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九十七學年第一學期 PHYS2310 電磁學 期末考試題(共兩頁)
[Griffiths Ch. 4-6]
2009/01/15, 10:10am–12:00am, 教師：張存續
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∂v ∂v
1 ∂vz ∂vφ
1 ∂ ( svφ ) ∂vs
]sˆ + [ s − z ]φˆ + [
]zˆ
◇ Useful formulas: Cylindrical coordinate ∇ × v = [
−
−
s ∂φ ∂z
s ∂s
∂z ∂s
∂φ
◇ Specify the magnitude and direction for a vector field.
1. (10%, 10%) The magnetic field on the axis of a circular current loop is far from uniform. We can
produce a more nearly uniform field by using two such circular loops a distance d apart.
(a) Find the total magnetic field B along the z-axis as a function of z.
(b) Show that ∂B/∂z is zero at the point midway between them.
(c) Determine d such that ∂B2/∂z2=0 at the midpoint, and find the resulting magnetic field at the
center (z=0). [Hint: This arrangement is known as a Helmholtz coil.]
2. (4% x 5) The space between the planes of a parallel-plate capacitor is filled with two slabs of
linear dielectric material. Each slab has thickness a, so the total distance between the plates is 2a.
Slab 1 has a dielectric constant of 9, and slab 2 has a dielectric constant of 4. The free charge
density on the top plate is +σ and on the bottom plate is －σ.
(a) Find the electric displacement D and the electric field E in each slab.
(b) Find the polarization P in each slab.
(c) Find the potential difference between the metal plates.
(d) Find the location and amount of all bound charges σb and ρb.
(e) Now that you know all the charges (free and bound), recalculate the field D and E in each
slab, and confirm your answer to (a).
-1-
3. (10%,10%)
(a) Write down and prove the electrostatic boundary conditions in terms of D, P, and σf.
(b) Write down and prove the magnetostatic boundary condition in terms of H, M, and Kf.
[Hint: write down the differential equations and their integral forms, and then use the divergence
theorem and Stokes’ theorem to prove them.]
4.
(7%, 7%, 6%) A large parallel-plate capacitor, with uniform surface charge +σ on the upper
plate and －σ on the lower, is moving with a constant speed v, as shown in the figure.
(a) Find the magnetic field between the plates and also above and below them.
(b) Find the magnetic force per unit area on the lower plate (attractive or repulsive force).
(c) At what speed v would the magnetic force balance the electric force?
5. (10%,10%) An iron rod of length L and square cross section (side a), is given a uniform
longitudinal magnetization M, and then bent around into a circle with a narrow gap (width w), as
shown in the figure. ( M = M 0φˆ )
(a) Find the Jb and Kb on the outer cylinder and the top surface.
(b) Find the magnetic field at the center of the gap, assuming w<a<<L.
Top surface
Outer cylinder
-2-
1.
(a) B(r )=
µ I dl′ × (r − r′)
. In the diagram only the z-component survives.
4π ∫ r − r′ 3
0
Upper coil: Bz ,up (z )=
µ I 2π a sin θ
a
zˆ , where sinθ =
2
2
4π ( ( z − d2 ) + a )
( z − d2 ) 2 + a 2
0
Total field: Bz (z ) = Bz ,up + Bz , down

µ 0 IR 2 
( z − d )2 + R 2
=
2


2
(
)
−
3
2
(
+ (z +
d )2
2
+R
2
)
−
3
2


in z -direction. ﹟
5
5

−
− 
∂Bz ( z ) µ 0 IR 2 
2
2 2
2
2 2
d
d
d
d
.
(b)
=
−3( z − ) ( z − ) + R
− 3( z + ) ( z + ) + R
2
2
2
2

2 
∂z


(
)
(
)
5
5

−
− 
∂Bz ( z = 0) µ 0 IR 2  d d 2
2 2
2
2 2
d
d
3( ) ( ) + R
=
− 3( ) ( ) + R
=0 ﹟
2
2

2  2 2
∂z


(
)
(
)
(c)
∂ 2 Bz ( z )
∂z 2
=

µ 0 IR 2 
−3 ( z − d ) 2 + R 2
2
2


(
)
−
5
5
7
7
−
−
− 
2 − 3 ( z + d ) 2 + R 2 2 + 15( z − d ) 2 ( z − d ) 2 + R 2 2 + 15( z + d ) 2 ( z + d ) 2 + R 2 2 
2
2
2
2
2

(
)
(
)
(


2  5( z − d ) 2 − ( z − d ) 2 + R 2 5( z + d ) 2 − ( z + d ) 2 + R 2 
3µ 0 IR 
2
2
2
2

=
+
7
7

2 


( z − d2 )2 + R 2 2
( z + d2 ) 2 + R 2 2


(


∂ Bz ( z = 0) 3µ 0 IR 2 
=
2 
∂z 2


)
2
(
(
)


−d 2 + R 2
−d 2 + R 2 
−d 2 + R 2
+
= 3µ 0 IR 2
7
7
7
( d2 ) 2 + R 2 2
( d2 ) 2 + R 2 2 
( d2 ) 2 + R 2 2

) (
)
(
)
⇒ d = R . The two coils are separated by a distance R. ﹟

µ 0 IR 2  R 2
Bz (z ) =
( ) + R2
2
2
(

)
−
3
2
+
(
( R )2
2
+R
2
)
−
3
2


=
8µ 0 I
53 / 2 R
in z -direction.
2.
(a)
∇ ⋅ D = ρf ⇒ v
∫ D ⋅ da = Q f enc
Draw a cylindrical rectangular Gaussian surface, of area a, and apply the Gauss’s law.
 σ ˆ
 − 9ε z
1
1

0
We find Da = aσ , D = −σ zˆ in slab 1 and slab 2. E = D =
D=
ε
ε 0ε r
 − σ zˆ
 4ε 0
-3-
slab 1
slab 2
)

 σ ˆ
 8σ ˆ
−
z
 − 9ε z slab 1

 9
0
(b) P = ε 0 χ e E, where E = 
⇒P=
 − σ zˆ slab 2
 − 3σ zˆ
 4
 4ε 0
(c) The potential difference between the metal plates
2a
a
2a
0
0
a
V = − ∫ E ⋅ dl = − ∫ E1 ⋅ dl − ∫ E2 ⋅ dl =
slab 1
slab 2
σ a σ a 13σ a
+
=
9ε 0 4ε 0 36ε 0
(d) ρb = −∇ ⋅ P = 0, because P is position independent.

8σ
8σ


 σ up = P1 ⋅ nˆ ↑ = − 9 zˆ ⋅ zˆ = − 9
 slab 1 
8σ
8σ

σ
= P1 ⋅ nˆ ↓ = −
zˆ ⋅ − zˆ = +
down


9
9
σ b = P ⋅ nˆ = 
3σ
3σ


σ up = P2 ⋅ nˆ ↑ = − zˆ ⋅ zˆ = −



4
4
slab 2 
3σ
3σ
σ

ˆ
zˆ ⋅ −zˆ = +
down = P2 ⋅ n ↓ = −

4
4

(e) Consider the Gauss’s law using the free charge and bound charge.
∇⋅E =
1
ε0
( ρ f + ρb ) ⇒
In slab 1, Ea =
In slab 1, Ea =
1
ε0
1
ε0
1
v∫ E ⋅ da = ε
(σ f + σ b1u )a = −
(Qf + Qb )
0
σ
zˆ ,
9ε 0
(σ f + σ b1u + σ b1d + σ b 2u )a = −
σ
zˆ ,
4ε 0
The results are the same as (a).
3. Textbook 4.3.3 and 6.3.3
(a)
Normal: ∇ ⋅ D = ρ f . Consider a wafer-thin pillbox. Gauss’s law states that
v∫
S
D ⋅ da = ∫ ρ f dτ .
v
The sides of the pillbox contribute nothing to the flux, in the limit as the thickness ε goes to zero.
⊥
⊥
( Dabove
− Dbelow
) =σ f .
Tangential: ∇ × D = ∇ × P . Consider a thin rectangular loop. The curl of the Ampere’s law states
that
v∫
//
above
(D
P
D ⋅ dA = v
∫ P ⋅ d A . The ends gives nothing (as ε→0), and the sides give
P
−D
//
below
//
//
)A = ( Pabove
− Pbelow
)A
//
//
//
//
⇒ ( Dabove
− Dbelow
) = ( Pabove
− Pbelow
) or
//
//
//
//
( Dabove
− D below
) = ( Pabove
− Pbelow
)
(b) Normal: ∇ ⋅ H = −∇ ⋅ M . Consider a wafer-thin pillbox. Gauss’s law states that
-4-
v∫
S
H ⋅ da = − v∫ M ⋅ da . The sides of the pillbox contribute nothing to the flux, in the limit as the
S
⊥
⊥
⊥
⊥
thickness ε goes to zero. H above
− H below
= −( M above
− M below
).
Tangential: ∇ × H = J f . Consider a thin rectangular loop. The curl of the Ampere’s law states that
v∫
P
H ⋅ d A = µ0 I f enc . The ends gives nothing (as ε→0), and the sides give
//
//
( H above
− H below
)A = µ 0 K f A
//
//
//
//
⇒ H above
− H below
= µ0 K f or H above
− H below
= K f × nˆ .
4. Prob. 5.16
(a) According to the boundary conditions, the top plate produces a parallel field µ0 K / 2 , pointing
out of the page for points above it and into the page for points below) The bottom plate produces a
parallel field µ0 K / 2 , pointing into the page for points above it and out of the page for points below).
Between the plates, the fields add up to B = µ0 K = µ0σ v .
Above and below both plates, the fields cancel B = 0 .
(b) dF = Id A × B = Kda × B = Jdτ × B
dF = Kda × B ⇒ dF = K × Bda
µ σ v µ σ 2v2
dF
= K ×B =σv 0 = 0
(repulsive force per unit area)
da
2
2
(c) The electric force of the plates is attractive
Balance:
dF d ( FB + FE ) µ0σ 2 v 2 σ 2
=
=
−
=0
da
da
2
2ε 0
dFE
σ σ2
=σE =σ
=
(attractive force per unit area)
ε0 ε0
da
⇒ v=
1
µ 0ε 0
= c the speed of light.
5.
∂M φ
1 ∂ ( sM φ )
M
(a) M( s,φ , z ) = M 0φˆ (A/m), J b = ∇ × M = [ −
]sˆ + [
]zˆ = 0 zˆ
∂z
s
∂s
s
K b = M × nˆ = M 0φˆ × nˆ ,
outer cylinder K b = − M 0 zˆ
outer cylinder nˆ = sˆ
⇒ 

 top surface nˆ = zˆ
 top surface K b = M 0sˆ
(b) Total magnet field is equal to a complete torus plus a square loop with reverse current.
Βtorus = µ0 ( H + M ) = µ0 M = µ0 M 0φˆ
L>>a, the contribution from the bound volume current can be omitted J b =
Bloop (r ) =
a/2
µ K b (r ′) × (r − r ′)
4µ
1
da ′ =
K b wzˆ ∫
dl
3
2
∫
−a / 2 ( a ) + l 2
4π
4π
r − r′
2
0
0
-5-
2π M 0
zˆ ≅ 0 .
L
K b = − M 0 , let l =
Bloop (r ) = −
a
tan θ ,
2
a
π /4
2µ
2 2µ
µ
π /4
2 sec θ
M 0 wφˆ ∫
dθ = −
M 0 wφˆ ⋅ sin θ −π / 4 = −
M 0 wφˆ
2
2
a
−π / 4 ( ) sec θ
a
a
π
π
π
2
0
B = Βtorus + Βloop = µ0 M 0φˆ −
0
0
2 2µ 0
2 2w ˆ
M 0 wφˆ = µ0 M 0 (1 −
)φ
aπ
aπ
-6-