# Homework Help: Reluctance in a cast steel ring

1. Mar 27, 2016

### DevonZA

1. The problem statement, all variables and given/known data
A coil of 450 turns is wound uniformly round a cast steel ring. The ring has a crosssectional area of 750mm2 and a mean circumference of 600mm. A flux of 0.825mWb is produced in the ring. (Given that H = 800At/m at that flux).
a.
Calculate the reluctance of the ring
b.
Calculate the current required to produce the given flux.

2. Relevant equations
Φ=BA
μ=B/H
S=l/μoμrA
Φ=F/S

3. The attempt at a solution
a.
Φ=BA
B=Φ/A
Flux density in ring: 0.825x10-3/750x10-6
= 1.1T

μ=B/H
=1.1/800
=1.375x10-3

S=l/μoμrA
=600x10-3/1.375x10-3x4πx10-7x750x10-6
=4.63x1011A/Wb

b. Φ=F/S
=0.825x10-3=mmf/4.63x1011
mmf=0.825x10-3x4.63x1011
=381975000A

These answers look very wrong to me but I am unsure where I have gone wrong?

2. Mar 27, 2016

### jwinter

If you are familiar with Ohm's law, then reluctance is analogous to resistance in a circuit, with electro-motive force (voltage) being analogous to magneto-motive force (ampere-turns) and resulting current flow (amperes) being analogous to magnetic flux (webers). The B-field that you may be more familiar with is flux density (webers/square meter), and the H-field (ampere-turns/meter) is analogous to the electric field (volts/metre).

a) So from ohms law R=V/I which for a magnetic loop becomes Reluctance = [Ampere.turns/meter * meters] / [resultant flux].

In your case this comes to (800At/m * 600mm / 0.825mWb) => 776*103 At/Wb (allowing for the millimeters and milliwebers).

b) You have been given the magnetising field intensity (800 ampere-turns/meter) required to produce the flux and you know the mean circumference of the core (600mm) so that multiplying these together gives the ampere-turns (as we already did in (a)). Dividing by the turns then gives you the amperes:

In your case this comes to (800At/m * 600mm / 450t) => 1.07 Amps

Rather easy when you understand the quantities that you are dealing with.

3. Mar 28, 2016

### DevonZA

Thank you jwinter.
There are so many different quantities that it becomes confusing, thank you for your thorough explanation.

I am not getting to the same answer as you have provided in a).
S=800x600x10^-3/0.825x10^-3
=581818.1818

What am I doing wrong?

4. Mar 28, 2016

### jwinter

Nothing - my calculator must have made a mistake

5. Mar 28, 2016

### DevonZA

My answer of 581818.18A/Wb is correct?

6. Mar 28, 2016

### jwinter

Yes it is correct. But don't keep so many significant figures - it looks stupid. You were given at most a 3 significant figure input so only 3 significant figures are useful in the output.

7. Mar 28, 2016

### DevonZA

I.E 0.582μA/Wb?

8. Mar 28, 2016

### jwinter

Gosh you need to think for yourself at least a tiny bit if you are ever going to pass an exam and actually become useful in a subject! If you are not then why bother studying at all?

Do you know the difference between micro (μ) and mega?

9. Mar 29, 2016

### DevonZA

micro = 10^-6
mega=10^6

10. Mar 29, 2016

### jwinter

So you noticed that you got it wrong?

11. Mar 29, 2016

### DevonZA

Yeah I went the wrong way my bad 0.582MA/Wb