6.641 Electromagnetic Fields, Forces, and Motion
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MIT OpenCourseWare http://ocw.mit.edu 6.641 Electromagnetic Fields, Forces, and Motion Spring 2009 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 2009 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 − → dv qµ0 = (vy H0 îx − vx H0 îy ) dt m m (i) dvx qµ0 vy = H0 dt m (ii) dvy −qµ0 vx = H0 dt m qµ0 H0 dvy d2 vx = dt2 m dt Substitute this into (ii): (i) ⇒ m d2 vx qµ0 vx =− H0 2 qµ0 H0 dt m d 2 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 � dvx �� (qvy0 îy ) × (H0 îz )µ0 îx = dt �t=0 m � � � qµ� qµ� � 0 H0 0 H0 B = vy0 m m � � B = vy0 1 Problem Set 5 6.641, Spring 2009 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 vx (t) = vx0 cos t + vy0 sin t 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 H0ˆiz ) − → E = −v0 µ0 H0ˆix V = −v0 µ0 H0 s ii d = 2r = d= 2mv0 V ; 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 2009 B i � � − d→ v V0 ˆ → − → − − → − → m = q E + q v × (µ0 H ) = −e − ix + v × (µ0 H0 îz ) dt s → − v × (µ0 H0 îz ) = vy H0 µ0 îx − vx H0 µ0 îy dvx V0 e = − eµ0 H0 vy dt s dvy m = eµ0 H0 vx dt � � d2 vy eµ0 H0 eV0 − eµ0 H0 vy →m 2 = dt m s m d 2 vy + dt2 m � eµ0 H0 m �2 vy = e2 µ0 H0 V0 m2 s (1) d 2 vx e2 µ20 H02 =− vx 2 dt m d 2 vx + dt2 � eµ0 H0 m �2 vx = 0 (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 � � � � �� V0 eµ0 H0 eµ0 H0 vx = c1 cos t − c2 sin t µ0 H0 s m m (vx from eµ0 H0 vx = m dvy dt ) vx (t = 0) = vy (t = 0) = 0 ⇒ c1 = 0, c2 = −1 � � � � � �� � V0 eµ0 H0 ˆ eµ0 H0 − → v (t) = sin t ix + 1 − cos t ˆiy µ0 H0 s m m � � �� � � �� � mV0 eµ0 H0 V0 mV0 eµ0 H0 → − → − ˆiy v (t)dt = d (t) = 2 2 1 − cos t îx + t− sin t 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 2009 Problem 5.2 A � − − → → H · dl = � C → − − → J · dA S 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 � = N I0 cos ωt + �0 � � �Jz da (all current in +z direction must return in −z direction) � �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 �× → → − − ∂H ∂J = ∂t ∂t �× → − ∂H → − = ωp2 ε E ∂t (4) (5) 4→ 5: 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 2009 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 → − 2 ∂H �(� · )−� = ωp2 ε(� × E ) ∂t ∂t − → Flux continuity for uniform µ → � · H = 0 and using Faraday’s Law → → − − ∂H 2 ∂H 2 � = ωp εµ ∂t ∂t �× Integrate and assume that integration constant is zero ωp2 → − → − → − �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 for Block A. ∂t − 2→ ∂ 2 Ĥy (x)ejωt = σµjω Ĥy (x)ejωt ∂x2 ∂ 2 Ĥy (x) = σµjω Ĥy (x) ∂x2 Assume Ĥy (x) = Ĥ0 ejkx . � = σµĤ�� � jkx ˆ� −k 2� H ejkx 0� �0 e jω � −k 2 = σµjω, k = ± −ωµjσ � σµω (1 − j) k=± 2 � σµω α≡ 2 5 Problem Set 5 6.641, Spring 2009 � � → − 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 e 0 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 = N I0 − Ĥ02 s � � N I0 N I0 αd jαd e e − H0 2 eαd ejαd − e−αd e−jαd = s s � � N I0 αd jαd 1−e e s Ĥ02 = −αd −jαd (e e − eαd ejαd ) � � N I0 N I0 1 − eαd ejαd 1 − Ĥ0 = s s e−αd e−jαd − eαd ejαd � � N I0 e−αd e−jαd − 1 = s e−αd e−jαd − eαd ejαd � � → − ˆ 01 eαx ejαx ejωt + Ĥ02 e−αx e−jαx ejωt ˆiy H = Re H Or � � N I0 cosh γ(x − d2 ) jωt − → H = Im e îy 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 ∂2 Ĥ0 ejkx ejωt = ωp2 εµĤ0 ejkx ejωt ∂x2 −k 2 = ωp2 εµ � √ k = ± ωp2 εµ = ±ωp εµj 6 N I0 s sin ωt. Problem Set 5 6.641, Spring 2009 � � − → 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 e−αd − s s � � N I0 � � Ĥ02 e−αd − eαd = 1 − eαd s � � N I0 1 − eαd 2 Ĥ0 = s e−αd − eαd � −αd � N I0 e −1 1 Ĥ0 = 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 1 H0 = 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 2009 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 − → 1 αx jαx jωt 2 −αx −jαx jωt J = Im Ĥ0 (α + jα)e e e + Ĥ0 (−α − jα)e e e ẑ = Im e îz s cosh( γd 2 ) � 1+j 2 γ= , δ= δ ωµσ For Block B ∂Hy (x) − → J = ẑ ∂x 8 Problem Set 5 6.641, Spring 2009 � � − → J = Im Ĥ01 αeαx ejωt − Ĥ02 αe−αx ejωt ẑ or � � → − J (x, t) = H01 αeαx − H02 αe−αx sin ωtẑ = N I0 k sinh k(x − d2 ) ωp sin ωtîz , k = kd s c cosh 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 ¯ix · 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 2009 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 īx − ∂y īy . Φ(x, y) = σ0 e−ax (cos(ay)īx + sin(ay)īy ), x > 0 2ε0 σ0 eax Ē = − (cos(ay)īx − sin(ay)īy ), x < 0 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 � � Boundary Condition 2: n̄ × H̄1 − H̄II = K0 cos(ay)īx . 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 H̄II = −aeax B sin(ay)īx − aeax B cos(ay)īy By BC1: µ0 A = −µB; By BC2: B − A = K0 a . 0 We can solve: A = − a(µµK and B = 0 +µ) 10 µ0 K0 a(µ0 +µ) . Problem Set 5 6.641, Spring 2009 Figure 3: Electric Field Lines and Equipotential Lines for Problem 5.3C (Image by MIT OpenCourseWare.) C The solution for H field: ¯1 = H � µK0 � −ax −e sin(ay)īx + e−ax cos(ay)īy (µ0 + µ) H̄II = µ0 K0 [−eax sin(ay)īx − eax cos(ay)īy ] (µ0 + µ) x>0 x<0 Problem 5.5 Line current I of infinite extent above a plane of material of infinite permeability, µ → 0. A � = µH � ⇒ for µ → ∞, in order to have B finite, we need H zero ⇒ continuity of normal B � and tangential B � H at the surface. B Using method of images to satisify because at y = 0 for medium where µ → ∞ because ⇒ Hx = Hz = 0 at y = 0 For a line current at origin (see Figure 6) 11 Problem Set 5 6.641, Spring 2009 Figure 4: A diagram showing a line current I of infinite extent above a plane of material of infinite permeability with field lines. Figure 5: A diagram showing how to apply the method of images with the image current in the same direction as the source current. (Image by MIT OpenCourseWare). � � · d�l = I ⇒ Hφ = I H 2πr C µ0 I � � =�×A � ⇒ − ∂Az = Iµ0 iφ , since B 2πr 2πr ∂r Iµ0 ln r → constant ⇒ Az = − 2π ⇒ for line currents I at z = d and I at z = −d � �� � �� �� 0 2 + (y − d)2 + ln 2 + (y + d)2 Az = − Iµ ln x x 2π � = ⇒B 12 Problem Set 5 6.641, Spring 2009 Figure 6: A diagram depicting a Gaussian Contour surrounding a line current (Image by MIT OpenCourseWare). ⇒ Az = − �� �� Iµ0 � � 2 2 2 ln x + (y − d) x2 + (y + d) 4π C � =�×A � = − ∂ Az i�y + ∂ Az i�x B ∂y ∂x � � � � � � � � 2 2 2 2 2 2 2 2 Iµ0 2x x + (y + d) + 2x x + (y − d) � Iµ0 2(y − d) x + (y + d) + 2(y + d) x + (y − d) � � �� � � �� � iy − ix = 2 2 2 2 4π 4π x2 + (y − d) x2 + (y + d) x2 + (y − d) x2 + (y + d) � = − Iµ0 B 2π � (y + d)i�x − xi�y (y − d)i�x − xi�y + 2 2 2 x + (y − d)2 x + (y + d) � D Force is applied on the line current due to the image line current Force per unit length: � ← field due to image charge at (x = 0, y = −d) F� = I� × B � � µ0 I 1 � � � = I iz × ix 2π 2d = − µ0 I 2 � iy 4πd so line current is attracted to the surface 13
