6.641 Electromagnetic Fields, Forces, and Motion

Transcription

1 MIT OpenCoureWare Electroagnetic Field, Force, and Motion Spring 009 For inforation about citing thee aterial or our Ter of Ue, viit:

2 6.64 Electroagnetic Field, Force, and Motion Spring 009 Proble Set 5 - Solution Prof. Marku Zahn MIT OpenCoureWare Proble 5. (i) v d = (q v ) (µ 0 H ); H = H0 ˆi z dt d v = qµ 0 dt d v qµ 0 = dt î x î y î z v x v y v z 0 0 H 0 (v y H 0 î x v x H 0 î y ) dv x dt qµ 0 v y = H 0 (ii) dv y dt qµ 0 v x = H 0 d v x qµ 0 H 0 dv y (i) = dt dt Subtitute thi into (ii): d v x qµ 0 v x = H 0 qµ 0 H 0 dt d v x q µ = 0H 0 ( v x dt ) ( ) qµ 0 H 0 qµ 0 H 0 v x = A co t + B in t ( ) ( ) qµ 0 H 0 qµ 0 H 0 v y = C co t + D in t v x (t = 0) = v x0 = A v y (t = 0) = v y0 = C Need two ore initial condition: I. Acceleration in x direction at t = 0 dv x = (qv y 0 î y ) (H 0 î z )µ 0 îx dt t=0 qµ 0 H 0 qµ 0 H 0 B = v y0 B = v y0

3 Proble Set , Spring 009 II. Acceleration in y direction at t = 0 dv y î y = (qv x 0 î x ) (µ 0 H 0 î z ) dt t=0 D = v x0 ( ) ( ) qµ 0 H 0 qµ 0 H 0 v x (t) = v x0 co t + v y0 in t ( ) ( ) qµ 0 H 0 qµ 0 H 0 v y (t) = v y0 co t v x0 in t v z (t) = v z0 Note: v x + v y + v z = contant in tie (for thi cae) (eay to check and verify) = vx + v + v 0 y 0 z 0 v xy = v + v x 0 y = velocity on xy plane 0 Fro 8.0, centripetal acceleration, a, i v xy a = r where r i radiu of circle. So: v xy = q v µ0h }{{ r }}{{} F =a Force due to B field v xy = q v xy µ 0 H 0 r v v + v xy x 0 y 0 r = = qµ 0 H 0 qµ 0 H 0 A i ii = 0 = q f E + q v µ0 H 0 = q E + q(v 0 î y ) (µ 0 H 0 ˆi z ) E = v0 µ 0 H 0ˆ ix V = v 0 µ 0 H 0 v 0 V d = r = ; v 0 = qµ 0 H 0 µ 0 H 0 d = v =0.50c for Mg 4 qb 0 0.5c for Mg c for Mg 6

4 Proble Set , Spring 009 B i ( ) d v = q E + q V0 v (µ0h ) = e ˆi x + v (µ 0 H 0 î z ) dt v (µ0 H 0 î z ) = v y H 0 µ 0 î x v x H 0 µ 0 î y dv x V 0 e = eµ 0 H 0 v y dt dv y = eµ 0 H 0 v x dt ( ) d v y eµ 0 H 0 ev 0 = eµ dt 0 H 0 v y d v y ( eµ0 H 0 ) e µ 0 H 0 + v y = V 0 () dt d v x dt e µ 0 H = 0 v x d ( ) v x eµ0 H 0 + v x = 0 dt () Solution to equation and (hoogenou + particular): ( ( ) ( )) V 0 eµ 0 H 0 eµ 0 H 0 v y = + c in t + c co t µ 0 H 0 ( ( ) ( )) V 0 eµ 0 H 0 eµ 0 H 0 v x = c co t c in t µ 0 H 0 (v fro eµ 0 H 0 v x = dvy x dt ) ii v x (t = 0) = v y (t = 0) = 0 c = 0, c = ( { } ) V 0 eµ 0 H 0 v (t) = in t ˆ eµ 0 H 0 i x + co t ˆi y µ 0 H 0 ( ) ( ) V 0 eµ 0 H 0 V 0 V 0 eµ 0 H 0 v (t)dt = d (t) = îx + t ˆ µ 0 H co 0 e t µ 0 H 0 eµ 0 H 0 in t i y V 0 d x (t) ax < µ0 H 0 e < V 0 H 0 > µ 0 e 3

5 Proble Set , Spring 009 Proble 5. A C H dl = J da S For contour through pace to left of block H L = NI = NI 0 co ωt y H L = NI 0 in ωt y For contour through pace to right of block 0 R H y = NI 0 co ωt + J z da (all current in +z direction ut return in z direction) Block H R = NI 0 in ωt y B H = J For Block A,l J = σ E. H = σ E For block B J = ω ε p E t (3) (4) ( H ) = ( J ) t t H J = t t (5) 45: C Block A H = ω p ε E t H = σ E H = σ E Aue unifor σ, and ue A = ( A ) A. ( H ) H = σ( E ) 4

6 Proble Set , Spring 009 Faraday Law E = µ H t Flux continuity: µ H = 0; for unifor µ H = 0. = σµ H H t Block B H ε = ω E t p H ε = ω E t p Uing A = ( A ) A and auing unifor propertie H H ( ) = ω p ε( E ) t t Flux continuity for unifor µ H = 0 and uing Faraday Law H H = ωp εµ t t Integrate and aue that integration contant i zero ω = ω pεµ p H H = H c c = εµ = peed of light in aterial D { } Aue H = I Hˆ y (x)ejωtˆ i y H H = σµ for Block A. t Ĥ y (x)e jωt = σµjω Ĥ y (x)e jωt x Ĥ y (x) = σµjω Ĥ y (x) x Aue Ĥ y (x) = Ĥ 0 e jkx. k H ˆ e jkx Ĥ jkx 0 0 = σµ e jω k = σµjω, k = ± ωµjσ σµω k = ± ( j) σµω α 5

7 Proble Set , Spring 009 { } H y = I Ĥ 0 e αx e jαx e jωt + H 0 e αx e jαx e jωt Apply B.C. to coplex H. At x = 0, K = 0 { } NI 0 = I e jωt Ĥ0 e 0 e 0 e jωt + Ĥ0 e 0 e jωt NI 0 jωt = e Ĥ 0 + Ĥ 0 = NI 0 At x = d, K = 0 H y (x = d, t) = NI a in ωt Ĥ 0 e αd e jαd e jωt + Ĥ 0 e αd e jαd e jωt = NI 0 e jωt Ĥ 0 e αd e jαd + Ĥ 0 e αd e jαd = NI 0 Fro B.C. at x = 0 Ĥ = NI 0 Ĥ 0 0 NI 0 e αd e jαd H [ e αd e jαd e αd e jαd] = NI 0 0 ( ) NI 0 αd jαd e Ĥ e 0 = (e αd e jαd e αd e jαd ) Ĥ NI 0 NI 0 e αd e jαd 0 = e αd e jαd e αd e jαd NI 0 e αd e jαd = e αd e jαd e αd e jαd { } H = Re Hˆ0 e αx e jαx e jωt + Ĥ0 e αx e jαx e jωt ˆi y Or NI 0 coh γ(x d H = I ) e jωt î y coh xd +j σµω with γ = δ, δ = ωµσ, α = = { δ. } For Block B: Aue H y (x, t) = I Ĥ 0 e jkx e jωt. H = ω εµ H p Ĥ 0 e jkx e jωt = ω p εµĥ 0 e jkx e jωt x k = ω p εµ k = ± ω p εµ = ±ω p εµj H y (x = 0, t) = NI0 in ωt. 6

8 Proble Set , Spring 009 { } (x) = I Ĥ αx e jωt + Ĥ 0 e αx e jωt H y α = ω p εµ 0 e 0 Apply B.C. at x = 0, K = 0 H + y = I0 N in ωt. Ĥ 0 e 0 e jωt + Ĥ 0 e 0 e jωt = I 0N e jωt Ĥ 0 + Ĥ0 = NI 0 d Apply B.C. at x = d, K = 0 H y = I0 N in ωt. αd jωt I 0 N jωt Ĥ αd jωt y e e + Ĥ 0 e e = e Ĥ αd + Ĥ 0 e αd NI 0 0 e = NI 0 NI 0 Hˆ0 e αd + Ĥ 0 e αd Ĥ [ 0 e αd e αd] NI0 = [ e αd] Ĥ NI 0 e αd 0 = e αd e αd Ĥ NI 0 e αd 0 = e αd e αd { } (x) = I Ĥ αx e jωt + Ĥ 0 e αx e jωt H y 0 e Or α = ω p εµ H NI 0 e αd 0 = e αd e αd H NI 0 e αd 0 = e αd e αd H y (x, t) = H 0 e αx + H 0 e αx in ωt H y (x, t) = NI 0 coh α(x d ) in ωt, with α = ω p = ω p εµ coh αd c 7

9 Proble Set , Spring 009 Figure : H y for block A and B for Proble 5. (Iage by MIT OpenCoureWare.) E For Block A H = J Since H = H y (x) H y (x) J = ẑ x { } J = I αx jαx jωt Ĥ0 + Ĥ0 e jαx jωt NI 0 γ inh γ(x d ) (α + jα)e e e ( α jα)e αx e ẑ = I jωt coh( γd e ) + j γ =, δ = δ ωµσ For Block B H y (x) J = ẑ x 8 î z

10 Proble Set , Spring 009 { } J = I jωt jωt Ĥ0 αe αx e Ĥ 0 αe αx e ẑ or [ J (x, t) = H0 αe αx H 0 αe αx] in ωtẑ = NI 0 k inh k(x d ) in ωtîz, k = ω p coh kd c Figure : J z for block A and B for Proble 5. (Iage by MIT OpenCoureWare.) Proble 5.3 A Fro the Boundary condition i ε 0 Ē(x = 0 + ) ε 0 E(x = 0 x ) = σ, we know σ E x (x = 0 + ) E x (x = 0 ) = ε0 9

11 Proble Set , Spring 009 and fro the yetry, we know E x (x = 0 + ) = E x (x = 0 ). So E x (x = 0 + ) = E x (x = 0 σ ) = σ 0 co(ay) ε 0. We can build the boundary condition for the calar potential: BC : Φ(x ) = 0, Φ(x ) = 0 ε 0 = BC : Φ(x, y) (x = 0 + ) = + Φ(x, y) (x = 0 ) = σ 0 co(ay) x x ε 0 Here we aue the general olution for the calar potential i given: Φ(x, y) = AX(x)Ψ(y). A BC iplied, X(x) = e ax, x > 0; and e ax, x < 0. With BC, we find the calar potential a: σ 0 e ax co(ay) Φ(x, y) =, x > 0 aε 0 σ 0 e ax co(ay) Φ(x, y) =, x < 0 aε 0 With Ē = Φ = Φ ī x Φ x y ī y. B C Ē = σ 0e ax (co(ay)ī x + in(ay)ī y ), x > 0 ε 0 Ē = σ ax 0e (co(ay)ī x in(ay)ī y ), x < 0 ε 0 For the electric field line, we have dy = Ey. At x > 0, dx E x At x < 0, dx = co(ay) dy e ax in(ay) = contant in(ay) dx = co(ay) dy e ax in(ay) = contant in(ay) See Figure 3 Proble 5.4 A At region x > 0, we have the general olution Ψ I = e ax A in(ay). At region x < 0, we have olution Ψ II = e ax B in(ay). We choe exponential in x for 0 potential at x = ±. We choe ine a H = Ψ will have to atify co(ay) and the boundary condition for x = 0. B Boundary Condition : n B [ I B II ] = 0 µ 0 H Ix x=0+ = µh IIx x=0. Boundary Condition : n H H II = K 0 co(ay)ī x. H Iy x=0+ H IIy x=0 = K 0 co(ay) Here H = Ψ, H I = ae ax A in(ay)ī x ae ax A co(ay)ī y H II = ae ax B in(ay)ī x ae ax B co(ay)ī y By BC: µ 0 A = µb; By BC: B A = K0. We can olve: A = µk 0 µ and B = 0K 0 a a(µ 0+µ) a(µ 0+µ). 0

12 Proble Set , Spring 009 Figure 3: Electric Field Line and Equipotential Line for Proble 5.3C (Iage by MIT OpenCoureWare.) C The olution for H field: µk 0 H = (µ0 + µ) e ax in(ay)ī x + e ax co(ay)ī y x > 0 H II = Proble 5.5 µ 0 K 0 [ e ax in(ay)ī x e ax co(ay)ī y ] x < 0 (µ 0 + µ) Line current I of infinite extent above a plane of aterial of infinite pereability, µ 0. A B = µh for µ, in order to have B finite, we need H zero continuity of noral B and tangential H at the urface. B Uing ethod of iage to atiify becaue at y = 0 for ediu where µ becaue H x = H z = 0 at y = 0 For a line current at origin (ee Figure 6)

13 Proble Set , Spring 009 Figure 4: A diagra howing a line current I of infinite extent above a plane of aterial of infinite pereability with field line. Figure 5: A diagra howing how to apply the ethod of iage with the iage current in the ae direction a the ource current. (Iage by MIT OpenCoureWare). C H d I l = I H φ = πr µ 0 I A z Iµ 0 B = πr i φ, ince B = A r = πr Iµ 0 A z = π ln r contant for line current I at z = d and I at z = d { [ ] [ ]} Iµ 0 A z = π ln x + (y d) + ln x + (y + d)

14 Proble Set , Spring 009 Figure 6: A diagra depicting a Gauian Contour urrounding a line current (Iage by MIT OpenCoure- Ware). { } Iµ 0 A z = 4π ln x + (y d) x + (y + d) C B = A = A z i y + A z i x y x x + (y + d) + x x + (y d) (y d) x + (y + d) + (y + d) x + (y d) = Iµ 0 i y Iµ 0 i x 4π x + (y d) x + (y + d) 4π x + (y d) x + (y + d) Iµ 0 (y + d)i x xi y (y d)i x xi y B = + π x + (y + d) x + (y d) x D Force i applied on the line current due to the iage line current Force per unit length: F = I B field due to iage charge at (x = 0, y = d) ( ) = I µ 0 I i z i x π d µ 0 I = i o line current i attracted to the urface 4πd y 3