Download Class 7 in Electrodynamics

Class 7 in Electrodynamics
Based on course by Yuri Lyubarsky and Edited By Avry Shirakov
Department of Physics, Ben-Gurion University, Beer-Sheva 84105, Israel
This exercise pool is intended for an undergraduate course in “Electrodynamics 1”.
Conservation Laws
Energy of the electromagnetic field:
Z 1
1 2
0 E 2 +
B dV
Uem =
2 V
µ0
Pointing’s theorem:
Z
I
d
1
1 2
1
dW
=−
0 E 2 +
B dV −
(E × B) ds
dt
dt V 2
µ0
µ0 S
Pointing’s vector (energy per unit time, per unit area):
S=
1
E×B
µ0
so, Sds -is energy per unit time through ds (energy flux) and we can call S as an energy flux density;
I
dUem
dW
=−
−
Sds
dt
dt
S
Maxwell’s stress tensor is defined by the following (the force per unit area acting on a surface, Tij is the force per
unit area in the ith direction acting on an element of surface in the jth direction):
1
1
1
Tij = 0 Ei Ej − δij E 2 +
Bi Bj − δij B 2
2
µ0
2
the force per unit volume can be written as:
f = ∇ · T − 0 µ0
∂S
∂t
and the total force on the charge in volume V is;
I
Z
d
F=
T · ds − 0 µ0
SdV
dt V
S
Conservation of momentum:
Z
I
dp
d
= F = −0 µ0
SdV +
T · ds
dt
dt V
S
R
the first integral represents the momentum stored in the em fields pem = 0 µ0 V SdV ,and the second one the
momentum flowing in through the surface. Density of em momentum can be written as:
p̌em = 0 µ0 S = 0 E × B
Angular momentum:
Z
L = r × p = 0 µ0
(r × S) dV
V
or by writing the angular momentum density,
l = 0 [r × (E × B)]
2
Problem 0331
~ enters a matter with a permeability constant µ . The angle between the surface and the magnetic
A magnetic field B
field is θ, inside the surface the magnetic field is perpendicular to the surface. Find the force acting on the surface.
Problem 0370
A long coaxial cable, of length l, consists of an inner conductor (radius a) and an outer conductor (radius b). It is
connected to a battery at one and and a resistor at the other(see figure). The inner conductor carries a uniform charge
per unit length lambda, and a steady current I to the right. the outer conductor have oppsite charge and current.
What is the em momentum stored in the fields?
Solution
We start by noting that B = B φ̂ and E = B ρ̂, due to the geometry of the problem. The em fields exist onlyin the
volume bounded by a < r < b. The electric field in this volume is (due to Gauss law)
E=
λ
ρ̂
2πr0
and the magnetic filed in the volume is (due to Amphere law)
B=
µ0 I
φ̂
2πr
Pointing vector for 0 < z < L; a < r < b
S=
1
1 λ µ0 I E×B=
ρ̂ × φ̂
µ0
µ0 2πr0 2πr
Z
P = µ0 0
V
µ0 λI
SdV =
4π 2
Z
L
Z
dz
0
2π
Z
dφ
0
a
b
µ0 λIL
1
rdrẑ =
ln
r2
2π
b
ẑ
a
What we have got here is that nothing is moving in the system, and yet it has a linear momentum. Something has to
move, so what is actually moving in this system? It turns out that there is a mechanical momentum associated with
the current flow, locating this momentum is not easy, and it is actually a relativistic effect.
Problem 0380
Given a long solenoid (R, n, I). Two insulating cylinders are coaxial with the solenoid: (λ, a) and (−λ, b). such that
a < R < b (see figure). When the current in the solenoid is gradually reduced, the cylinders begin to rotate. where
does the angular momentom come from?
Tip: calculate the angular momentum stored in the fields, and compare with the angular momentum due to the
turque acting on the cylinders (since an electric field is induced by the coil)
3
Solution
The angular momentum is stored in the fields.
As in the previous question, the aim is to calculate the Pointing vector
R
since the angular momentum is L = 0 µ0 V (r × S) dV . The electric field at the a < r < b volume is
E=
λ
r̂
2π0 r
The magnetic field at the r < R volume is (here we assume that the length of the solenoid is much larger than b)
B = µ0 nI ẑ
so the Pointing vector at a < r < R
S=
1
nIλ
nIλ
(r̂ × ẑ) =
− φ̂
E×B=
µ0
2π0 r
2π0 r
Momentum density
p̌ = µ0 0 S
Angular momentum density
ľ = r × p̌ =
µ Inλ
µ0 Inλ 0
r̂ × −φ̂ =
− ẑ
2π
2π
and the total angular momentum is
Z
L=
ľdV =
V
µ0 Inλ
2π
Z
2π
Z
dφ
0
L
Z
dz
0
R
rdr = ... = µ0 InλL
a
(R2 − a2 )
− ẑ
2
this is the angular momentum stored in the em fields, which is constant.
4
When the current is turned off, the changing magnetic field induces a circumferential (φ̂) electric field given by
Faraday’s law:
Z
∂B
∇×E=−
/ ds
∂t
Z
(∇ × E) ds = −
S
∂
∂t
Z
I
Edl = −
Bds = stokes =
S
∂
∂t
Z
Bds
S
for r < R
∂
2πrE = −
∂t
E = −µ0 n
Z
r
2π
Z
µ0 nIrdr
0
0
r2
µ0 nI2π
2
∂
dφ = −
∂t
∂
dφ = −
∂t
r ∂I
φ̂
2 ∂t
for r > R
∂
2πrE = −
∂t
E = −µ0 n
Z
R
Z
2π
µ0 nIrdr
0
0
R2
µ0 nI2π
2
R2 ∂I
φ̂
2r ∂t
So the moment force (torque) on the cylinders:
τ = r × F = r × qE
for an outer cylinder
τb = b(−λL)E|b ẑ =
λLµ0 nR2 ∂I
ẑ
2
∂t
for an inner cylinder
τa = b(λL)E|a ẑ = −
λLµ0 na2 ∂I
ẑ
2
∂t
Angular momentum temporal change of the cylinders (angular mitkaf)
0
Z
τb dτ = −
Lb =
I
Z
La =
τa dτ =
I
Ltot
0
λLnµ0 R2
I ẑ
2
λLnµ0 a2
I ẑ
2
R2 − a2
= µ0 IλLn
− ẑ
2
The angular momentum lost by the em fields is exactly what is gained by the cylinders and the total angular momentum
is conserved.
Problem 0331
~ enters a matter with a permeability constant µ . The angle between the surface and the magnetic
A magnetic field B
field is θ, inside the surface the magnetic field is perpendicular to the surface. Find the force acting on the surface.
5
Solution
Soon