 2017

I'm finding the OP's homeworks to be very educational=(in the physics curriculum in my college days many years ago, transformer problems were not given much emphasis.) Anyway, here is a "link" to another transformer problem from the same OP. Absolute Value of Magnetization $\\$@marcusl and @rude man might find this one of interest as well=it looks like you both have expertise in this area. I think I gave this second one a better answer than my responses above. (I also learned from @marcusl response above). This one I think is also a better "first" gap problem, and the one above is better understood after having worked this other one.

12. May 21, 2017

### rude man

13. May 31, 2017

### jim hardy

I'd say responses have encircled the solution but not nailed it.

.....................................................................................
I would calculate flux Φ1 by integrating voltage, as OP did

and since the secondary is shorted, ie zero volts as several folks mentioned , fluxΦ3 is zero (or near enough zero to ignore)
so all his flux flows through the air gap

then calculate amp-turns required to push that much flux through his 2mm air gap

then observe that many amp-turns are required of both windings , else there'd be a flux Φ3 and voltage in secondary
so whatever is I1
I2 must be greater by ratio 210 / 40

sound reasonable ?
What's the "st" denominator in cos argument? Is that 500 volts peak cosinewave at 300 hz ?
If so,
volts per turn = 500/210 X cos (300 X 2π X t) = dΦ1/dt
Φ1 = ∫volts per turn and ∫cos(ωt) = 1/(ω) X sin(ωt)
Φ1 = (500) / (210 ) X 1/(2π X 300) sin (2π X 300 X t)
Φ1 = 1.2631X10-3 sin(600π t) Webers, real close to 400π microWebers peak.
That's peak flux density of 1.2631 X10-3 Webers / 0.004m2 = 0.3158 Teslas ? That's reasonable.

We're not supposed to work the problem for homework posters, so i'll stop here.

See if you get 4π amps for I2.

old jim

Last edited: May 31, 2017
14. Jun 1, 2017

### jim hardy

PS i once had the good fortune of a big three phase transformer core in my garage to play with.
I performed almost that exact experiment , at 60 hz of course.
Shorting the secondary as shown indeed pushes (virtually) all the flux into center leg, only enough secondary current flows to make the required mmf.
I was at first surprised by the absence of sparks when i touched the wires together. It was an eye opener to magnetic circuits.

old jim