6.642 Continuum Electromechanics

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6.642 Continuum Electromechanics
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Fall 2008
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Fall 2008
6.642 — Continuum Electromechanics
Problem Set 4 - Solutions
Prof. Markus Zahn
MIT OpenCourseWare
Problem 1
a)
� × H = 0 ⇒ H = −�Φ
� · H = 0 ⇒ �2 Φ = 0
�
�
C
r>R
Φ(r, θ) = Dr + 2 cos θ
r
⎛
0⎞
1 ∂Φ
1 ∂Φ� ⎟
⎜ ∂Φ
H = −�Φ = − ⎝
ir +
iθ +
� iφ ⎠
∂r
r ∂θ
r sin θ�∂φ
��
�
�
�
�
2C
1
C
= − D − 3 cos θir −
Dr + 2 sin θiθ
r
r
r
r>R
B.C. H(r → ∞, θ)
= H0 iz = H0 (ir cos θ − iθ sin θ) = −(D cos θir − D sin θiθ )
D = −H0
�
�
1
C
Hθ (r = R, θ) = 0 =
DR + 2 sin θ ⇒
C = −DR3 = H0 R3
R
R
�
�
�
�
��
R3
2R3
H = H0 1 + 3 cos θir − 1 − 3 sin θiθ
r > R
r
r
B = µ0 H
b)
�
�
2R3
cos θ
1
+
3
r
dr
µ0 Hr
=
= �
3�
R
rdθ
µ0 Hθ
− 1 − r3 sin θ
�
�
3
− 1 − Rr3 dr
cos θ
�
=
dθ
3�
sin θ
r 1 + 2R
r3
�
cos θ
dθ = ln(sin θ) + C1
sin θ
�
�
� 1 − R33 dr �
r
ln r 1
(r 3 − R3 )dr
�
� =
=−
+ ln(r 3 + 2R3 ) + C2
2R3
r(r 3 + 2R3 )
2
2
r 1 + r3
ln(sin θ) + C1 =
ln r 1
− ln(r 3 + 2R3 ) + C2
2
2
ln[sin2 θ(r 3 + 2R3 )/r] = 2(C2 − C1 )
1
Problem Set 4
6.642, Fall 2008
�
� � �
1 r 2 R
+
= constant
sin θ(r + 2R )/r = constant ⇒ sin θ
2 R
r
equation of magnetic field lines for r > R
2
3
2
3
c)
From Melcher’s Continuum Electromechanics, Appendix A with A(r, θ) = A(r, θ)iφ .
�
�
Aφ
�2 A(r, θ))iφ = �2 Aφ − 2 2
iφ
r sin θ
�
�
�
�
�
�
1 ∂
∂Aφ
1
∂
∂Aφ
Aφ
= 2
iφ
r2
+ 2
sin θ
− 2 2
r ∂r
∂r
r sin θ ∂θ
∂θ
r sin θ
=0
�
�
1
∂(sin θAφ )
1 ∂(rAφ )
� × A(r, θ) = � × (A(r, θ)iφ ) = ir
− iθ
r sin θ
∂θ
r ∂r
= Br ir + Bθ iθ
�
�
1 ∂(rAφ )
R3
Bθ = −µ0 H0 1 − 3 sin θ = −
r
r ∂r
�
� 2
�
�
3
1 ∂(rAφ )
R
r
R3
sin θ + f (θ)
= µ0 H0 r − 2 ⇒ rAφ = µ0 H0
+
sin θ ∂r
r
2
r
�
�
2R3
1 ∂(sin θAφ )
Br = +µ0 H0 1 + 3 cos θ =
r
r sin θ
∂θ
�
�
�
�
3
1
sin θf (θ)
∂
r
R
2
=
µ0 H0 sin θ
+ 2 +
r sin θ ∂θ
2
r
r
�
�
�
�
�
�
1
r
R3
1 d
2R3
µ0 H0
+ 2 2 sin θ cos θ +
(sin θf (θ))
µ0 H0 1 + 3 cos θ =
r
r sin θ
2
r
r dθ
�
�
r
d
R3 2 cos θ
1
= µ0 H0
+ 2
+ 2
(sin θf (θ))
2
r
r
r sin θ dθ
�
�
2R3
1
d
= µ0 H0 1 + 3 cos θ + 2
(sin θf (θ))
r
r sin θ �dθ ��
�
0⇒f (θ)=0
Aφ = µ0 H0
�
�
3
R
r
+ 2
2
r
sin θ
d)
�2 Aφ (r, θ)iφ =
1 ∂
r 2 ∂r
�
r2
∂Aφ
∂r
�
�
1 ∂
r 2 ∂r
�
�
�
�
�
∂Aφ
1
∂
∂Aφ
Aφ
r2
+ 2
sin θ
− 2 2
iφ
∂r
r sin θ ∂θ
∂θ
r sin θ
� �
��
µ0 H0 sin θ ∂
1 2R3
2
r
−
r2
∂r
2
r 3
�
�
2R3
µ0 H0 sin θ
r+ 2
=
2
r
r
=
2
Problem Set 4
6.642, Fall 2008
�
�
�
� µ H r + R3
2
0
0
2
r
1
∂Aφ
∂
∂
sin θ
=
(sin θ cos θ)
r 2 sin θ ∂θ
∂θ
r 2 sin θ
∂θ
�
�
3
µ0 H0 2r + R
r2
cos 2θ
=
r 2 sin θ
�
�
r
R3
�
�
µ
H
3
+
2
0
0
2
r
Aφ
R
µ0 H0
r
= 2 2
+ 2 sin θ =
r
r 2 sin θ
r 2 sin2 θ
r sin θ 2
� Aφ (r, θ)iφ = iφ
�
µ0 H0 sin θ
r2
��
�
r
R3
+ 2
2
r
�
= iφ
�
µ0 H0 sin θ
r2
�
�
r
R3
+ 2
2
r
�
2
=0
2R3
r+ 2 +
r
⎡
3
⎢
⎢r + 2R +
⎣
r2
3
r
+ R2
cos 2θ
− 2 2r
2
sin θ
sin θ
⎤
cos 2θ − 1 ⎥
⎥
2
⎦
sin
θ
� �� �
−2
e)
Magnetic field line: r sin θAφ (r, θ) = constant
�
�
R3
r
+ 2 sin θ = constant
r sin θµ0 H0
2
r
� � �
�
1 r 2 R
3
π
2
sin θ
+
= constant = when r = R, θ =
2 R
r
2
2
x = r sin θ cos φ = D for φ = 0 at r = ∞, θ = π
lim
r→∞,θ=π
r sin θ = D
√
1 D2
3
= ⇒ D = 3R
2 R2
2
3
�
Problem Set 4
6.642, Fall 2008
f)
Problem 2
F = (P · �)E + ρf E,
P = D − �0 E,
ρf = � · D
4
Problem Set 4
�×E=0⇒
6.642, Fall 2008
∂Ej
∂Ei
=
∂xj
∂xi
F = (D − �0 E) · �E + (� · D)E
∂Dj
∂Ei
+
Ei
∂xj
∂xj
∂Ei
∂Dj
∂Ei
= Dj
+ Ei
− �0 Ej
∂xj
∂xj
∂xj
� �� �
Fi = (Dj − �0 Ej )
�0 Ej
∂Ej
∂xi
�0 ∂
∂
(Dj Ei ) −
(Ej Ej )
∂xj
2 ∂xi
�
�
∂
1
=
(Dj Ei ) − δij �0 Ek Ek ,
∂xj
2
=
=
�
0 i �= j
δij =
1 i=j
1
∂Tij
⇒ Tij = Dj Ei − δij �0 Ek Ek
∂xj
2
Problem 3
F = µ0 (M · �)H + J × µ0 H,
µ0 M = B − µ0 H,
1
(� × H) × H = (H · �)H − �(H · H)
2
� × H = J,
F = (B − µ0 H) · �H + µ0 (� × H) × H
1
= B · �H − µ0 (H · �)H + µ0 ((H · �)H − �(H · H))
2
µ0
= (B · �)H −
�(H · H)
2
∂Hi
µ0 ∂(Hk Hk )
−
∂xj
2
∂xi
�
∂(Bj Hi )
∂Bj
∂ � µ0
=
− Hi
−
Hk Hk
∂xj
∂xj
∂xi 2
� �� �
Fi = Bj
�·B=0
�
∂(Bj Hi )
∂ � µ0
1
− δij
Hk Hk ⇒ Tij = Hi Bj − δij µ0 Hk Hk
=
∂xj
∂xj 2
2
5
�·B=0