Wb/ Μ. /Α Ua-, / / Βζ * / 3.3. Ηλεκτρομαγνητισμός Ι Μ. 1. Β = k. 3. α) Β = Κ μ Π 2. B-r, 2 10~ ~ 2 α => I = ~ } Α k M I = 20Α

ΛΥΣΗ ΠΡΟΒΛΗΜΑΤΩΝ 3.3 39 3.3. Ηλεκτρομαγνητισμός 1. Β = k 21 9 1Π 2 β = 10 " ίιτκ τ^β = 2 10 " τ 3. α) Β = Κ μ 21 B-r, 2 10~ 5 20 10~ 2 α => I = ~ } Α k M -2 2-10 I = 20Α ϊ)β 2 2Ι = Κ ψ- _ 10' 10^40 7 2 40 = 5 2 20 10 >0 10" -22 Ίυ 4 '=ΊΓ ζ ι ε, 90 Λ -» l = = ea i r B = k 21 n 10 _7 2 6 B " 10 10" έ τ Β = 1,2 10 _5 Τ 5. α) Ομόρροπα: l Β ολ = Β 1 -Β 2 = Κ μ - Κ μ Βολ=-^(Ιΐ-"2) Β» = < - 1 0 > τ ~ /Α Ua-, / / Βζ * / Ι Μ Β 0 Λ = "10~ 5 Τ" β) αντίρροπα: Β ολ = B^Bg 4k Βολ =-Γ-(ΙΙ+'2) =* Β ολ = 4 10" 5 Τ. III ΐ2 Wb/ Μ.

40 ΛΥΣΗ ΠΡΟΒΛΗΜΑΤΩΝ 3.3 6. α) ομόρροπα Πρέπει Β, = Β 2 = r\i r 21, Κ,-βΙ, χ d-x 3 d-x ^ d-x = 3χ / (' / I, 12=31, / Bl *-xhik-d-x-> ' / /Β 2 / \ / \ / / / => d=4x =>χ = =7,5 cm. β) αντίρροπα Πρέπει = Β 2 => Κ., 2 - Κ,2 31 : d+x j\_ χ d+x d = 2χ => χ = d+x = 3χ d χ = 15 cm. 7. Πρέπει B 1 = Β 2 +Β 3 => k -2L. = k ^ - + k -!*~ χ r-x 2r-x li_ = Ji_ + _2.Sk χ r-x 2r-x (r-x)x = x(2r-x)+2,5x(r-x) ==> x 2-10x+16 = 0 => χ = 8cm ή χ = 2cm

ΛΥΣΗ ΠΡΟΒΛΗΜΑΤΩΝ 3.3 41 8. Β ολ = VB^BI (1) = Γ, => _ 1CT 7 Β ι 215 _ 4 Τ " 3 10~ 2 Τ _ 1 Τ ο, 2U Β 2 - k M ^ => 0 _ 10-7 2 20 4 8 2 " 4-10 Τ " 1 0 Τ (1) =* Β ολ = V(10~ 4 ) 2 +(10~ 4 ) 2 => Β ολ = 10" 4 Υ2 Τ 9. Πρέπει Β, = Β 2 => k _2lL- k Jk. μ ψ ~ Κμ χ => _[ι_ Ι Λ/3 1 * = ψ Χ εφθ = χ => θ = 30 : JL χ V3 1 I 2 X V } } y - I f με τον αγωγό που διαρρέεται από ρεύμα Ι,. 10. Β = k,, 2πΙ Br 2π 10 5-10-10 2 2nk,, 2π 10-7 Α => I = 10Α. 11. Β = k -^-Ν =* r = k r = 10 2 nm => 2πΙ Ν Β r = ncm - -m-7 2π 5 3 r = 10 3 10~ 4 m

42 ΛΥΣΗ ΠΡΟΒΛΗΜΑΤΩΝ 3.3 12 ι => ι = 20 Α ι Α Χ " 1 ~ R!+R 2 1 20 Β = k M =» Β = 1 ' 7,ο^Ο-» "* Β = 210_6Τ 13. B = k, 21,1 Γ ί Β = Κ,» Γ B. 1 0 -, J ^ 0 ± χ Β = 2 10~ 4 Τ 14. B 1 = B 2 = k M -^- =* Β, = Β 2 = 1 " 2 7."^1 1 2 Τ=10-4 Τ Β ολ = VBf+B => Β ολ = V(10~ 4 ) 2 +(10~ 4 ) 2 = 10" 4 V2T 15. α) Β = Β,+Β 2 => Β = k M -ψ- +k M => Β = k M (1+Π) β) Β = VB 2 +B 2 =* β οκ = J B 0 = k M -f-vi^ 2.

ΛΥΣΗ ΠΡΟΒΛΗΜΑΤΩΝ 3.3 43 16. Β, = k M ^ => Β, = 1CT 7 3 2 q 1 1 5 0 2 2 = 2-10" 5 Τ Β 2 = Β, = 2 10~ 5 Τ 2π ' 3 R 3 - Κk μ ρ-* Βο = 10~ 7 2π 30 π 15 10~ 2 ΒΟ = 4 10~ 5 Τ Άρα Β ΟΛ = Β 3 = 4-10 5 Τ α) Β, 2 = Β, Β 2 = 0 β) Β 1Ι2 = Β 1 +Β 2 = 4 10" 5 Τ Β ΟΛ = VB^I2 +B1 =* Β ΟΛ = V(4 10~ 5 ) 2 +(4 10~ 5 ) 2 Β ΟΛ = 4-10~ 5 V2 Τ 17. B 1 = k M^-,B 2 = k M 2nlj _ 2r ' Β 3 - --il ο Β, - i V 18. Πρέπει Β, = Β 2 k Κ -2b Μ k 5R ~ ΚΜ 2πΙ Ρ _L_ = _!2lL 5r r Ιι = 5 5 π π = 25A 19. Β = k π2ι I = 5 10~ 5 0,2 10~ 7 2π 50 π

44 ΛΥΣΗ ΠΡΟΒΛΗΜΑΤΩΝ 3.3 R = -f - Β = - ^ = 2πΩ π λ R = 2π Ω 2nr ~ 2π 0,2 ~ m ΟΠ ι 4 ^ϋ 'ι ~ _3R_ - 3R 4,2 ~ R η 3Β η 3k.2nli 3k u 2π 4Ε ι = e\r Βι = Τ - ~ Βι * 4r =* Βι = ==> V ~3^Γ Βι=Κ-~- ΌμοιαΒ 2 = - - =Κ~^~ Άρα: Β ολ = Β,-Βζ = 0 O i CD I Λ Ν D Η / 1-7 4 π 1 Ο 2 2 0 -Γ D yi Η/-ι~3-Γ B = 4 1 0 Τ 21. Β = k M -4n I => Β = 10 20-10-2 ~~ΓΓ ^ 22. Β = k M 4π -γ- I => I = Β = 8π 10~ 4 ι ρ Α V4n-f 10-' 4Π 10 3 23. Β ολ = Β;+Β 2 Ρ - Β! Κ, 4π Ν ι-ι 10" 7 4π 10 3 1 η- 4χ ~ 2? - 2π ' 1 Τ

ΛΥΣΗ ΠΡΟΒΛΗΜΑΤΩΝ 3.3 45 Β 2 = ^ = V y - I = 10~ 7, 4π 4 10 3 1 = 8π. 10 _ 4τ Β ολ = 10π 10~ 4 Τ => Β ολ = π 107 3 Τ 24. R z = NR = 20 Ω I = ε => = = 1Α Ρ ολ 20+20 D.. Ν. D 10~ 7 4π 10 3 Β = k -4n 7Γ I => Β = ί,=^d- 40π ΊΟ 2 Β = 10-3 Τ. 25. Πρέπει Β, = Β 2 => Ι< μ 4π I, = Ι< μ 2 ' 2 Ν 2 Λ Ν,. Ι 2 Ν 2 2Ν.Ι-, 101^2 ' 1 r r ^ 2.500 = 10 10 => r = = 0,1 m r 10 10 26. Β ς = k M -4n -7- >2- l 2 "Σ" Β Ζ = 10" - 7 4π 100 - Β ς = 4 10~ 4 Τ β ι = Κ μ^- =* Β1 = ΙΟ' 7 2 2^ 2 =* Β 1= 3.10-4 Τ Βολ = λ/βΐ+β^ = 5-10 4 Τ 27. α) Fi = Β Ι ημ90 = 2-10-0,2 = 4Ν β) F 2 = Β Ι ημ30 = 2-10-0,2 = 2Ν

46 ΛΥΣΗ ΠΡΟΒΛΗΜΑΤΩΝ 3.3 γ) F L = ΒΊ ημ90 = ON 28. Πρέπει; ιτ^ημ30 = F L = ιτ^ ημ30 = Β Ι => Β = ηπρ ημ30 ^ Β = 100-10~ 3-10 - 5-40-10-2 = 0,25 Τ 29. F L = Β-Ι- => F l = 0,4-10-20-10 -2 Ν F l = 0,8 Ν W Fl = F-S = F-^-at 2 = 0,8-^-2-10 2 = 80 J 30< α) Πρέπει F L = mg => B l,- = mg => ^. - JH9 0 1 10 = 2 5A ^ ' 1 ~ B - 2-0,2 ^ β) Πρέπει: mg-f L = ma => mg-b-l 2 = mg. jra_ = B,. t «- B\ 2 e - I 2 =4? 3B. 0,1-10-2, 5 2 2-0,2-3 3 γ) Πρέπει: F L -mg = ma => B-l 3 - -mg = m => B-l 3 - = mg+m -*J-

ΛΥΣΗ ΠΡΟΒΛΗΜΑΤΩΝ 3.3 47 ΒΊ 3 = 5mg 4 => l 3-5mg 4Β. 5 0,1 10 25_. 3 4-2 0,2 8 31. Φ\ y - L 4 F, Ι; α i Β <&> V ^F2 I Ρ 2 ημφ Γ ^F 2 L / 1 jk<p F? Ρ2συνφ x I ι 'F 3 Β \ R F 3 ' r ZF = 0 * α = 0 = 0 32. Για να ισορροπεί πρέπει: F l = mg => k M 211 χ ' 2 ' = mg - ι 2 = ^ χ k M -2li. _ 5-10~ 3-10-2-10~ 2 2 10~ 7-2-10 2-1 A => U = 50 A. Ν 33. Β ς = ^ 4π -J- \-I=> Β Σ = 10~ 7 4π 2,5 Τ => 10 Β ς = π 10~ 3 Τ

48 ΛΥΣΗ ΠΡΟΒΛΗΜΑΤΩΝ 3.3 Ρ Ρ.Ι C Ι FL Ι 2.ΤΤ. 10 Σ 2 2 2 By-e π 10" 3 40 10 2 => Ι 2 = 50 Α I ^ => R _ go 2 Ο " ROΛ Λ " 50 ~ 2 Ω R OK - R+ R => R = 2-0,5 = 1,5 Ω 34. Όταν δε διαρρέεται από ρεύμα mg = F e, άρα mg = 0,4 Ν F L = FI F 2 = 0,2N Αλλά: F l = Β Ι => Β = => β ' Ί ο Τ =210 ~ 2t 35. Όταν τα ελατήρια έχουν το φυσικό τους μήκος, ισχύει: F l = mg => Β Ι = mg => mg = 3Ν. Όταν έχουν επιμηκυνθεί κατά χ θα έχουμε: F L +mg = 2kx => ^. _ 2kx-mg 1 Β- ^ Ι 1 = 2-10-4,5 10 3-3 ^ 1,5 40 10-2 => h = 10 Α 36. Η επιπλέον ένδειξη οφείλεται στη δύναμη Laplace. Άρα: F L = 3Ν. Όμως ZF X = 0, άρα F 0)SL = ΣΡ Ψ =>

ΛΥΣΗ ΠΡΟΒΛΗΜΑΤΩΝ 3.3 49 => Ρ ολ = ΣΒΨΔ ημθ F L = Β Ι ΣΔ ημθ = Β Ι 2Γ Β = FL 2-l-r Β = 10-2-0,15 Τ = 1Τ 37. F 0, = F 1 -F 2 ==> ρ. - k 2Ι 1 -Ι 2 α - k 2I 1 -l 2 >-0λ ~ *μ d *μ d + p F«= V 2,l 2a(-^^ F 0 x = 10~ 7 2 10 5 0,1 40-10~ 2 10 10~ 2 (40 10~ 2 +10 10~ 2 ) F = k = -5^ = 8-1 0 _ 6 N OO, 2l r lp- 10" 7 2 10 50 10 38. F L = k *- 2 - F L = r j F, = 5-10 N. 39. Πρέπει: F, = F 2 => μ 2Ιι Ι 3 2I p -I 3 χ μ r-x _LL = χ '? r-x 51, r-x =*> r-x = 5x ==> r = 6x => χ = 12 = 2cm. 40. Β-, = Κ Μ = 10 4^Q2 :F => Β, = 2 10~ 4 Τ mg = By\ 2 - => mg = 2-10 -50-2 => mg = 2-10 ' d N.

50 ΛΥΣΗ ΠΡΟΒΛΗΜΑΤΩΝ 3.3 F L +mg = 2k-Ax => B 2 -l 2 - +mg = 2k-Ax => k " itvto +m9 = 2kAx 10~ 7 2 40 50 2 5-10~ 2 +2-10~ 2 = 2k-10~ 2 10~ 2 1,6+2 10" 2 = 2k-10~ 2 => 3,6-10 ~ 2 = 2k-10~ 2 => k = 1,8 Ν m 41. FA,k = ~^ k" < -» 10~ 7 2-10-20 Fa.K" 10-10" 2 1 F a,k = 4-10" 4 Ν Ρδκ = α = 4-10" 4 Ν F A, r X Τ l\45 F Ajk Ιλ,Δ F,. FΑ,Γ 2k / => F Ar = 10" 7, 2 η ' 1 Α - 1 2 Λ / 0 1 = V2-10" 4 Ν αυ2 ' 10-10 V2 Fi = V(4-10~ 4 ) 4 +(4-10~ 4 ) 2 = 4-10" 4 V2 Ν F ox = F^Fat = 4-10~ 4 V2 V2-10" 4 = 3V2-10~ 4 Ν 42. B 0 = k M 4π -γ-1 => B 0 = 10~ 7 4π 10 2 10 ^ B 0 = 4π 10~ 4 Τ Β = μβ 0 = 4π 10" 1 = 0,4 π Τ

ΛΥΣΗ ΠΡΟΒΛΗΜΑΤΩΝ 3.3 51 43. α) Φ = B-S-συνΟ" = 2-20-10" 4 = 4-10" 3 wb β) Φ = B S-auv90 = Owb γ) Φ = B S-auv60 =* Φ = 2 20 10-4 -^- = 2-10 -3 wb. ΛΛ ΔΦ Κ1 10-2-10 2 44. ε = it- Ν = = 5ν 0,2 45. S = π-r 2 Φαρχ^Β-S Φτελ 0 >=> ΔΦ= I Φτελ-ΦαρχΙ =B"S e = JmL n^ e = 4P-N 0,1 π(10 10~ 2 ) 2. Α λ 2 = h : L 1 => ε = π 10 ν. U, ι 46. S = π r 2 = π(20 10 2 ) 2 => S = π 400 10 4 = 4π 10 2 m 2 α) Φ αρχ = B-S ' => ΔΦ = I Φ τελ Φ αρχ Ι = 3B S Φ τελ = 4BS ε = I Δ* I Ν ^ ε = _ 3 ρ _ Ν ε = 3 2 4π 10~ 2 π 20 => ε = 4,8 Ν. ε = 4,8 ν

52 ΛΥΣΗ ΠΡΟΒΛΗΜΑΤΩΝ 3.3 β) Φ αρχ B-S Φτ«= τ s ΗΔΦΙ = I Φ τελ Φ, αρχ! -f- S-B-Sl = 4" B-S 4 4 e = -t#l Ν = m - N - e = 3-2 1 n : 1 2 4-4 π 20-1.2v Υ) Φ αρχ B-S. ΑΦ = IΦ τελ φ αρχ I = l-bs-bsl = 2BS Φ τελ = BS συν 180 ε = I ΑΦ I 2BS K1 2 2 4π 10~ 2 ηλ Ν =*> ε = γγ- Ν = ζ 20 = 3,2 ν π 47. R n = N'R, = 90Ω Βολ = Βπ+^2 = 100 Ω α) ε = ΔΦ Ν = ^ φτελ~ α; b φ α Ρ χ_ι Ν = 12BS-BSI IN =* t Δί Ν e=jg-n=*e = 2-100-10' 4 1 10 2 = 2v Rc* 100 = 2 10~ 2 A β) ε = ΑΦ Ν» e = ν ^ ε = -4 2100-10 1 10 2 => ε = 2ν R ολ 100 = 2 10~ 2 Α.

ΛΥΣΗ ΠΡΟΒΛΗΜΑΤΩΝ 3.3 53 48. α) Φ αρχ = BS φ τελ ~ 0 ΔΦ= Φ τελ -Φ αρ χι ΔΦ = 10-BS = k M -4n -y- I S ΔΦ = 10~ 7 4π 500 2 20 10~ 4 ΔΦ = 8π 10~ 7 wb ε = I ΔΦ Ι Ν = 4π 10 2 ν Q = Ν = 8π 1 7 500 = n10" 5 C Rox 40 β) <»apx=bs Φ τελ = MBS ΔΦ = BS(M-1) ε = Ν Κ μ 4π y i (μ-1) Ν S ε = 0,8 πν Q = ΔΦ R Ν = 2n-10~ 2 C 49. α) ε = ε = nr 2 ΙΔΦΙ ΔΒ Δί => ε = ABS ε = 2-n(10Vn-10~ 2 ) 2

54. ΛΥΣΗ ΠΡΟΒΛΗΜΑΤΩΝ 3.3 ε = 2π 10 2 π 10~ 4 => ε = 2π 2 10 2 => ε = 0,2ν Q = Οε =*> Q = 2-0,2 = 0,4μ0 β) U = ~Ct 2 => U = - -2-10 H3 (0,2) 2 =>U = 10" 6-4 10~ 2 = 4-10" 8 J 50. Φ αρχ = BS Φ-Α = 0 ' =» ΔΦ = I Φτελ - Φαρχ I = BS 0 = ψ-μ ^Q = - S-N =Φ "ολ 'ολ Β = ' Βρλ => S.N Β = 5 10~ 3 50 3 0,2-20 ^ Β = 10 3 Τ 51 ' 61 = ^ Ε1 = - ^ = ' 2ν α) Από (1 εως 2)s ε 2 = = ' 2 ~ 0,2 = 0V Από (2 εως 3)3 ε 3 = = ( ~» 2 ) = -0,2V

ΛΥΣΗ ΠΡΟΒΛΗΜΑΤΩΝ 3.3 55 = =0,02A Ι 2 = OA Ι 3 = -0,02Α 52. ε = Βυ => ε = 0,2-10-2 = 4V 53. α) ε = Βυ => ε = 0,8 5 0,5 => ε = 2 V -Ι--^--0,2 Α β) Ρ = l 2 R ox =* Ρ = (0,2) 2 10 = 0,4 w γ) F ee r = F L = BL = 0,8-0,2-0,5 = 0,08 Ν δ) VKA = IR => = 0,2-2 = 0,4 V. 54. Πρέπει F = F L = 0,4 Ν Fl = ΒΙ => FL = Β ^ ί => R ~R~' Βυ = υ = F,-R 55. R 0 * - +Βκλ =* Ro\ - "4r~ + 2-4Ω ολ " Βυ ~Ε =* 'ολ = 0,2-10-1 7 = ' 5Α Μ ολ 4 VKA = U-RI,2 = 0,5-2 = 1V

56 ΛΥΣΗ ΠΡΟΒΛΗΜΑΤΩΝ 3.3 Ιι = 12 = < 2 > 1 RI 6 VKA 1 R 2 3 56. Πρέπει: F L = 0 => I = 0 f 10 Αρα: ε = ε επ => ε = Buf => υ = => υ =. " = 25 m/s Be 0,4-1 57. Πρέπει F L = ητ^ημφ => ΒσυνφΙί = ιτ^ημφ => Ρ Βσυνφ ( = mgημφ => Η 0 λ Βσυνφυί Βσυνφ 5 ^ = m 91M9 ολ U = rngmj^r, ολ Β συν φ 20-10 3 1 010 υ = 1 2 1 (V3/2) 2 ^ U = 3 m/s r_ i ΔΦ B-AS για =T Β πί 2 f= ι\τ 2 ε ~ Δί ~ ΔΙ Ε ~ Τ ^ ε - Β π ε = 0,5 π(0,15) 2 60 =* ε = 2,12V. 59 ε = ΔΦ = BAS v =T ^ = Β π(κμ) 2 Βπ(ΚΛ) ζ

ΛΥΣΗ ΠΡΟΒΛΗΜΑΤΩΝ 3.3 57 ε = Βπί W-(KA2) ε = Βπ(ΚΜ 2 -ΚΛ 2 )ί ε = 10~ 4 π(3 2-2 2 )-^- =Φ ε = 10" 2 V ' π β. ε = = ^ ψ - => ε = Βίπί 2 ε = 0,2 π 3 2 π ε = 18V 4ιλ = φ => ί ΜΛ = y 3 = nm. Έχουμε 9Ω για πί RMA =! Π R = ^ = 3 Q Ιι = R, ΜΛ = 6Α - 1 Ο Λ 2 R-FW 9-3 'ολ - 'ι + '2-9Α.