6.641 �� Electromagnetic Fields, Forces, and Motion
advertisement
MIT OpenCourseWare http://ocw.mit.edu 6.641 �� Electromagnetic Fields, Forces, and Motion Spring 2005 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 6.641 — Electromagnetic Fields, Forces, and Motion Spring 2005 Problem Set 5 - Solutions Prof. Markus Zahn MIT OpenCourseWare Problem 5.1 → d− v − → → − → = (q − v ) × (µ0 H ); H = H0ˆiz dt îx îy îz → d− v m = qµ0 vx vy vz dt 0 0 H0 − → qµ0 dv = (vy H0 îx − vx H0 îy ) dt m m (i) qµ0 vy dvx = H0 dt m (ii) dvy −qmu0 vx = H0 dt m d2 vx qµ0 H0 dvy = dt2 m dt Substitute this into (ii): (i) ⇒ qµ0 vx m d2 vx =− H0 2 qµ0 H0 dt m d2 vx q 2 µ20 H02 = − vx dt2 m2 � � � � qµ0 H0 qµ0 H0 vx = A cos t + B sin t m m � � � � qµ0 H0 qµ0 H0 vy = C cos t + D sin t m m vx (t = 0) = vx0 = A vy (t = 0) = vy0 = C Need two more initial conditions: I. Acceleration in x direction at t = 0 � (qvy0 îy ) × (H0 îz )µ0 dvx �� = îx dt �t=0 m � � � qµ� qµ� � 0 H0 0 H0 = vy0 B m m � � B = vy0 1 Problem Set 5 6.641, Spring 2005 II. Acceleration in y direction at t = 0 � dvy �� (qvx0 îx ) × (µ0 H0 îz ) îy = dt �t=0 m D = −vx0 � � � � qµ0 H0 qµ0 H0 t + vy0 sin t vx (t) = vx0 cos m m � � � � qµ0 H0 qµ0 H0 vy (t) = vy0 cos t − vx0 sin t m m vz (t) = vz0 � Note: vx2 + vy2 + vz2 = constant in time (for this case) (easy to check and verify) � = vx20 + vy20 + vz20 � ⇒ vxy = vx20 + vy20 = velocity on xy plane From 8.01, centripetal acceleration, a, is 2 vxy r Where r is radius of circle. So: a= 2 vxy |m |= � ��r � F =ma − → → |q − v × µ0 H | � �� � → − Force due to B field 2� vxy = q� vxy �µ0 H0 r � m vx20 + vy20 mvxy r= = qµ0 H0 qµ0 H0 m A i − → − → − → → f = 0 = q E + q− v × µ0 H − → 0 = q E + q(v0 îy ) × (µ0 H0 îz ) − → E = −v0 µ0 H0ˆix V = −v0 µ0 H0 s ii d = 2r = d= V 2mv0 ; v0 = qµ0 H0 −µ0 H0 s 2m|v | =0.50cm for 12Mg24 qB02 s 0.52cm for 12Mg25 0.54cm for 12Mg26 2 Problem Set 5 6.641, Spring 2005 B i � � → V0 d− v − → − → − → − → = q E + q v × (µ0 H ) = −e − îx + v × (µ0 H0 îz ) m dt s − → v × (µ0 H0 îz ) = vy H0 µ0 îx − vx H0 µ0 îy V0 e dvx = − eµ0 H0 vy dt s dvy m = eµ0 H0 vx dt � � eµ0 H0 eV0 d2 vy − eµ0 H0 vy →m 2 = dt m s � �2 e2 µ0 H0 eµ0 H0 d2 vy v = + V0 y dt2 m m2 s m (1) e2 µ20 H02 d2 vx = − vx dt2 m �2 � d2 vx eµ0 H0 vx = 0 + dt2 m m (2) Solution to equations 1 and 2 (homogenous + particular): � � � � �� V0 eµ0 H0 eµ0 H0 vy = 1 + c1 sin t + c2 cos t µ0 H0 s m m � � � � �� eµ0 H0 eµ0 H0 V0 c1 cos t − c2 sin t vx = µ0 H0 s m m (vx from eµ0 H0 vx = m dvy dt ) vx (t = 0) = vy (t = 0) = 0 ⇒ c1 = 0, c2 = −1 � � � �� � � � eµ0 H0 ˆ V0 eµ0 H0 − → sin t ix + 1 − cos t ˆiy v (t) = µ0 H0 s m m � � �� � �� � � eµ0 H0 mV0 eµ0 H0 mV0 V0 − → − → 1 − cos t ˆix + t− sin t îy v (t)dt = d (t) = 2 2 µ0 H0 se m µ0 H0 s esµ20 H02 m ii |dx (t)|max < s ⇒ H0 > � 2mV0 <s µ20 H02 se 2mV0 µ20 s2 e 3 Problem Set 5 6.641, Spring 2005 Problem 5.2 A � C → − → − H · dl = � S → − → − J · dA For contour through space to left of block HyL S = N I = N I0 cos ωt N I0 sin ωt s For contour through space to right of block HyL = HyR s �0 � � J da (all current in +z direction must return in −z direction) = N I0 cos ωt + �� z �Block � HyR = N I0 sin ωt s B − → − → �×H = J − → − → For Block A,l J = σ E . (3) − → − → � × H = σE For block B − → ∂J − → = ωp2 ε E ∂t − → ∂ − ∂(� × H ) → = (J ) ∂t ∂t �× 4→5: (4) − → − → ∂H ∂J = ∂t ∂t (5) − → ∂H − → �× = ωp2 ε E ∂t C Block A − → − → � × H = σE − → − → � × � × H = � × σE − → − → − → Assume uniform σ, and use � × � × A = �(� · A ) − �2 A . − → − → − → �(� · H ) − �2 H = σ(� × E ) 4 Problem Set 5 6.641, Spring 2005 Faraday’s Law − → ∂µ H − → �× E =− ∂t − → − → Flux continuity: � · µ H = 0; for uniform µ ⇒ � · H = 0. − → ∂H − → �2 H = σµ ∂t Block B − → ∂H − → = ωp2 ε E ∂t − → ∂H − → �×�× = � × ωp2 ε E ∂t − → − → − → Using � × � × A = �(� · A ) − �2 A and assuming uniform properties �× − → − → ∂H ∂H − → ) − �2 = ωp2 ε(� × E ) ∂t ∂t − → Flux continuity for uniform µ → � · H = 0 and using Faraday’s Law �(� · �2 − → − → ∂H ∂H = ωp2 εµ ∂t ∂t Integrate and assume that integration constant is zero 2 − → − → ωp − → �2 H = ωp2 εµ H = 2 H c 1 c = √ = speed of light in material εµ D � � − → ˆ y (x)ejωtˆiy Assume H = Im H − → ∂H � H = σµ ∂t → 2− for Block A. ∂2 Ĥy (x)ejωt = σµjω Ĥy (x)ejωt ∂x2 ∂2 Ĥy (x) = σµjω Ĥy (x) ∂x2 Assume Ĥy (x) = Ĥ0 ejkx . � = σµH �jω ˆ� ˆ� −k2� H ejkx ejkx 0� 0� � 5 Problem Set 5 6.641, Spring 2005 � −k2 = σµjω, k = ± −ωµjσ � σµω k=± (1 − j) 2 � σµω α≡ 2 � � − → H y = Im Ĥ01 eαx ejαx ejωt + H02 e−αx e−jαx ejωt Apply B.C. to complex H. At x = 0, K = 0 ⇒ Hy (x = 0, t) = � � N I0 jωt = Im e s Ĥ01 e0 e0 ejωt + Ĥ02 e0 ejωt = N I0 jωt e s Ĥ01 + Ĥ02 = N I0 s At x = d, K = 0 ⇒ Hy (x = d, t) = N Ia s sin ωt Ĥ01 eαd ejαd ejωt + Ĥ02 e−αd e−jαd ejωt = ⇒ Ĥ01 eαd ejαd + Ĥ02 e−αd e−jαd = N I0 jωt e s N I0 s From B.C. at x = 0 Ĥ01 = Or N I0 − Ĥ02 s � � N I0 N I0 αd jαd e e − H02 eαd ejαd − e−αd e−jαd = s s � � N I0 αd jαd 1−e e s Ĥ02 = −αd −jαd (e e − eαd ejαd ) � � 1 − eαd ejαd N I0 N I0 1 − Ĥ0 = s s e−αd e−jαd − eαd ejαd � � e−αd e−jαd − 1 N I0 = s e−αd e−jαd − eαd ejαd � � − → H = Re Ĥ01 eαx ejαx ejωt + Ĥ02 e−αx e−jαx ejωt ˆiy � � N I0 cosh γ(x − d2 ) jωt ˆ − → H = Im e iy s cosh xd 2 1+j δ ,δ � 2 ωµσ , � σµω = 1δ . � � For Block B: Assume Hy (x, t) = Im Ĥ0 ejkx ejωt . with γ = = α= 2 − → − → �2 H = ωp2 εµ H 6 N I0 s sin ωt. Problem Set 5 6.641, Spring 2005 ∂2 Ĥ0 ejkx ejωt = ωp2 εµĤ0 ejkx ejωt ∂x2 −k2 = ωp2 εµ � √ k = ± ωp2 εµ = ±ωp εµj � � − → H y (x) = Im Ĥ01 eαx ejωt + Ĥ02 e−αx ejωt √ α = ωp εµ + Apply B.C. at x = 0, K = 0 ⇒ Hy0 = Ĥ01 e0 ejωt + Ĥ02 e0 ejωt = 1 2 → Ĥ0 + Ĥ0 = I0 N s sin ωt. I0 N jωt e s N I0 s Apply B.C. at x = d, K = 0 ⇒ Hyd = Ĥy1 eαd ejωt + Ĥ02 e−αd ejωt = I0 N s sin ωt. I0 N jωt e s N I0 Ĥ01 eαd + Ĥ02 e−αd = s � � 2 N I0 N I0 − Hˆ02 eαd + Ĥ0 eαd − s s � � N I0 � � Ĥ02 e−αd − eαd = 1 − eαd s � � αd 1 − e N I 0 Ĥ02 = s e−αd − eαd � � N I0 e−αd − 1 Ĥ01 = s e−αd − eαd � � − → H y (x) = Im Ĥ01 eαx ejωt + Ĥ02 e−αx ejωt √ α = ωp εµ Or � � N I0 1 − eαd s e−αd − eαd � � N I0 e−αd − 1 H01 = s e−αd − eαd � 1 αx � Hy (x, t) = H0 e + H02 e−αx sin ωt H02 = Hy (x, t) = N I0 cosh α(x − d2 ) sin ωt, s cosh αd 2 with α = 7 √ ωp = ωp εµ c Problem Set 5 6.641, Spring 2005 Figure 1: |Hy | for blocks A and B for Problem 5.2 (Image by MIT OpenCourseWare.) E For Block A − → − → �×H = J − → − → Since H = H y (x) ∂Hy (x) − → J = ẑ ∂x � � � � N I0 γ sinh γ (x − d2 ) jωt ˆ − → −αx −jα jωt 1 αx jαx jωt 2 ˆ J = Im Ĥ0 (α + jα)e e e + H0 (−α − jα)e e e zˆ = Im e iz s cosh( γd 2 ) For Block B ∂Hy (x) − → J = ẑ ∂y 8 Problem Set 5 or 6.641, Spring 2005 � � − → J = Im Ĥ01 αeαx ejωt − Ĥ02 αe−αx ejωt ẑ � � − → J (x, t) = H01 αeαx − H02 αe−αx sin ωtẑ = N I0 k sinh k(x − d2 ) sin ωtîz s cosh kd 2 Figure 2: |Jz | for blocks A and B for Problem 5.2 (Image by MIT OpenCourseWare.) Problem 5.3 A � � From the Boundary condition īx · ε0 Ē(x = 0+ ) − ε0 E(x = 0− ) = σs , we know Ex (x = 0+ ) − Ex (x = 0− ) = σs ε0 9 Problem Set 5 6.641, Spring 2005 and from the symmetry, we know Ex (x = 0+ ) = −Ex (x = 0− ). So Ex (x = 0+ ) = −Ex (x = 0− ) = σ0 cos(ay) . 2ε0 σs 2ε0 = We can build the boundary condition for the scalar potential: BC1 : Φ(x → ∞) = 0, Φ(x → −∞) = 0 ∂Φ(x, y) ∂Φ(x, y) σ0 cos(ay) (x = 0+ ) = + (x = 0− ) = ∂x ∂x 2ε0 Here we assume the general solution for the scalar potential is given: Φ(x, y) = AX(x)Ψ(y). As BC1 implied, X(x) = e−ax , x > 0; and eax , x < 0. With BC2, we find the scalar potential as: BC2 : − Φ(x, y) = σ0 e−ax cos(ay) ,x > 0 2aε0 σ0 eax cos(ay) ,x < 0 2aε0 ∂Φ ¯ ¯ With Ē = −�Φ = − ∂Φ ∂x ix − ∂y iy . Φ(x, y) = σ0 e−ax (cos(ay)īx + sin(ay)īy ), x > 0 2ε0 σ0 eax (cos(ay)īx − sin(ay)īy ), x < 0 Ē = − 2ε0 Ē = B For the electric field line, we have dx = dy dx = Ey Ex . At x > 0, cos(ay) dy ⇒ e−ax sin(ay) = constant sin(ay) At x < 0, dx = − cos(ay) dy ⇒ eax sin(ay) = constant sin(ay) C See Figure 3 Problem 5.4 A At region x > 0, we have the general solution ΨI = e−ax A sin(ay). At region x < 0, we have solution ΨII = eax B sin(ay). We chose exponentials in x for 0 potential at x = ±∞. We chose sines as H̄ = −�Ψ will have to satisfy cos(ay) and the boundary condition for x = 0. B � � ¯I − B ¯II = 0 ⇒ µ0 HIx |x=0+ = µHIIx |x=0− . Boundary Condition 1: n̄ · B � � ¯1 − H ¯ II = K0 cos(ay )īx . Boundary Condition 2: n̄ × H HIy |x=0+ − HIIy |x=0− = K0 cos(ay) Here H̄ = −�Ψ, H̄I = ae−ax A sin(ay)īx − ae−ax A cos(ay)īy ¯ II = −aeax B sin(ay)īx − aeax B cos(ay)īy H By BC1: µ0 A = −µB; By BC2: B − A = K0 a . 0 We can solve: A = − a(µµK and B = 0 +µ) 10 µ0 K 0 a(µ0 +µ) . Problem Set 5 6.641, Spring 2005 Figure 3: Electric Field Lines and Equipotential Lines for Problem 5.3C (Image by MIT OpenCourseWare.) C The solution for H field: H̄1 = � µK0 � −ax −e sin(ay)īx + e−ax cos(ay)īy (µ0 + µ) ¯ II = H µ0 K0 [−eax sin(ay)¯ix − eax cos(ay)īy ] (µ0 + µ) 11
