Understanding Magnetic H Field Strength: MMF, Effective Length & Reluctance

  • Context: Undergrad 
  • Thread starter Thread starter bootsnbraces
  • Start date Start date
  • Tags Tags
    Field Magnetic
Click For Summary
SUMMARY

The discussion focuses on the relationship between magnetic H field strength, magnetomotive force (MMF), effective length, and reluctance in magnetic circuits. It establishes that H field strength is calculated using the formula H = MMF / effective length, but emphasizes that reluctance must also be considered, particularly when comparing different geometries like a toroid versus a straight iron bar. The conversation highlights the mathematical equivalence between magnetic and electric fields, noting that while the equations are similar, the physical interactions differ significantly due to material properties.

PREREQUISITES
  • Understanding of magnetomotive force (MMF)
  • Familiarity with magnetic reluctance and permeance
  • Knowledge of the SI units for magnetic and electric fields
  • Basic concepts of electromagnetism and circuit analysis
NEXT STEPS
  • Study the principles of magnetic reluctance in different materials
  • Learn about the mathematical equivalence between electric and magnetic circuits
  • Explore the effects of material permeability on magnetic field strength
  • Investigate the applications of magnetic circuits in engineering design
USEFUL FOR

Students and professionals in electrical engineering, physicists, and anyone involved in the study of electromagnetism and magnetic circuit design will benefit from this discussion.

bootsnbraces
Messages
13
Reaction score
0
Hi all, I am trying to get my head around some experiment results but I am struggling a bit.
According to my notes H field strength is equal to MMF / effective length

However this doesn't take into account the effects of reluctance? Eg a toroid made from 2m of iron bar and a straight 1m length of iron bar will require the same mmf to generate the same H field regardless that in one case 50% of the path is through the low permitivitty of air?

can anyone shed some light on this for me please:confused:

thanks in advance guys and girls
 
Physics news on Phys.org
please -- someone's got to have a better understanding of this than me lol
 
Last edited by a moderator:
It's kind of similar to the way electric fields work. If you apply e.g. 1V to a parallel plate capacitor with a gap of 1mm you get 1000V/m independent of the permittivity of the dielectric.
Understanding the equations of electromagnetism becomes easier if you realize that the units and equations used in the SI system are defined such that the magnetic H field behaves mathematically equivalent to the electric E field. And B equivalent to D. That's just mathematically though and only in the SI system. The physical reality is basically the opposite of this, which can cause confusion. That has to do with the fact that matter interacts differently with magnetic fields than with electric fields. A dielectric material will weaken an E field. A ferromagnetic material will strengthen a B field.

Let's compare a few equations here.

Electric capacitance: C = q/V, measured in As/V
But you could also write: C = \Phi_D/V
Magnetic permeance: P = \Phi_B/NI, measured in Vs/A

So the magnetic permeance is mathematically equivalent to the electric capacitance.

E = EMF/l

H = MMF/l

D = \epsilon E

B = \mu H

In fact when calculating a magnetic circuit you can usually look at the mathematically equivalent electric circuit ( by replacing B with D, H with E, C with P, V with A, etc. ), do all the calculations with that and in the end transform the results back.
 

Similar threads

  • · Replies 25 ·
Replies
25
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 20 ·
Replies
20
Views
5K
  • · Replies 42 ·
2
Replies
42
Views
3K
  • · Replies 43 ·
2
Replies
43
Views
7K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K