6. THE STATIC MAGNETIC FIELD 6-1 LORENTZ FORCE EQUATION
Transcript of 6. THE STATIC MAGNETIC FIELD 6-1 LORENTZ FORCE EQUATION
Chapter 6 6-1 Proprietary of Prof. Lee, Yeon Ho
6. THE STATIC MAGNETIC FIELD 6-1 LORENTZ FORCE EQUATION
Lorentz force equation ( )e m q= + = + ×F F F E v B
Example 6-1 An electron has an initial velocity o o yv=v a in a uniform magnetic flux density o zB=B a . (a) Show that the electron moves in a circular path.
(b) Find the radius or of the circular path in terms of electron charge e , electron mass m and the initial speed ov .
Solution (a) The magnetic force on the electron is m e= ×F v B . Because mF is perpendicular to v , it
only changes the direction of v , and corresponds to the centripetal force on the electron. (b) The magnetic force on the electron m o oev B ρ=F a ( )0e <
The force on the electron, mass× acceleration d d dm mdt d dt
φ=
φv v
where tφ = ω and ω is angular velocity. Rewrite it using ov φ=v a and /d dφ ρφ = −a a
( ) ( ) ( )oo
d v d tm mv
d dtφ
ρ
ω= − ω
φ
aa
Chapter 6 6-2 Proprietary of Prof. Lee, Yeon Ho
From two equations
oeBm
−ω =
Using o ov r= ω
oo
o
mvre B
= −
where 0e < . 6-2 BIOT-SAVART LAW
Biot-Savart law
24
Idd
π′×
=l a
H R
R : ′= −r rR
For a steady current flowing in a wire
24C
Id′
′ ×= ∫
l aH R
Rπ (6-7)
Chapter 6 6-3 Proprietary of Prof. Lee, Yeon Ho
Id ′l in terms of J and sJ sId dv ds′ ′ ′= =l J J
For a current density J
24V
dvπ′
′×= ∫
J aH R
R (6-9)
For a surface charge density sJ
24s
S
dsπ′
′×= ∫
J aH R
R (6-10)
All the unit vectors in the unprimed coordinates The integration in the primed coordinates.
Example 6-2 H of a steady current in a straight wire
Determine H at a point ( ); , ,P x y z due to an infinitely long straight conducting wire of negligible thickness carrying a steady current I and lying along z -axis.
Solution The line current has cylindrical symmetry, and Biot-Savart law only gives dH in φa direction. So we expect ( )H φ φ= ρH a The translation vector, : ' zzρρ ′= − = −r r a aR The vector differential length, : zd dz′ ′=l a
Chapter 6 6-4 Proprietary of Prof. Lee, Yeon Ho
From the definition
( )3 3/22 2
( )44
z zz z
z z
Idz z dzI
z
′ ′=+∞ =+∞ρ φ
′ ′=−∞ =−∞
′ ′ ′× ρ − ρ= =
ππ ′ρ +∫ ∫
a a a aH
R
(6-11)
or
2I
φ=πρ
H a
(6-12)
H also has cylindrical symmetry, being independent of φ and z .
• For a finite line current
( )
1
1
2
2
3/2 2 22 24 4
z zz
zz z
dzI I zzz
′ ′=′ φ
φ′′ ′=
′ ′ρ= =
π πρ ′ρ +′ρ +∫
aH a
Referring to Fig. 6-4
[ ]1 2cos cos4I
φ= θ + θπρ
H a (6-13)
This formula can be applied even for 1θ
or 2θ larger than 90o .
Chapter 6 6-5 Proprietary of Prof. Lee, Yeon Ho
Example 6-3 Find H , at a point ;( , ,0)p a b in the first quadrant, due to an infinite right-angled wire carrying a steady current I .
Solution
For 0y < : 1θ = α , 2 0θ = and aρ =
[ ] ( )1 cos 1
4 zIa
= α + −π
H a
For 0x < : 1 0θ = , 2θ = β and bρ =
[ ] ( )2 1 cos
4 zIb
= + β −π
H a
In Fig. 6-5
( )2 2
cos sin 90o ba b
α = − α − = −+
( )2 2
cos sin 90o aa b
β = − β − = −+
.
The magnetic field intensity at p is
( ) ( )
( )
1 2 2 2 2 2
2 2
1 14 4
4
z z
z
I b I aa ba b a bI a b a bab
⎡ ⎤ ⎡ ⎤= + = − + − + − −⎢ ⎥ ⎢ ⎥π π+ +⎣ ⎦ ⎣ ⎦
⎡ ⎤= + − +⎣ ⎦π
H H H a a
a
Chapter 6 6-6 Proprietary of Prof. Lee, Yeon Ho
Example 6-4 Determine H at a point ( ); 0,0,p b due to a circular wire of radius a centered at the origin in xy -plane, and carrying a steady current I.
Solution
In cylindrical coordinates : zb a ρ= −a aR
d ad φ′ ′= φl a . From Eq. (6-7)
2
2 30
22 2
3 30 0
( )4 4
4 4
z
C
z
Ia d b aI d
Iab Iad d
′φ = π φ ρ
′ ′φ =
′ ′φ = π φ = πρ
′ ′φ = φ =
′′ φ × −×= =
π π
′ ′= φ + φπ π
∫ ∫
∫ ∫
a a al aH
a a
R
R R
R R (6-14)
The first integral cancels to zero due to ( ) ( )ρ ρ′ ′φ = φ = − φ = φ + πa a .
2
3/22 22z
Ia
a b=
⎡ ⎤+⎣ ⎦H a
(6-15)
Chapter 6 6-7 Proprietary of Prof. Lee, Yeon Ho
6-3 AMPERE’S CIRCUITAL LAW
The closed line integral of H is equal to the total current enclosed by the path of integration.
Cd I=∫ H li : Right-hand rule for C and I
• The total enclosed current I
The continuity equation is
0S
d =∫ J si (6-17)
The total current enclosed by C is equal to the total current across the surface bounded by C .
• Ampere’s circuital law
C Sd d=∫ ∫H l J si i
: Right-hand rule for C and S
Applying Stokes’s theorem
S Sd d∇× =∫ ∫H s J si i
(6-19)
The point form of Ampere’s circuital law ∇× =H J (6-20) To determine H by Ampere’s circuital law, the closed path C should be chosen such that the magnetic field intensity H is at least constant on the path C .
Chapter 6 6-8 Proprietary of Prof. Lee, Yeon Ho
Example 6-5 Find H , by Ampere’s circuital law, due to an infinitely long straight wire lying along z -axis and carrying a steady current I.
Solution
The cylindrical symmetry → H independent of φ and z Biot-Savart law. → dH due to Id ′l only in φa direction We expect ( )Hφ φ= ρH a : H φ depends only on ρ On C of a radius ρ
( )2 2
0 0Cd H d H d I
π π
φ φ φ φφ= φ== ρ φ = ρ φ =∫ ∫ ∫H l a ai i
2IHφ φ φ= =πρ
H a a
The same result with that by Biot-Savart law
Chapter 6 6-9 Proprietary of Prof. Lee, Yeon Ho
Example 6-6 An infinitely long coaxial cable carries a uniform steady current I . The radii of the cylindrical surfaces are , a b and c , respectively. Find H in the region 0ρ > by Ampere’s circuital law.
Solution The uniform current → Cylindrical symmetry Current elements at 1φ and 1−φ → Only H φ -component We can expect ( )H φ φ= ρH a : H φ is constant on the circle 1C In the region 0 a< ρ ≤ , ( )
1
2
02
Cd H d H
π
φ φ φ φφ== ρ φ = π ρ∫ ∫H l a ai i
The current enclosed by 1C , 2
2 Iaπρπ
The result
22IHaφ φ φ
ρ= =
πH a a
: ( 0 a< ρ ≤ )
The enclosed current I : (a b≤ ρ ≤ )
2 2
2 2
cIc b
− ρ−
: (b c≤ ρ ≤ )
0 : ( cρ ≥ ) The result
2I
φ=πρ
H a
(a b≤ ρ ≤ ) (6-23b)
Chapter 6 6-10 Proprietary of Prof. Lee, Yeon Ho
2 2
2 22I cc b φ
− ρ=
πρ −H a
(b c≤ ρ ≤ ) (6-23c)
0=H ( cρ ≥ ) (6-23d) Example 6-7
Find H of a toroidal core having N turns of a wire carrying a steady current I . The toroid has the inner radius of a and the outer radius of b .
Solution
Rotation symmetry about z -axis → Constant H on 1C I across 1φ = φ plane → Hρ and zH , samll Large enclosed current NI → H φ , large An ideal toroidal coil → H φ φ=H a inside 0=H outside. In the region a b< ρ <
1
2
02
C Cd H d H d H
π
φ φ φ φ φ= ρ φ = ρ φ = π ρ∫ ∫ ∫H l a ai i
The total enclosed current isNI
2NIH φ = πρ
2NIHφ φ φ= =πρ
H a a : (a b< ρ < )
0=H (No current enclosed) : aρ < and bρ >
Chapter 6 6-11 Proprietary of Prof. Lee, Yeon Ho
6-4 MAGNETIC FLUX DENSITY
In free space o= μB H : 74 10o
−μ = π × , permeability of free space The total magnetic flux
SdΦ = ∫ B si .
An infinite line current I → H , concentric circles An arbitrary distribution of I → H , always closed. ↑ Linear property Gauss’s law for the magnetic field
0S
d =∫ B si Point form of Gauss’s law 0∇ =Bi
• Maxwell’s equations for the static electric and magnetic fields
00
v∇ = ρ
∇ × =∇ =∇ × =
DEBH J
i
i
In integral forms
0
0
vS V
C
S
C S
d dv
d
d
d d
= ρ
=
=
=
∫ ∫∫∫∫ ∫
D s
E l
B s
H l J s
i
i
i
i i