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8.2.3 Conservation of Momentum
The force on a charged particle:
r dpr mech
F
F=
dt
fem=(d/dt)(µεS)
T
The force equation in the system can be written as
r ⎞
r dpr mech
t r
t r
d⎛
d r
= ∫ T ⋅ da − ε 0 µ0 ∫ Sdτ = − ⎜⎜ ∫ ε 0 µ0 Sdτ ⎟⎟ + ∫ T ⋅ da .
F=
dt ⎝ V
dt V
dt
S
⎠ S
The momentum of the electromagnetic fields stored in the volume of the system is
r
r
pem = ∫ ε 0 µ0 Sdτ .
V
The electromagnetic momentum per unit volume is
r
r
r S
r
pem
Pem =
= ε 0 µ0 S = 2 .
V
c
(
r 1 r r
E×B
The Poynting vector: S =
µ0
1.
2.
)
For the energy of a volume system, the Poynting vector is the energy per unit
time, per unit area, transported through bounding syrfaces by the fields.
r r
d
(
)
u
+
u
d
τ
+
S
mech
em
∫S ⋅ da = 0
dt V∫
(The direction of Poynting vector represents the way that energy is transported
through.)
For the momentum of a volume system, the Poynting vector divided by square
of the electromagnetic field speed is the electromagnetic momentum per unit
volume.
r
r S
r
Pem = ε 0 µ0 S = 2
c
The force equation can be given as
r
r ⎞
t r
dpmech d ⎛
+ ⎜⎜ ∫ ε 0 µ0 Sdτ ⎟⎟ = ∫ T ⋅ da .
dt
dt ⎝ V
⎠ S
r
r ⎞
r ⎞
t r
t
dpmech d ⎛
d⎛ r
⎜ ∫ Pmech + ε 0 µ0 S dτ ⎟ = ∫ ∇ ⋅ T dτ
+ ⎜⎜ ∫ ε 0 µ0 Sdτ ⎟⎟ = ∫ T ⋅ da Æ
⎟
dt
dt ⎝ V
dt ⎜⎝ V
⎠ S
⎠ V
(
)
(
)
Be written in differential form:
r
t
r
t
d r
d r
Pmech + ε 0 µ0 S = ∇ ⋅ T Æ
Pmech + Pem = ∇ ⋅ T
dt
dt
(
)
(
)
t
The negative Maxwell stress tensor − T is the momentum flux density.
− Tij is the momentum in the i direction crossing the surface oriented in the j
direction.
1
1
⎞
⎞ 1 ⎛
⎛
The Maxwell stress tensor: Tij = ε 0 ⎜ Ei E j − δ ij E 2 ⎟ + ⎜ Bi B j − δ ij B 2 ⎟
2
2
⎠
⎠ µ0 ⎝
⎝
t
1. For the force of a volume system, the Maxwell stress tensor T is the
electromagnetic stress (force per unit area) acting on the bounding surface.
r
t r 1 ∂ r
F = ∫ T ⋅ da − 2 ∫ S dτ
c ∂t V
S
2.
For the momentum of a volume system, the negative Maxwell stress tensor
t
− T is the flow of momentum (the momentum current density) transported by
the fields.
r
t
d r
Pmech + Pem + ∇ ⋅ − T = 0
dt
E
b
Example: A long coaxial cable, of length l, carries a uniform
B
charge per unit length λ and a current I (the current flows a
down the surface of the inner cylinder, radius a, and back along
the outer cylinder, radius b). Find the electromagnetic
momentum stored in the fields.
(
r
E=
)
( )
r
1 r r
λI
r µI
λ
E×B =
zˆ
ˆr , B = 0 φˆ Æ S =
4π 2 ε 0 r 2
µ0
2πε 0 r
2πr
The power transported is
λI
λI
2πrdr =
ln (b / a ) = IV .
2
2
4π ε 0 r
2πε 0
a
b
r r
r
P = ∫ S ⋅ da = ∫ S ⋅ zˆ 2πrdr = ∫
The power is delivered from the battery to the resistor (provided by the battery and
consumed to the resistor).
The momentum in the field is
⎞
µ λIl
zˆ ⎟⎟2πrldr = 0 ln(b / a )zˆ
2π
⎠
r
⎛ λI
r
pem = µ0ε 0 ∫ Sdτ = µ0ε 0 ∫ ⎜⎜ 2 2
⎝ 4π ε 0 r
The momentum is balanced by a momentum when the current is turned on.
Exercise: 8.6
8.2.4 Angular Momentum
As the electromagnetic fields exhibit in the space, they carry energy
uem =
ε0E2
2
+
B2
2µ0
and momentum
r
r
r
r r
S
Pem = 2 = µ0ε 0 S = ε 0 E × B .
c
According to Newton’s rule, the electromagnetic fields should carry angular
momentum
r
r r
r r r
lem = r × Pem = ε 0 r × E × B .
z
Example: Imagine a very long solenoid with radius R, n
solenoid
turns per unit length, and current I. Coaxial with the
solenoid are two long cylinder shells of length l – one,
inside the solenoid at radius a, carries charge +Q,
uniformly distributed over its surface; the other, outside
the solenoid at radius b, carries charge –Q. When the
current in the solenoid is gradually reduced, the
cylinders begin to rotate, where does the angular
momentum come from?
r
Q
E=
rˆ a < r < b
2πε 0 rl
(
)
[ (
)]
r
B = µ0 nIzˆ
(
)
r
r r
r
r
µ nIQ ˆ
1 r
φ Æ lem = rr × Pem = − µ0 nIQ zˆ
Pem = 2 S = ε 0 E × B = − 0
2πrl
c
2πl
r
r
µ nIQ
µ nIQ 2
πR 2 − πa 2 lzˆ = − 0
R − a 2 zˆ
Lem = ∫ lem dτ = − 0
2πl
2
(
)
(
)
The angular momentum of the outer cylinder (-Q) is
r
⎛ µ nIQR 2
⎛ µ nIQ ⎞
⎛ µ nIQ ⎞ 2
Lb = ∫ ⎜ − 0
zˆ ⎟dτ = ⎜ − 0
zˆ ⎟ πR l = ⎜⎜ − 0
2πl ⎠
2πl ⎠
2
⎝
⎝
⎝
(
)
⎞
zˆ ⎟⎟ .
⎠
Considering the negative angular momentum of -Q (in comparison with positive
angular momentum of +Q) and the total angular momentum −
µ 0 nIQ
2
(R
2
)
− a 2 zˆ , the
angular momentum of the inner cylinder (+Q) is
r ⎛ µ nIQa 2
La = ⎜⎜ 0
2
⎝
⎞
zˆ ⎟⎟ .
⎠
When the current is turned off, the changing magnetic field induces a circumferential
electric field, given by Faraday’s law:
2πrE = −
(
)
(
)
d
(µ0nI )πr 2
dt
1
dI
Æ E = − µ0 n rφˆ r < R
dt
2
d
dI R 2 ˆ
1
2
2πrE = − (µ0 nI )πR Æ E = − µ0 n
φ r>R
dt
dt r
2
The torque on the inner cylinder is
r
r r r
1
1
dI ˆ ⎞
dI
d
L
⎛
τ a = r × F = (arˆ ) × ⎜ − Q µ0 n aφ ⎟ = − µ0 na 2Q zˆ = − a .
2
2
dt
dt
dt
⎝
⎠
The torque on the outer cylinder is
r
⎛
r r r
dI R 2 ˆ ⎞ 1
dI
dLb
1
2
τ b = r × F = (brˆ ) × ⎜⎜ − (− Q ) µ0 n
φ ⎟⎟ = µ0 nR Q zˆ = −
.
dt
b
dt
dt
2
2
⎝
⎠
Exercise: 8.8(a)