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Static Magnetic Fields (2 Weeks)
5.2 Fundamental postulates of magnetostatics in free space
5.4 Biot-Savart Law and Applications
5.10 Inductances and Inductors
Intro.1
V. STATIC MAGNETIC FIELD
5.1 Law of conservation of magnetic flux
∇ .B = 0
Point form
∫ B. d S = 0
Integral form
B is called magnetic flux density (unit: tesla T) . The law
states that the net magnetic flux leaving a closed surface is
equal to zero (compare with the case in electric field).
Intro.2
5.2 Biot-Savart Law
• Source of magnetic field: current
used for calculating the magnetic field intensity dH at a
point P due to a current I flowing in an element dl.
dH =
Id l × r
4π r
3
I
dl
r
P
The resultant magnetic field intensity H at P is found by
integrating dH for all current elements. (Unit of H is
ampere/metre)
Intro.3
Relation between B and H:
B = µ o µ r H = µH
µo is permeability of free space (vacuum), µ r is the relative
permeability of the medium material, and µ is the
permeability of the medium material. (Unit of µ o and µ are
Henry/metre)
In free space,
µr = 1
B = µoH
Intro.4
5.3 Ampere’s Law (Ampere’s circuital law)
• an equivalent form of Biot-Savart Law
• Integral form: The line integral of H around any closed
path C is equal to the current Ienc enclosed by the path (in a
right-handed sense).
Ienc
∫ H. dl = I
C
enc
dl
• Differential form of Ampere’s law:
∇×H = J
Intro.5
The above form is very convenient in calculating the
magnetic field around cylindrically symmetric current
distributions.
Example: Show that the field at a point distance r from the
centre of a solid cylindrical conductor of radius a carrying a
current Io is:
rI o
aφ
2
2π a
Io
H =
aφ
2π r
H =
r<a
r>a
Intro.6
z
path of radius
r
y
x
Io
Solution: (a) For a path of radius r >a, applying Ampere’s law,
2π r H = I o
Io
H =
aφ
2π r
Intro.7
(b) For a path of radius r < a, the current enclosed
2
r
is equal to
I
2 o
a
(for uniform current density)
Applying the Ampere’s law,
r 2Io
2π r H =
a2
Ior
H =
aφ
2
2π a
Hφ
b
r
Intro.8
Example
Example 5-2
– Determine the magnetic flux density inside a closely wound
toroidal coil with an air core having N turns of coil and
carrying a current I. The toroid has a mean radius of b, and
the radius of each turn is a
I
I
Intro.9
Summary
Static Electric field
Static Magnetic field
Source
Charges
Current
Divergence equation
(Gauss’s law)
∇ ⋅ D = ρv
∫ D ⋅ ds = Q
s
Curl equation
∇⋅B = 0
∫ B ⋅ ds = 0
s
∇×E = 0
∇×H = J
∫ E ⋅ dl = 0
∫ H ⋅ dl = I
c
c
Intro.10
5.4 Inductance
Lij= flux linking
Φ ij due to current in Si
current in Si
L11 is called the self-inductance; L12 is called the mutualinductance
Φ12 = ∫ B1 ⋅ ds 2
s2
S2
S1 I
Φ12
L12 =
I1
Intro.11
Example: The coaxial cable shown consists of inner and
outer conductors of radii b and a separated by a nonmagnetic medium of permeability εo. Find the inductance
per unit length of the cable.
Solution: Let the inner conductor
carries a current I. Apply Ampere’s law
to a closed path of radius r where the
magnetic field magnitude is B(r).
r
B(r)
a
2π r
B (r )
=I
µo
µo I
B (r ) =
2π r
Intro.12
The total magnetic flux between the inner and outer conductor
(per unit length) is:
a
φ = ∫ B ( r ) dr
b
a
=
∫
b
µ o Idr
µoI  a 
ln  
=
2π r
2π
b
The self-inductance L per unit length is:
µo  a 
L= =
ln  
I 2π  b 
φ
Intro.13
Example
Example 5-8
I
I
Intro.14