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6.641 Electromagnetic Fields, Forces, and Motion
Spring 2009
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Spring 2006
6.641 — Electromagnetic Fields, Forces, and Motion
Final- Solutions
Prof. Markus Zahn
MIT OpenCourseWare
Problem 1
A
Question: What are the boundary conditions necessary to solve for the magnetic fields for
x < 0 and for 0 < x < s?
Solution:
y
H = -� χ
σ = 0,� 0
σ
σ =0,�
z
∞
� 2χ = 0
x
s
Kz (x = 0, y) = K0 cos ky
Figure 1: A sheet of surface current at x = 0. (Image by MIT OpenCourseWare.)
B.C. I
B.C. II
B.C. III
B.C. IV
�
�
n × H II − H I = K ⇒ Hy (x = 0+ , y) − Hy (x = 0− , y) = K0 cos ky
χ → 0 as x → −∞
�
�
n · µII H II − µI H I = 0 ⇒ Hx (x = s− , y) = 0
�
�
�
�
�
�
n · µII H II − µI H I = 0 ⇒ µ0 Hx � − = µHx � +
x=0
1
x=0
Spring 2006 Final Exam
6.641, Spring 2005
B
Question: What are the magnetic scalar potential and magnetic field distributions for x < 0
and 0 < x < s?
Hint: The algebra will be greatly reduced if you use one of the following forms of the potential
for the region 0 < x < s
I. sin(ky) cosh k(x − s)
II. cos(ky) cosh k(x − s)
III. sin(ky) sinh k(x − s)
IV. cos(ky) sinh k(x − s)
Solution:
�2 χ = 0 ⇒
∂2χ ∂2χ
+ 2 =0
∂x2
∂y
General solution is of the form
χ(x, y) = ekx (C1 sin ky + C2 cos ky) + e−kx (C3 sin ky + C4 cos ky)
For x < 0 only ekx term is relevant from B.C. II lim χ → 0; also from surface current boundary condition
x→−∞
I only sin ky term is needed. Therefore, χ(x, y) = C1 ekx sin ky for x < 0. For x > 0, we can work with the
general solution in sinh, cosh, due to surface current condition in y, only sin ky term is needed; also due to
boundary condition at x = 0, x = s we can use the form
χ(x, y) = A1 sin ky cosh(x − s)
0<x<s
�
C1 sin kyekx
x<0
⇒ χ(x, y) =
A1 sin ky cosh k(x − s) 0 < x < s
�
x<0
−C1 kekx (sin kyix + cos kyiy )
⇒ H = −�χ =
−A1 k(sin ky sinh k(x − s)ix + cos ky cosh k(x − s)iy ) 0 < x < s
�
�
B.C. III ⇒ Hx �
= 0 ⇒ satisfied
x=s
�
�
µ
�
�
B.C. IV ⇒ µ0 Hx � − = µHx � + ⇒ −µ0 C1 k� = −µA1 k� sinh(−ks) ⇒ C1 = − A1 sinh ks
µ0
x=0
x=0
�
�
�
�
B.C. I ⇒ Hy � + − Hy � − = K0 cos ky ⇒ −A1 k cosh(−ks) + C1 k = K0
x=0
x=0
K0
µ
K0
⇒ −A1 cosh ks + C1 =
⇒ −A1 cosh ks −
A1 sinh ks =
k
µ0
k
⇒ A1 (µ0 cosh ks + µ sinh ks) = −
⇒ A1 = −
⇒ C1 =
µ0 K0
k
µ0 K0
k(µ0 cosh ks + µ sinh ks
µK0 sinh ks
k(µ0 cosh ks + µ sinh ks)
2
Spring 2006 Final Exam
6.641, Spring 2005
�
µ sinh ksekx sin ky
x<0
K0
⇒χ=
k(µ0 cosh ks + µ sinh ks) −µ0 cosh k(x − s) sin ky 0 < x < s
�
−µ sinh ksekx (sin kyix + cos kyiy )
K0
H=
(µ0 cosh ks + µ sinh ks) +µ0 (sin ky sinh k(x − s)ix + cos ky cosh k(x − s)iy )
x<0
0<x<s
C
Question: What is the surface current distribution on the x = s surface?
Solution:
y
x
s
Figure 2: Surface current boundary condition at x = s. (Image by MIT OpenCourseWare.)
�
�
�
�
�
�
From magnetic field boundary condition n × H II − H I = K S . Therefore, −Hy �
= Ksz � .
x=s−
K(x = s, y) = −
x=s
K0 µ0 cos kyiz
µ0 cosh ks + µ sinh ks
D
Question: Use the Maxwell Stress Tensor to find the total force, magnitude and direction,
on a section of the perfect conductor at x = s that extends over a wavelength 0 < y < 2kπ and
0 < z < D. Assume that µ in the region 0 < x < s does not depend on density so that dµ
dρ = 0.
Hint:
�
cos2 (ky)dy = 12 y +
1
4k
sin(2ky)
Solution:
� ��
1 � 2
µ Hx − Hy2 − Hz2 �
2
x=s−
�
�
=0
(Hz (x, y) = 0)
= µHz Hx �
Txx =
Tzx
x=s−
3
Spring 2006 Final Exam
y
6.641, Spring 2005
σ=0
�
σ = 0 �0
σ
Tyx
∞
y
Txx
n
Tzx
x
Txx
s
x
z
n = -ix
Figure 3: Maxwell Stress tensor used to find force on perfect conductor. (Image by MIT OpenCourseWare.)
Tyx
�
�
= µHy Hx �
x=s−
Txx =
=0
⎡
� �
�
Hx �
x=s−
⎤
�
=0
�2
�
�
�
⎥
1 ⎢
1
K0 µ0
2�
2�
⎥
H
−H
=
−
cos2 ky
µ⎢
µ
�
y�
2 ⎣ x x=s−
2
µ0 cosh ks + µ sinh ks
x=s− ⎦
� �� �
0
Txy = Tyx = µ(Hy Hx )x=s− = 0
fx =
�
Txj nj da ⇒ fx = −D
S
�
�
�
because Hx �
x=s−
2π
k
=0
Txx dy
0
�
�2 � 2 π
k
K0 µ0
Dµ
=
cos2 kydy
2 µ0 cosh ks + µ sinh ks
0
��
�
�
� 2π
k
�
)� = πk
( y2 + sin(2ky)
4k
0
fx =
�
Dµπ
K0 µ0
2k µ0 cosh ks + µ sinh ks
�2
Problem 2
A
Question: What is the electric field in the free space region, 0 < r < R1 as a function of time?
Solution: Charge relaxation.
⎫
∂ρ
Conservation of charge�
� · J f + ∂tf = 0
⎬σ
∂ρf
ρ
ρf +
=0
Gauss’ Law � · E = �f
ρf
� · J f = σ� · E = σ ε = 0⎭ ε
∂t
Linear Media Jf = σE
4
Spring 2006 Final Exam
6.641, Spring 2005
ε, σ
ε 0, σ = 0
σ
∞
R1
R2
ρf (t = 0) = ρ0
i(t)
Figure 4: Cylindrical shell of uniform volume charge. (Image by MIT OpenCourseWare.)
t
ε
where τ =
⇒ ρf (t) = ρf (t = 0)e− τ
σ
�
0
0 < r < R1
⇒ ρf (r, t) =
− τt
ρ0 e
R1 < r < R2
E(r, t) = 0 for 0 ≤ r ≤ R1
B
Question: What is the volume charge density and electric field within the cylindrical shell,
R1 < r < R2 as a function of radius and time?
Solution: Using Gauss’ Law
�
�
εE · da = ρf dV
S
0
t
π(r 2 − R12 )dρ0 e− τ
0 < r < R1
R1 < r < R2
2πrdε0 Er = 0
t
2πrdεEr = π(r 2 − R12 )dρ0 e− τ
0 < r < R1
R1 < r < R2
Charge enclosed
�
�
5
Spring 2006 Final Exam
6.641, Spring 2005
R2
R1
r
Figure 5: Using Gaussian surfaces (dashes) with Gauss’ law to determine the electric field in each region.
(Image by MIT OpenCourseWare.)
Er =
⎧
⎪
⎨0
0 < r < R1
t
r 2 −R12
ρ e− τ
⎪ 2εr 0
⎩
0
R1 < r < R2
r > R2
The final case (r > R2 ) is given by the problem statement.
C
Question: What is the surface charge density on the interface at r = R2 ?
Solution:
�
�
n · DII − DI = σsf
�
�
at r = R2 : ε0 Er �
r=R2+
�
�
− εEr |�
r=R2−
�
�
= σsf �
r=R2
=−
D
Question: What is the ground current, i(t)?
Solution: From Ampere’s Law
6
R22 − R12
ρ0 e−t/τ
2R2
Spring 2006 Final Exam
6.641, Spring 2005
R2
R1
i(t)
Figure 6: Using conservation of charge at r = R2 to determine the terminal current i(t). (Image by MIT
OpenCourseWare.)
�×H =J +
∂D
⇒�·[
∂t
J
����
conduction
current
⇒
+
∂D
∂t
����
]=0
displacement
current
∂Er ��
)�
i(t) = 2πR2 d(σEr + ε
∂t r=R2
��
� 2
�
R2 − R12
R22 − R12 1 − t
− τt
= 2πR2 d σ
+ε −
ρ0 e
ρ0 e τ
2εR2
2εR2
τ
� 2
�
2
2
R2 − R12
t
R
−
R
σ
t
1
= 2πR2 d σ
ρ0 e− τ − ε 2
ρ0 e− τ = 0
2εR2
2εR2
ε
i(t) = 0
Problem 3
A
Question: Prove that the magnetic scalar potential χ obeys Laplace’s equation for 0 < r < Rp ,
and Rp < r < R where H = −�χ.
Solution: Since there are no free currents in the regions 0 < r < Rp and Rp < r < R, from Ampere’s Law
� × H = J f = 0 ⇒ H = −�χ
for 0 < r < Rp
B = µ0 (H + M ), � · B = 0 ⇒ � · µ0 (H + M ) = 0 ⇒ � · H = −� · M
7
Spring 2006 Final Exam
6.641, Spring 2005
z
θ
ir
iθ
R
�0
Rp
M = M0 iz
σ=∞
Figure 7: A magnetized sphere. (Image by MIT OpenCourseWare.)
H = −�χ ⇒ �2 χ = � · M =
∂Mz
=0
∂z
Because Mz is constant, �2 χ = 0.
For Rp < r < R :
B = µ0 (H + M ), M = 0
From Gauss’ Law
� · B = 0 ⇒ � · (µ0 H) = 0 ⇒ � · (µ0 (−�χ)) = 0
⇒ �2 χ = 0
because µ0 is constant.
B
Question: What are the boundary conditions required to determine the magnetic field in
regions 0 < r < Rp and Rp < r < R?
Solution: B.C. I
�
�
�
�
n · B II − B I = 0 ⇒ Hr �
r=R−
=0
B.C. II
Since there are no surface currents on r = Rp surface.
�
�
�
�
n × H II − H I = K = 0 ⇒ Hθ �
+
r=Rp
�
�
= Hθ �
−
r=Rp
8
Spring 2006 Final Exam
6.641, Spring 2005
B = µ0 (H + M ), � · B = 0 ⇒ � · (µ0 (H + M )) = 0
�
�
�
�
⇒ � · (µ0 H) = −� · (µ0 M ) ⇒ n · µ0 (H II − H I ) = −n · µ0 (M II − M I )
�
�
�
�
�
�
−
H
=
+M
⇒ Hr �
�
�
r
r
−
+
−
r=Rp
r=Rp
r=Rp
B.C. III
�
�
Hr �
+
r=Rp
�
�
= Hr �
−
r=Rp
+ M0 cos θ
C
Question: Find the magnetic field H(r, θ) in regions 0 < r < Rp and Rp < r < R.
Solution:
�
χ=
Ar cos θ
Br cos θ +
C
r2
cos θ
r < Rp
Rp < r < R
∂χ
1 ∂χ
ir −
iθ
∂r
r ∂θ
�
r < Rp
−A cos θir + A sin θiθ
⇒ H = −�χ =
2C
C
−(B cos θ − r3 cos θ)ir + (B sin θ + r3 sin θ)iθ Rp < r < R
�
�
2C
BR3
= 0 ⇒ B cos θ − 2C
B.C. I: ⇒ Hr �
R3 cos θ = 0 ⇒ B = R3 ⇒ C = 2
r=R�−
�
�
�
⇒ A sin θ = B sin θ + RC3 sin θ
B.C. II ⇒ Hθ �
=
H
�
θ
+
−
H = −�χ = −
r=Rp
A=B+
r=Rp
p
2Rp3 + R3
1 R3
R3
B
=
B(1
+
)
=
B=A
Rp3 2
2Rp3
2Rp3
B.C. III
�
�
Hr �
+
r=Rp
�
�
= Hr �
−B cos θ +
−
r=Rp
+ M0 cos θ
2C
cos θ = −A cos θ + M0 cos θ
Rp3
2 BR3
R3
= −B −
B + M0
3
Rp 2
2Rp3
�
�
R3
3 R3
R3
=
M
⇒
B = M0
B −1 + 3 + 1 +
0
Rp
2Rp3
2 Rp3
� �3
2Rp3 + R3
Rp
BR3
1
2
⇒C=
⇒ C = M0 Rp3 ⇒ A = M0
B = M0
3
R
2
3
3R3
⎧
3
�
2Rp
+R3 �
⎪
⎨M0 3R
+
sin
θi
−
cos
θi
3
r
θ
�� �
� � �
� �3 �
� �3 �
3
3
H=
Rp
Rp
Rp
Rp
M0
⎪− 2M0
−
cos
θi
+
2
+
sin θiθ
⎩
r
3
R
r
3
R
r
−B +
9
r < Rp
Rp < r < R
Spring 2006 Final Exam
6.641, Spring 2005
D
Question: Find the free surface current density K on the r = R surface.
Solution:
�
�
n × H II − H I = K
�
�
⇒ −Hθ �
= Kφ
r=R−
� �3
Rp
⇒ K = −M0
sin θiθ
R
Problem 4
I1
a
�0
r
T, m
I2
x
ξ(x,t)
g
L
Figure 8: A current carrying string in a magnetic field from a line current. (Image by MIT OpenCourseWare.)
A
Question: To linear terms in membrane displacement ξ(x, t), find the magnetic force per unit
length on the string centered at r = 0.
Solution: From Ampere’s Law,
�
C
H · dl = Ienclosed ⇒ Hφ 2πr = I1 ⇒ Hφ =
Hφ
r
I
2πr .
Lorentz Force on the
I1
φ
Figure 9: The magnetic field from a line current is determined by using Ampere’s circuital law. (Image by
MIT OpenCourseWare.)
conducting strings
�
F
µ0 I1
�
= (I2 ix ) × (µ0 H �
) = I2
ir
l
2π(a − ξ)
r=a−ξ
10
Spring 2006 Final Exam
⇒
6.641, Spring 2005
1
1
µ0 I1 I2
µ0 I1 I2
µ0 I1 I2
F
=
=
≈
ξ
l
2π (a − ξ)
2πa (1 − a )
2πa
�
�
ξ
1+
a
B
Question: What is the governing linearized differential equation of motion of the membrane?
Solution: The equation of motion for the
�
∂2ξ
∂2ξ
µ0 I1 I2
1+
m 2 = T 2 − mg +
∂t
∂x
2πa
strings is given by
�
ξ
a
C
Question: What must I1 be in terms of I2 , m and other relevant parameters so that the mem­
brane is in static equilibrium with ξ(x, t) = 0?
Solution: In static equilibrium at ξ = 0,
0 = −mg +
d
dt
→ 0, therefore
µ0 I1 I2
2πamg
⇒ I1 =
2πa
µ0 I2
D
�
�
ˆ j(ωt−kx) find the ω − k
Question: For small membrane deflections of the form ξ(x, t) = Re ξe
dispersion relation. Plot the ω − k relationship showing significant intercepts on the axes and
slope asymptotes. Assume that k is real and that ω can be pure real or pure imaginary.
Solution: For small deflections from equilibrium point at ξ = 0, the perturbation equation is
m
∂2ξ�
d2 ξ �
µ0 I1 I2 �
=
T
+
ξ
∂t2
dx2
2πa2
⇒
T ∂2ξ�
µ0 I1 I2 �
∂2ξ�
ξ
=
+
2
2
∂t2
m
∂x
2πa
����
� �� m�
vp2
vp2 =
T
µ0 I2 I1
g
, w2 =
=
m c
2πma2
a
ωc2
∂2ξ�
∂2ξ�
= vp2 2 + ωc2 ξ �
2
∂t
∂x
�
�
ˆ j(ωt−kx)
For deflections of the form ξ � (x, t) = Re ξe
−ω 2 ξˆ = −vp2 k2 ξˆ + ωc2 ξˆ ⇒ ω 2 = vp2 k2 − ωc2
dispersion relationship: for I1 I2 > 0, ω 2 = vp2 k2 − ωc2 , with ωc2 > 0.
11
Spring 2006 Final Exam
6.641, Spring 2005
ω
for
ωc
I1 I2 >0
ωc
vp
ωc
vp
k
ωc
real ω
vp
vp
imaginary ω
Figure 10: ω − k dispersion relation with I1 I2 > 0. (Image by MIT OpenCourseWare.)
E
Question: What are the allowed values of k that satisfy the zero deflection boundary condi­
tions at x = 0 and x = L?
Solution: From dispersion relationships
�
2
2
ω
+
ω
ω 2 + ωc
2
c
ω 2 = vp2 k2 − ωc2 ⇒ k2 =
⇒
k
=
±
vp2
vp2
For k0 =
form
�
ω 2 +ω02
vp2
ξ(x, t) = Re
where k0 is the positive root of solution for k the displacement will have solution of the
��
�
�
ξˆ1 e−jk0 x + ξˆ2 ejk0 x ejωt
with boundary conditions B.C. I ξ(x = 0, t) = 0, ⇒ ξˆ1 + ξˆ2 = 0 ⇒ ξˆ1 = −ξˆ2 and
B.C. II ξ(x = L, t) = 0 ⇒ ξˆ1 e−jk0 L + ξˆ2 ejk0 L = 0.
⇒ ξˆ1 e−jk0 L − ξˆ1 ejk0 L = 0
nπ
ξˆ1 (−2j sin k0 L) = 0 ⇒ k0 =
, n = 1, 2, 3
L
12
Spring 2006 Final Exam
6.641, Spring 2005
F
Question: Under what conditions will the membrane equilibrium with ξ(x, t) = 0 first become
unstable?
Solution: First unstable condition happens when k =
� π �2
g m � π �2
ωc2
=
⇒
=
2
vp
L
aT
L
13
π
L
=
ωc
vp