Magnetostatic Field:

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CHAPTER 3
ELECTROMAGNETIC FIELDS THEORY

• Magnetostatic Field
• Magnetic force
• Magnetic circuit

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In this chapter you will learn

• Biot-Savats Law
• Ampere Circuital Law
• Magnetic flux density vector and magnetic
  potential vector
• Magnetic force
• Magnetic circuit (Not cover)
• Faradays Law
• Maxwells Equation

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Test 2

• 19th March 2009 (Thursday)
• 8.00 9.30 p.m.
• Chapter 3 Magnetostatic
• Biott-savart law,
• Ampere circuit law.

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Forces Due to Magnetic Fields

• Three ways in which force due to magnetic fields
  can be experienced
• Due to a moving charged particle in a B field
• On a current element in an external B field
• Between two current elements

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Force on a Charged Particle

• Electric force,
• Fe QE
• Magnetic field can only exert force on a moving
  charge
• Magnetic force, Fm, experienced by a moving
  charge Q with velocity u in a magnetic field B is
• Fm is perpendicular to both u and B
• Fe is independent of the velocity of the charge
  and can perform work on the charge and change its
  kinetic energy

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Force on a Charged Particle

• Fm is depend on the charge velocity and cannot
  perform work because it is at right angles to the
  direction of motion of the charge (Fm .dl 0)
• Fm does not cause an increase in kinetic energy
  of the charge
• Magnitude of Fm is generally small compared Fe
  except at high velocities

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Force on a Charged Particle

• So the total force on a moving charge in the
  presence of both electric and magnetic field is
  given by
• F Fe Fm
• Or
• If the mass of the charged a prticle moving in E
  and B field is m, by Newton's second law of
  motion,

Lorentz force equation
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Force on a Charged Particle
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Example

• A 1.0 nC charge with velocity u 100. m/sec in
  the y direction enters a region where the
  electric field intensity is E 100. V/m az and
  the magnetic flux density is
• B 5.0 Wb/m2 ax. Determine the force vector
  acting on the charge.

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Solution
Thus
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Exercise

• A 10. nC charged particle has a velocity
• v 3.0ax 4.0ay 5.0az m/s as it enters a
  magnetic field
• B 1000. T ay (recall that a tesla T Wb/m2).
• Calculate the force vector on the charge.

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Force on a Current Element

• Alternatively,
• Thus
• Which shows that an elemental current dQ moving
  with velocity u, producing convection current
  element dQu is equivalent to a conduction current
  Idl.

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Force on a Current Element

• Thus the force on a current element Idl in a
  magentic field B is found by replacing Q by Idl
• The B field that exerts force on Idl must be due
  to another element. B is external to the current
  element Idl.
• If the current I is through a closed path L or
  circuit, the force on the circuit is given by

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Force on a Current Element

• For surface current, the force exerted by an
  external B field is given by
• For a volume current element Jdv
• The magnetic field B is defined as the force per
  unit current element

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Force between Two Current Elements

• Consider two current elements I1dl1 and I2dl2.
• According to Biott-Savarts law, both current
  elements produce magnetic fields.

2
1
I2dl2
d(dF1 )
R21
I1dl1
I1
I2
I1
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Force between Two Current Elements

• From equation (3) the force on element I1dl1,
• From Biott-Savarts law,

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Force between Two Current Elements

• Hence,
• Then force on loop 1 due to current loop 2 is

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Example
A rectangular loop of wire in free space joins
points A(1, 0, 1) to B(3, 0, 1) to C(3, 0,
4) to D(1, 0, 4) to A. The wire carries a current
of 6 mA, flowing in the az direction from B to C.
A filamentary current of 15 A flows along the
entire z axis in the az direction.
a) Find F on side BC
b) Find F on side AB
c) Find Ftotal on the loop
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Solution
z

• Sketch the problem

D(1, 0, 4)
15 A
C(3, 0, 4)
A(1, 0, 1)
6mA
y
B(3, 0, 1)
x
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Solution (a)
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Solution (b)
The field from the long wire now varies with
position along the loop segment. We include that
dependence and write
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Solution (c)
by symmetry, the forces on sides AB and CD will
be equal and opposite, and so will cancel. This
leaves the sum of forces on sides BC (part a) and
DA, where
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Exercise

• A square loop of wire in the z 0 plane carrying
  2 mA in the field of an infinite filament on the
  y-axis, as shown below. Find the total force on
  the loop.

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Magnetic Materials
where
Is magnetic susceptibility of a medium
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Magnetic Materials
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Solenoid
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Solenoid
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Solenoid
At the center of the solenoid,
If
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Inductance

• Consider a circuit carrying current I
• Magnetic field B is produced which causes a flux
  to pass through each turn of the
  circuit.
• If the circuit has N identical turn, flux linkage
  ?, is define as

• flux linkage ? is define as the total magnetic
  flux linking a given circuit or conducting
  structure

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Inductance

• flux linkage ? is proportional to current I
• Where L is inductance of the circuit. Unit of L
  is henry (H) or webers/ampere.
• Since linkages are produced by the inductor
  itself, Self inductance
• A circuit that has inductance is called inductor

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Inductance

• Inductance measures how much magnetic energy is
  stored in an inductor (analogous to capacitor
  which store electric energy)
• Magnetic energy in an inductor
• And thus

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Inductance

• When we have two current carrying circuit,
• Four components of fluxes are produced

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Inductance

• Flux is flux passing through circuit 1
  due to current I2 in circuit 2
• If B2 is the field due to I2 and S1 is the area
  for circuit 1, then

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Inductance

• And mutual inductance, which is the ratio of flux
  linkage on circuit 1 to current I2, is
• Similarly,
• So

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Inductance

• Self inductance for both circuits are
• Total energy

and
where
and
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Exercise

• (1) Find the inductance of an ideal solenoid with
  200 turns,
• 0.20 m, and a circular cross section of radius
  0.01 m.
• (2) If the solenoid in Q(1) is placed inside a
  second coil of
• N2 400, r2 0.02 m, and 0.20 m. Find the
  mutual inductance assuming free-space
  conditions.

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Magnetic Energy

• Magnetostatic energy density wm (J/m3) is given
  by
• Hence
• Thus the energy in magnetostatic field in a
  linear medium is
• or

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Magnetic Circuits
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Magnetic Circuits

• Magnetomotive force (mmf) (in ampere-turns)
• Reluctance (in ampere-turns/weber)
• Mean length, S cross sectional area of the
  magnetic core
• reciprocal of reluctance is called permeance
• Ohms law

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Magnetic Circuits

• Magnetomotive force (mmf) (in ampere-turns)
• Reluctance (in ampere-turns/weber)
• Mean length, S cross sectional area of the
  magnetic core
• reciprocal of reluctance is called permeance
• Ohms law

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Kirchhoffs current and voltage laws

• for n magnetic circuit elements in series
• And
• For n magnetic circuit elements in parallel
• And

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Kirchhoffs current and voltage laws

• Unlike an electric circuit where current 1 flows,
  magnetic flux does not flow.
• Conductivity s is independent of current density
  J in an electric circuit whereas permeability µ
  varies with flux density B in a magnetic circuit.

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Force on magnetic materials

• Useful in electromechanical systems.
• Ignoring fringing, the magnetic field in air gap
  is the same as that in iron,
• To find the force between the two pieces of iron,
  calculate the change in the total energy that
  would result were the two pieces of the magnetic
  circuit separated by a differential displacement
  dl.

iron
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Force on magnetic materials

• The work required to effect the displacement is
  equal to the change in stored energy in the air
  gap
• where
• S is the cross-sectional area of the gap
• 2 accounts for the two air gaps
• negative sign indicates that the force acts to
  reduce the air gap

The force is exerted on the lower piece, not on
the current-carrying upper piece, giving rise to
the field
Thus
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Force on magnetic materials

• The attractive force across a single gap
• Attractive pressure (in N/m2) in a magnetized
  surface is
• which is the same as the energy density wm in the
  air gap.