MIT OpenCourseWare http://ocw.mit.edu 6.642 Continuum Electromechanics �� Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 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