Lecture14 - Magnetic Circuits

Lecture 14 Magnetic Circuits An electromagnetic-mechanical system is formed with the following blocks. H i

Ampere’s Law

Material properties

B, Φ

Faraday’s Law

e

Ohm’s Law, KCL, and KVL

Lorenz’s Law and energy conservation

We know that the integral form of Ampere’s Law is

Newton’s and Lagrange’s Laws

L∫ H idl = ∫ J id S . C

Speed, displacement

Considering the

S

next figure, an iron core with cross-sectional area A and with mean length of perimeter l, several assumptions were made:

1. H is tangential to the path. 2. H
H is constant throughout the path. 3. There is no field outside the iron. Since there are N turns in the coil then Hl = Ni . Consider now an iron core with an air gap.

Figure 1 - Iron core with air gap

Let lc be the mean length of the iron core and g the length of the air gap, then by Ampere’s Law,


∫ H idl = H l

c c

+ H g g = Ni

C

We need to relate the field in the core and in the gap with a common variable in order to solve the equation. The magnetic field density B is related to the magnetic field intensity H in magnetically linear materials by, B = µH

Where µ is the permeability of the material. A typical curve is shown in Figure 2. The permeability of the material is defined as, µ = µr µ0 Where µ 0 = 4π × 10−7 Wb / A ⋅ turn ⋅ m or in henrys per meter and µr is the relative permeability with ranges from 2000 to 80,000 for materials used in transformers and rotating machines.

Figure 2 - B-H curve

The magnetic flux is defined as, Φ = ∫ Bid S S

If Bc = µcHc is uniformly distributed in the core then Φ = BcAc. In the air gap the flux tends to “fringe” outwards.

Figure 3 - Fringing fields

The cross-section in the air gap is therefore bigger than in the core Ag > Ac. This is called the fringing effect. But there is continuity of flux, Φ core = Φ gap

Hence,

Φ = Bc Ac = µ c H c Ac Φ = Bg Ag = µ 0 H g Ag Then we can express Ampere’s Law as,

Ni = H c lc + H g g =

lc g Φ+ Φ µ c Ac µ 0 Ag

And Φ=

Ni lc g + µ c Ac µ 0 Ag

Let’s consider now an iron core with two coils as shown in Figure 4.

Figure 4 - Iron core with two coils

Now current i1 is entering the surface created by the contour C and i2 is going out, therefore,

∫ J id S = N i − N i 11

S

2 2

It follows that,

Φ


∫ H idl = Hl = µ A l C

Then,

N1i1 − N 2i2 =

lΦ µA

For complicated magnetic circuits we need a more systematic approach. To do this let us look an analogy between electric and magnetic circuits. Electric Circuit i Vs = voltage (emf) Resistance along path ( R) Ohm’s law (V=iR) KCL ∑ ik = 0 k

KVL

∑V

k

k

=0

Magnetic Circuit Φ Ni = mmf Reluctance along path ( ℜ ) Mmf drop across path Flux continuity ∑ Φ k = 0 k

Ampere’s law

∑ mmf k

k

=0