MAGNETOSTATICS The magnetic field produced by steady currents or permanently magnetized material is called

magnetostatics. On top of the electrostatic force experience by charges in an electric field, moving

charges in an external magnetic field also experience a magnetic force:

Proportional to the charge

Perpendicular to both the magnetic field density and velocity of the charge

The total electromagnetic force on a charge q is then given by Lorentz’ force equation:

FUNDAMENTAL POSULATES OF MAGNETOSTATICS IN FREE

SPACE There are two fundamental postulates of magnetostatics:

Where, is the permeability of free space.

These two postulates are analogous to the electrostatic ones, and by a similar nature we can apply the

divergence and stokes’ theorem to find the integral form of the postulates.

The first postulate is the law of conservation of magnetic flux which indicates that magnetic flux lines

always close upon themselves – lines of flux start at the north pole and end at a south pole. This means

that isolated magnetic poles cannot exist.

The second postulate leads to Ampere’s law in which the circulation of the magnetic flux density in free

space around any closed path is equal to the permeability of free space multiplied by the total current

flowing through the surface bounded by the path:

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Note that if the problem had been a thin tube carrying a surface current, then inside the tube, and

outside the tube, the total surface current is giving a magnetic field density of:

Example

An infinitely long, straight conductor, with a circular cross section of radius b carries a steady

current I. Determine the magnetic flux both inside and outside the conductor

a) Inside the conductor:

Using Ampere’s law and noting that the magnetic flux is parallel to the contour and also the fact

that :

b) Outside the conductor:

The magnetic field lines

around a wire form circles.

Hence the magnetic field is

the same everywhere on a

circular path centred on the

wire and lying in a plane

perpendicular to the wire. If

we vary the current and

distance from the wire, we

find that B is proportional to

the current and inversely

proportional to the distance

from the wire.

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Example

Determine the magnetic flux density inside a closely wound toroidal coil with an air core having

N turns and carrying a current I. The toroid has a inner radius of a and an outer radius of b.

Cylindrical symmetry ensures that has only a component and is constant along any circular

path about the axis of the toroid.

Example

Determine the magnetic flux density inside an infinitely long solenoid with air core having n

closely wound turns per unit length and carrying a current I.

Note that there is no magnetic field outside of the solenoid. We construct a rectangular contour

of length L that is partially inside and partially outside the solenoid.

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VECTOR MAGNETIC POTENTIAL Since is solenoidal, the vector algebra null identity gives us:

Where the vector field A is the vector magnetic potential in (Wb/m). Thus, if we know the magnetic

vector potential of a current distribution, we can find . However, the definition of a vector requires us to

know the curl and its divergence. We must thus choose what is by using the second postulate to

derive Poisson’s equation

By the vector identity:

Choosing yields the divergence of a gradient of the magnetic vector potential:

Which has a general solution of:

Which is analogous to the solution to the scalar Poisson’s equation

, which gives

Example

A thin, infinitely large sheet lying in the yz plane carries a current of linear current density . The

current is in the y direction, and represents the current per unit length measured along the z

axis. Find the magnetic field near the sheet.

In this situation, an infinitely large sheet will form an almost continuous field parallel to the

sheet, hence we use a small rectangular path. We note that the sides of w length have an element

of perpendicular to the field, and do not contribute to the line integral. On the two other

sides however, the line integral is non zero and are the same so:

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MAGNETIC FLUX We define the magnetic flux (Wb) as the total of the magnetic flux density over a given area:

Proof:

Since and we can apply Stokes’ theorem:

BIOT-SAVART LAW In many applications we are interested in determining the magnetic field due to a current

carrying circuit where we cannot exploit the symmetry required for Ampere’s law. In this case

we use Biot Savart Law:

Proof:

For a thin wire of length dl, . This means that the general solution of the vector magnetic

potential is given by:

Now:

The first term goes to zero since the curl of a vector in the primed coordinates is independent of the

unprimed coordinates. Also:

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Example

A direct current I flows in a straight wire of length 2L. Find the magnetic flux density at a point

located at a distance r from the wire in the bisecting plane.

Use the alternative form of Biot Savart’s law:

Now:

Let

Note that

Example

Calculate the magnetic field at O for the current carrying wire segment shown in the figure. The

wire consists of two straight portions and a circular arc of radius a, which subtends an angle .

The straight segments have an element ds which is parallel to . Hence we only worry about the

curved segment whereby the vectors d and perpendicular:

Applying the right hand thumb rule, the magnetic field goes into the page.

Edmund Li

MAGNETIC DIPOLE A magnetic dipole is a closed circulation of electric current, such as a loop of wire. It can be shown that

from the diagram:

Where is the dipole moment given by:

These equations are of the same form as that seen for the electric dipole where

and

.

Example

Consider a circular wire loop of radius a located in the yz plane and carrying a steady current I as

shown. Calculate the magnetic field at an axial point P a distance x from the centre of the loop.

Due to the symmetry of the ring, the y components of the magnetic fields are cancelled.

Using geometry:

We see that x I, a, are constant thus:

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We find that the magnetic dipole moment is a vector:

Whose magnitude is the product of the current in and area of the loop

Whose direction is the direction of the thumb when the finger of right hand is the direction

of current.

MAGNETISATION Orbiting electrons of atom cause circulating currents and form microscopic magnetic dipoles with

associated magnetic dipole moments . In the absence of an external magnetic field, these magnetic

dipoles have random orientation so the net magnetic moment is zero.

When an external magnetic field is applied, the magnetic dipole moments are aligned and induce an

additional magnetic moment due to a change in the orbital motion of the electrons. We define the

magnetization vector for a material with magnetic dipole moments of :

From this, we also can produce the vector magnetic potential:

Which is more commonly expressed as:

We thus find that the magnetization vector is equivalent to both a volume current density and surface

current density:

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MAGNETIC FIELD INTENSITY The macroscopic effect of magnetization can be studied by incorporating the equivalent volume current

density:

We define the magnetic field intensity as:

This means that:

Where is the volume density of free current. By taking the surface integral and Stokes’s theorem:

Which is another form of Ampere’s Law – the circulation of the magnetic field intensity around any closed

path is equal to the free current flowing through the surface bounded by the path.

Edmund Li

PERMEABILITY OF MAGNETIC MATERIAL When the magnetic properties of the medium are linear and isotropic, the magnetization is directly

proportional to the magnetic field intensity:

Where is known as the magnetic susceptibility, and is dimensionless. Now substitution this equation

into

yields:

Where is defined as the relative permeability of the medium while is the absolute permeability of

the medium measured in H/m. Depending on the value of we can say that:

: the materials are diamagnetic and the magnetic moment is zero in the absence of an

external field e.g Gold, silver, copper

: The magnetic moments in the paramagnetic material are non zero but small

: The magnetic moments in the ferromagnetic material are aligned in the magnetic

domains.

HYSTERESIS LOOP When an external field is applied to a ferromagnet, the dipoles in the domains align themselves with the

external field. Even when the external field is removed, party of the alignment will be retained and the

material will become magnetized. To demagnetize the material, it would be necessary to apply a

magnetic field in the opposite direction.

When a strong field is magnetized up to point a along the dotted line, the domain wall movements

become irreversible, and domain rotatation toward the direction of the applied field will occur. When the

applied field is reduced to zero from point a, the B-H relationship will follow along the solid line to path b.

If we apply the applied field in the opposite direction, we then follow the path to cd.

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MAGNETIC CIRCUITS In this section, we focus out approach on the following sets of equations:

is known as he magnetomotive force (mmf) and is analogous to the emf in an electric circuit. Thus it is

not a force but rather a measure of strength of the magnetic source. It is measured in A-t.

Example

Assume an N turns of wire wound around a toroidal core of ferromagnetic material with permeability

. The core has a mean radius of and a circular cross section of radias , and a narrow air gap

of length . A steady current flows in the wire. Determine:

a) The magnetic flux density

b) The magnetic field intensity in the core

c) The magnetic field intensity in the air gap

a) We start with Ampere’s Law:

Since the magnetic field intensity vector and the contour are in the same direction then we

simplify the equation to:

Now ignoring fringing effects, the flux in the air gap must be the same as in the core, so we

can say :

Since the permeability in the core and air gap are different, then the magnetic field intensity

will also be different:

Thus we write ampere’s law as:

b)

c)

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Note that if the cross section of the core is much smaller than the mean radius of the toroid, then:

And we can write:

Where is the reluctance ( ) of the ferromagnetic core which represents the dc resistance of

straight piece of homogeneous material with a uniform cross section area S. Thus, we may compare these

results to that of electric circuits and make analogies:

Moreover, there are equivalents to the KCL and KVL:

1. Around a closed path in a magnetic circuit, the algebraic sum of ampere turns is equal to the

algebraic sum of the products of the reluctances and fluxes

2. The algebraic sum of all the magnetic fluxed flowing through a junction in a magnetic circuit is

zero.

Using these concepts, we can better solve magnetic circuits with delving necessarily into the fundamental

formulas e.g the postulates and Ampere’s Law.

Edmund Li

Note that it is not always possible to use these methods to solve for and using the relationship

and . This occurs when is not the same or when is a non linear function of . A graphical

approach would then be better. The next example shows how such an approach can be taken.

Example

Consider the magnetic circuit shown which has steady currents flowing in the windings of

turns respectively. The core has cross sectional and permeability . Determine the magnetic flux,

in the centre leg.

We draw the magnetic circuit in terms of an electric circuit model and note that:

For loop 1:

For loop 2:

Solving simultaneously yields:

Example

Consider now that the windings are on the centre leg. Determine the magnetic flux density.

For loop 1:

Solving simultaneously then:

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Example

By determining the load line, find the intersection of the load line with the BH curve so that and

can be found for the core material in the setup shown.

By Ampere’s Law and noting that

This is in the form of the straight line, with the slope –

and y intercept of NI. The intersection of

this load line with the BH curve can now be found.

Example

A toroidal magnetic circuit contains an air gap of length 1.5 mm. The mean length of the steel core is

0.3 m with a cross sectional area of with a mmf=NI=1000.

Assuming no fringing and leakage, find the flux density in the steel.

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BOUNDARY CONDITIONS Consider the pillbox. For a linear and isotropic media and using the first postulate with the divergence

theorem:

Now as then:

If we consider the tangential component and

Note that is the surface urrent density through the interface normal to the closed contour, and for

materials with finite conductivity, .

Example

An electromagnet consists of three uniform sections:

a) Length of 8 cm and cross sectional area of 0.5

b) Length of 6 cm and a cross sectional area of

c) Air gap of length 0.5 mm and cross sectional area of

Find the current in a 3000 turn coil to produce a flux density of 0.3 T in the airgap. Use the BH curve

of cast steel and neglect any fringing.

Since the cross sectional areas are not the same, then . However, flux is constant so starting

from the air gap:

In Section A:

Using the BH curve, corresponds with . Hence

In section B:

Edmund Li