Magnetic Circuits

Magnetic circuits are analogous to electric circuits:

Basic Materials:

! In electric circuits, copper is

$$ 10^{22}

$$ more conducting than air, thus we can say air is an insulating material. However, in magnetic circuits magnetic materials are just 4000-5000 times more "conducting", thus there is not a real insulator in magnetic circuits, and air-gaps should be considered in calculations.

Magnetic Field and Magnetic Force:

When electric charges are in motion(e.g. current conducting through a wire), they experience some forces apart from the electrostatic forces, which are called magnetic forces.

Flux, $$\Phi$$, (Wb=Volt.s)

A magnetic field characterized by lines of force called flux lines.

Flux Density, B ($$Wb/m^2 = T$$)

It's the density of magnetic flux per unit area. If the magnetic flux is uniform:

$$B=\frac{\Phi}{A} \quad (Wb/m^2)$$

Magneto-motive Force, MMF ($$A.t$$)

It is the excitation force in a magnetic circuit. The unit of MMF is Ampere turns(

$$ A.t

$$ ), or just Amperes (

$$ A

$$ ).

Magnetic Circuits & Maxwell Equations

$$\begin{align} \textrm{Gauss' Law}\quad &\nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0}\ \textrm{Gauss' Law ($\vec{B}$ Fields)} \quad &\nabla \cdot \vec{B} = 0\ \textrm{Faraday's Law} \quad &\nabla \times\quad \vec{E} = - \frac{\partial \vec{B}}{\partial t}\ \textrm{Ampere's Law} \quad &\nabla \times \vec{B} = \mu_0 \vec{J} + \mu_0\varepsilon_0\frac{\partial \vec{E}}{\partial t} \end{align}$$ In quasi-static fields:

1. Ampere's Law

The line integral of the magnetic field around a closed path is equal to the surface integral of current density of the same loop (or to total Ampere.turns).

$$\oint_C {\vec{H}.d\vec{\ell}} = \bigcirc!!!!!!!!!\iint_S \vec{J}.\vec{n}dA$ = \sum I_n$$ Magnetic field intensity(H) is independent of the material properties!

Calculate the magnetic field around a current-carrying conductor (as a function of the radius,r):

For r, H is constant due to symmetry:

$$\oint_C {\vec{H}.d\vec{\ell}} = \sum I_n$$$$H \oint_C {d\vec{\ell}} = H 2\pi r = I \rightarrow H=\frac{I}{2\pi r}$$ Calculate the MMF in the circuit below:

Assume, H is constant inside the magnetic core.

$$MMF= \oint_C {\vec{H}.d\vec{\ell}} = \sum I_n =NI$$

2. Gauss Law (for Magnetism)

In a specific region of magnetic field, the number of flux lines entering and leaving the region is zero.

$$\nabla \cdot {B} =0$$ or in integral form, net flux lines in a closed surface is zero

$$\bigcirc!!!!!!!!!\iint_S \vec{B}.\vec{n}dA = 0$$ Practical meaning: There are no net magnetic flux sources. (If you break a magnet into two smaller magnets, you don't have one south-pole and one north-pole magnet. Each magnet will have its own north and south pole) You can watch Sheldon Cooper on magnetic di-poles. You can also watch Richard Feynman on magnetic forces.

Ohm's Law (for a magnetic circuit)

In electric circuits, we have:

$$V=IR$$ In magnetic circuits(Magneto-motive-force:

$$ \mathcal{F}

$$ ):

$$\mathcal{F} = \Phi \mathcal{R}$$ **Magnetic Reluctance (

$$ \mathcal{R}

$$ )** Similar to the calculation of electric resistance (instead of conductvity use permeability)

$$\mathcal{R} = \frac{l}{\mu A}$$ where

$$ l

$$ is the length of the material,

$$ A

$$ is the cross-sectional area and **

$$ \mu

$$ is the magnetic permeability**.

Flux Density - Magnetic Field Relation ($$B=\mu H$$)

Magnetic flux density is found dividing the magnetic flux by area:

$$B=\frac{\Phi}{A}$$ MMF is the

$$ H

$$ multiplied by length (

$$ l

$$ ), (or similarly in the integral form):

$$\mathcal{F} = H l$$ Thus,

$$ \mathcal{F} = \Phi \mathcal{R}

$$ can be written as:

$$Hl = BA \frac{l}{\mu A}$$ which gives:

$$\vec{B} = \mu \vec{H}$$ It is similar to the relation between current density, electric field and conductivity (i.e.

$$ \vec{J}=\rho \vec{E}

$$ ).

Magnetic Permeability

The magnetic permeability of free space(vacuum) is:

$$\mu_0 = 4 \pi 10^{-7} \quad H/m$$ Instead of defining exact permeability of materials, it is common practice to define permeability relative to permeability of vacuum, which is called relative permeability (

$$ \mu_r

$$ ).

$$\mu = \mu_o . \mu_r$$ For example typical relative permeability of iron is around 4000 (

$$ \mu_r=4000

$$ ), which means iron condurc magnetic fields 4000 times easier compared to vacuum. The exact permeability of iron is:

$$\mu_{iron} = 4000\mu_0 = 16\pi10^{-4} H/m$$ You can also come across with Permeance, which is the inverse of permeability(

$$ \frac{1}{\mu}

$$ ).

(Permeability = conductivity, permeance = resistivity).

Exercise:

• Calculate the reluctance of the iron core in terms of $$ \mu_0 $$ , assuming $$ \mu_r=4000 $$ , $$ l=500 mm $$ , $$ A_c=2500mm^2 $$ .

!! Don't forget to convert into SI units (e.g. mm to m)

$$\mathcal{R}=\frac{l}{\mu0\mu_rA}$$$$\mathcal{R{core}}=\frac{0.5}{\mu_0.4000.2500.10^{-6}}=\frac{0.05}{\mu_0}$$

• Now, calculate the reluctance again with the air-gap added as below. Relative permeability of air is 1 ( $$ \mu_r=1 $$ ), and assume the air-gap is 3 mm.

!git s

We can think the circuit as two series resistances connected together.

Core reluctance

$$ \mathcal{R}_{core}

$$ almost stays same:

$$\mathcal{R}_{core}=\frac{0.499}{\mu_0.4000.2500.10^{-6}} \sim \frac{0.05}{\mu_0}$$ Air-gap reluctance (

$$ \mu_r=1

$$ ):

$$\mathcal{R}_{ag}=\frac{0.003}{\mu_0.2500.10^{-6}}= \frac{1.2}{\mu_0}$$ Total reluctance becomes:

$$\mathcal{R}{eq}=\mathcal{R}{core}+\mathcal{R}{ag}$$$$\mathcal{R}{eq}=\frac{0.05}{\mu_0}+\frac{1.2}{\mu_0}=\frac{1.25}{\mu_0}$$ We increased the equivalent reluctance 25 times, by adding just a small air-gap. Air-gap reluctances dominates the magnetic circuit, so it is usually acceptable to neglect core reluctance by assuming

$$ \mu_r = \infty

$$ .

• Calculate and compare the flux density (B) in the core, for each case assuming $$ N_{turns}=7 $$ and $$ i=5A $$ .

The equivalent circuit diagrams are shown above. Let's calculate the total flux in each case:

$$\mathcal{F} = \Phi \mathcal{R}$$$$\Phi = \frac{NI}{\mathcal{R}{eq}}$$$$\Phi{without-gap} = \frac{NI \mu0}{0.05}$$$$\Phi{with-gap} = \frac{NI \mu_0}{1.25}$$ putting in the numbers:

$$\Phi{without-gap} \sim 0.88 mWb$$$$\Phi{with-gap} \sim 0.035 mWb$$ Flux density can be found by dividing the magnetic flux by area.

$$B=\frac{\Phi}{Area}$$$$B{without-gap} =\frac{0.88mWb}{2500.10^{-6} m^2}=0.352 \; T$$$$\Phi{with-gap} \frac{0.035mWb}{2500.10^{-6} m^2}=0.014 \; T$$ Increasing reluctance (by adding air-gap_ reduces the magnetic flux.