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Phys 4610, Fall 2004 Exam #
- In the calculation of the energy stored in an electrostatic system, we first derived the expression W = 20
∫ E^2 dτ. Later when considering linear dielectric media we found the expression W = (^12)
∫ D · E dτ. What is the difference between these two expressions? What parts(s) of the energy is accounted for in each one?
A polarized medium conists of positive and negative charges which have been stretched away from their “normal” positions so as to give a net buildup of charge on the surfaces of the dielectrics and/or in their interiors. There may also be free charges in the system. The expression W = 20
∫ E^2 dτ (to be used with continuous charge distributions!) gives the energy required to assemble the free and bound charge densities and nothing else. The expression W = (^12)
∫ D·E dτ gives the energy to assemble the free and bound charge densities and also to pull the charges of the medium apart, even if the stretching results in no volume charge density. The second one does give the complete system energy when linear dielectrics are present.
a
a a
a
I
I
I
s
- A planar loop with six 90◦^ bends in it is coplanar with a very long wire which carries a current I. The loop also carries a current I, as shown. (Three sides of the loop are parallel to the long wire.)
Find the magnitude and direction of the net force on the loop.
I
I
F
F F
F
y
1 x
3
2
First we note that there is no net force on the parts of the loop that run in the y direction. This is because for corre- sponding current elements at the same distance from the long wire the forces have the same magnitude but are in opposite directions, as shown here. Summing over the elements for the “y” legs gives zero net force. The force on side 1 of the loop is
Il 1 × B 1 = I(axˆ) ×
μ 0 I 2 πs
ˆz = −
μ 0 aI^2 2 πs
yˆ
The force on side 2 of the loop is
Il 2 × B 2 = I(axˆ) ×
μ 0 I 2 π(s + a)
ˆz = −
μ 0 aI^2 2 π(s + a)
ˆy
The force on side 3 is
Il 3 × B 3 = I(− 2 axˆ) ×
μ 0 I 2 π(s + 2a)
ˆz = +
μ 0 aI^2 2 π(s + 2a)
ˆy
Adding these, the total force is
Fnet = −
μ 0 aI^2 2 π
( 1 s
(s + a)
(s + 2a)
) y^ ˆ
R s
J
- A long straight wire of radius R carries a total current I, but the current density J is nonuniform; it falls off linearly to zero at the surface, i.e. J = k(R − s)
a) Find k.
The integral of J(s) over the x-sec area must give I, hence ∫ J(s) da = 2 π
∫ (^) R
0
k(R − s)s ds = 2πk
[ R
s^2 2
s^3 3
] ∣ ∣∣R 0
= 2 πkR^3
)
πkR^3 3
= I
So that
k =
3 I
πR^3
b) Find the magnitude of magnetic field inside and outside the wire. Give its direction, though you don’t need to repeat the reasons for this.
s
J
B B
B
B
For s < R take an Wmperian loop of radius s. With the current coming out of the page, the B field is tangential as shown here, with ∮ B · dl = (2πs)B = μ 0 Ienc
= μ 0
∫ (^2) π
0
∫ (^) s
0
J(s′)s′^ ds′^ dφ′^ = μ 02 πk
∫ (^) s
0
(R − s′)s′^ ds′
= μ 02 πk
[ R
s′^2 2
s′^3 3
] ∣∣ ∣∣ ∣
s
0
= μ 02 π
3 I
πR^3
s^2
[ R 2
s 3
]
6 μ 0 Is^2 R^3
(3R − 2 s) 6
μ 0 Is^2 (3R − 2 s) R^3 Divide by 2πs and include the direction of B, then
B =
μ 0 Is(3R − 2 s) 2 πR^3
φφφ^ ˆ
for s ≤ R. For s > R, an Amperiain loop encloses all the curent so the result is the same as for a long thin wire,
B
μ 0 I 2 πs
ˆφφφ
Note, the s ≤ R and s > R results agree at s = R.
Since B = ∇ × A, that is,
B =
∣∣ ∣∣ ∣∣ ∣
ˆx yˆ ˆz ∂x ∂y ∂z Ax Ay Az
∣∣ ∣∣ ∣∣ ∣
We see that we can get the given B field with
Ay = Az = 0,
∂Ax ∂z
= B ⇒ Ax = zB, so A = zBxˆ
But we can also get it with
Ax = Ay = 0, −
∂Az ∂x
= B ⇒ Az = −xB, so A = −xBˆz
Both of these solutions give ∇ × A = B for the given B, and also ∇ · A = 0. So the vector potnetial has a remaining (non-trivial!) ambiguity even after imposing the condition ∇ · A = 0.
w
R
s = ks
- A rotating disk of radius R has a charge density given by σ = ks. The disk rotates with angular speed ω.
Find the magnetic moment of the rotating disk.
w
ds s
Consider a thin ring of the disk at radius s of width ds. We can reason out the current dI for this ring as follows:
Charge contained: dq = (2πs)dsσ = 2πks^2 ds
The linear charge density of the rotating ring is:
dλ =
dq Circ
2 πks^2 ds 2 πs
= ks ds
Speed of the moving charge is v = ωs so that
dI = vdλ = kωs^2 ds Area of the loop is πs^2 so dm = (πs^2 )dI = kπωs^4 ds Now integrate from s = 0 to s = R:
m =
∫ a dI = kπω
∫ (^) R
0
s^4 ds = kπω
R^5
- What do we mean when we say a medium acquires a magnetization M?
The quantity M gives the magnetic dipole moment per unit volume of the material. The currents in the material coordinate themselves so as to give a net magnetic dipole for a volume dτ of the material. The magnetic moment of this volume is then
dm = Mdτ
The vector function M can depend on location.
a b
M
M
M
M M
- A very long cylindrical shell has outer radius b and inner radius a. The material carries a uniform magnetic polarization M which points along the direction of the cylinder’s axis.
a) Find the volume and surface bound currents Jb and Kb. (Be careful with the directions for Kb.)
Within the volume of the material, M is uniform, so
Jb = ∇ × M = 0
a^ b
K K K K K K
K K K K K K
On the inner surface, ˆn = −ˆs, so
Kb = M × nˆ = M(ˆz × (−ˆs)) = −M φφφˆ
On the outer surface, ˆn = ˆs, so
Kb = M × nˆ = M(ˆz × ˆs) = +M φˆφφ
b) Find the direction and magnitude of the magnetic field B for 0 < s < a, a < s < b and b < s.
The surface currents create a B field equivalent to solenoids with currents going in op- posite directions! Use the equivalence K = nI for a solenoid. For s < a the fields from the solenoids cancel, so B = 0. For a < s < b we get only the field from the outer solenoid, so
Bz = μ 0 nIˆz = μKˆz = μ 0 Mˆz
For b < s we get no field from either solenoid, so B = 0.
More Math:
x
z
y
z
q
r
r
Jan
In the figure at the right,
r =
√ r^2 + z′^2 − 2 rz′^ cos θ
If x < 1 then
(1 + x)α^ = 1 + αx +
α(α − 1) 2!
x^2 +
α(α − 1)(α − 2) 3!
x^3 + · · ·
sin 2θ = 2 sin θ cos θ cos 2θ = 2 cos^2 θ − 1 ∫ sin^2 x dx =
x −
sin 2x ∫ cos^2 x dx =
x +
sin 2x ∫ (^) a
0
sin(nπy/a) sin(n′πy/a) dy =
{ 0 , if n′^6 = n a 2 if^ n
′ (^) = n
1
r
r
∑^ ∞ n=
( r′ r
)n Pn(cos θ′) V (r, θ) =
∑^ ∞
l=
( Alrl^ +
Bl rl+
) Pl(cos θ)
P 0 (x) = 1 P 1 (x) = x P 2 (x) = (3x^2 − 1)/ 2 P 3 (x) = (5x^3 − 3 x)/ 2 ∫ (^1)
− 1
Pl(x)Pl′ (x)dx =
∫ (^) π
0
Pl(cos θ)Pl′ (cos θ) sin θ dθ =
{ 0 if l′^6 = l 2 2 l+1 if^ l
′ (^) = l
Physics:
F =
4 π 0
qQ
r^2
ˆrrr F = QE E =
4 π 0
q ˆrrr
r^2
V (r) = −
∫ (^) r
O
E · dl
∇ × E = 0 E = −∇V − ∇^2 V = ∇ · E =
ρ 0
V (r) =
4 π 0
∫ (^) ρ(r′)
r
dτ ′
E above⊥ − E below⊥ =
σ 0
E‖ above = E‖ below W =
8 π 0
∑^ n i,j i 6 =j
qiqj
rij
W =
∫ ρV dτ =
∫ E^2 dτ f =
σ^2 nˆ P =
E^2 C ≡
Q
V
p ≡
∫ r′ρ(r′) dτ ′^ Vdip(r) =
4 π 0
p · ˆr r^2
Edip(r, θ) =
p 4 π 0 r^3
(2 cos θ ˆr + sin θ ˆθθθ)
p = αE N = p × E F = (p · ∇)E U = −p · E σb = P · nˆ ρb = −∇ · P P = 0 χeE
D = 0 E + P = E ∇ · D = ρf
∮ D · da = Qf, enc
Fmag = Q(v × B) Fmag =
∫ I(dl × B) K ≡
dI dl⊥
J ≡
dI da⊥
∇ · J = −
∂ρ ∂t
B(r) =
μ 0 4 π
∫ I × ˆrrr
r^2
dl′^ =
μ 0 4 π
I
∫ dl′ × ˆrrr
r^2
μ 0 = 4π × 10 −7 N A 2 1 T = 1 (^) AN·m
∇ · B = 0 ∇ × B = μ 0 J
∮ B · dl = μ 0 Ienc B = ∇ × A
∇ · A = 0 ∇^2 A = −μ 0 J A(r) =
μ 0 4 π
∫ (^) J(r′)
r
dτ ′
B above⊥ = B below⊥ Babove−Bbelow = μ 0 (K×nˆ) Aabove = Abelow
∂Aabove ∂n
∂Abelow ∂n
= −μ 0 K
Adip(r) =
μ 0 4 π
m × ˆr r^2
where m ≡ I
∫ da = Ia
Adip(r) =
μ 0 4 π
m sin θ r^2
φφφ^ ˆ^ Bdip(r) = ∇ × A = μ^0 m 4 πr^3
(2 cos θ ˆr + sin θ θˆθθ)
N = m × B F = ∇(m · B)
A(r) =
μ 0 4 π
∫
V
Jb(r′)
r
dτ ′+
μ 0 4 π
∮
S
Kb(r′)
r
da′^ where Jb = ∇×M and Kb = M×nˆ
J = Jb + Jf H ≡
μ 0
B − M
Specific Results:
L
z
L
P
Linear charge dens. l
Ez =
4 π 0
2 λL z
z^2 + L^2
q (^1) q 2
I
s
B =
μ 0 I 4 πs
(sin θ 2 − sin θ 1 )
z
R I
B =
μ 0 I 2
R^2
(R^2 + z^2 )^3 /^2
