פתרון בגרות חשמל — שאלון 36371

d = 4 m | E(4) = 1810 V/m | k = 9·10⁹ N·m²/C²
Q > 0 | σ > 0 | V(∞) = 0

4.0 m

3.0·10⁻⁸ C/m²

2.0·10⁻⁷ C

112.5

1696

1809

93.8

7.0

E_Q(x) = kQx²

E_σ = σ2ε₀ = 2πkσ

E(x) = kQx² + σ2ε₀

E_∞ = σ2ε₀ ≈ 1696 V/m

σ = 2ε₀ · E_∞ = E_∞2πk = 16962π · 9·10⁹ = 3.0·10⁻⁸ C/m²

kQd² = 1810 − 1696 = 114 V/m Q = 114 · d²k = 114 · 169·10⁹ = 2.0·10⁻⁷ C

E(4) = 9·10⁹ · 2·10⁻⁷16 + 2π · 9·10⁹ · 3·10⁻⁸ = 112.5 + 1696.5 = 1809 V/m

F = qE(x) → a = qE(x)m → a_∞ = qσ2ε₀m = const

V_d,2d = V(d) − V(2d) = [V_Q(d) − V_Q(2d)] + [V_σ(d) − V_σ(2d)]

V_Q(d) − V_Q(2d) = kQd − kQ2d = kQ2d = 9·10⁹ · 2·10⁻⁷8 = 225 V

V_σ(d) − V_σ(2d) = σ2ε₀ · d = 1696 · 4 = 6786 V

V_d,2d = 225 + 6786 = 7011 ≈ 7.0·10³ V

d = 4 m → d2 = 2 m < d = 4 m < 2d = 8 m

E_σ = σ2ε₀

V_d,2d = kQ2d + σ2ε₀ · d

R₁ = 4 Ω | λ = 0.8 Ω/cm | ℓ_MN = 30 cm | R_MN = 24 Ω
I = 1.5 A → V = 18 V ; I = 2.5 A → V = 16 V

10.0 cm

8.0

12.0

1.50

18.0

20.0

V = ε − I·r , I = εR + r

R ↑ → I ↓ → I·r ↓ → V ↑

ε = Wq → [ε] = JC = V

18 = ε − 1.5r 16 = ε − 2.5r

18 − 16 = (ε − 1.5r) − (ε − 2.5r) = r r = ΔVΔI = 21 = 2 Ω

ε = 18 + 1.5 · 2 = 21 V

R_כוללת = εI = 211.5 = 14 Ω

R_MP = 14 − r − R₁ = 14 − 2 − 4 = 8 Ω

ℓ_MP = R_MPλ = 80.8 = 10 cm

ℓ_PN = 30 − 10 = 20 cm

R_{מקבילית} = R_MP · R₁R_MP + R₁ = 8 · 412 = 83 ≈ 2.67 Ω

I = εr + R_{מקבילית} = 212 + 83 = 21143 = 4.5 A

V = ε − I·r = 21 − 4.5 · 2 = 21 − 9 = 12 V

R_MP = 24 Ω → R_{מקבילית} = 24 · 428 = 3.43 Ω I = 215.43 = 3.87 A → V = 21 − 7.74 = 13.3 V

R_MP = 0 → R_{מקבילית} = 0 · 40 + 4 = 0 I = εr = 212 = 10.5 A V = ε − I·r = 21 − 10.5 · 2 = 0

כל נורה: 3 V / 2 W | V₁ = 9⅓ V | V₂ = 4 V
מעגל 1: שלוש נורות בטור | מעגל 2: שלוש נורות במקביל

0.50 Ω

0.67

9.33

8.00

1.78

77.8

I_נורה = PV = 23 ≈ 0.667 A R_נורה = V²P = 92 = 4.5 Ω

I₁ = 23 ≈ 0.667 A

I₂ = 3 · 23 = 2 A

V_נורות = 3 · 3 = 9 V V_R₁ = 913 − 9 = 13 V R₁ = V_R₁I₁ = 1323 = 0.5 Ω

V_נורות = 3 V → V_R₂ = 4 − 3 = 1 V R₂ = 12 = 0.5 Ω

913 = ε − 23r , 4 = ε − 2r → r = 4 Ω , ε = 12 V

P_מושקע = ε · I

P₁ = 12 · 23 = 8 W P₂ = 12 · 2 = 24 W P₂P₁ = 248 = 3

P_מבוזבז = I²·r

P_r,1 = (23)² · 4 = 169 ≈ 1.78 W P_r,2 = 2² · 4 = 16 W P_r,2P_r,1 = 16169 = 9

P_נורות = 3 · 2 = 6 W (בשני המעגלים)

η = P_נורות + P_RP_מושקע

P_R₁ = I₁²·R₁ = 49 · 0.5 ≈ 0.22 W → η₁ = 6 + 0.228 = 77.8 % P_R₂ = I₂²R₂ = 4 · 0.5 = 2 W → η₂ = 6 + 224 = 33.3 %

η = V·Iε·I = Vε

η₁ = 91312 = 0.778 = 77.8 % η₂ = 412 = 0.333 = 33.3 %

m_p = 1.67·10⁻²⁷ kg | q_p = 1.6·10⁻¹⁹ C | B ⊗ אל תוך הדף
v₁ = 2.0·10⁵ m/s | v₂ = 2.4·10⁵ m/s | Δt_C = 3 ns

4.7 T

1.4·10⁵ V/m

0.0

2.00

0.44

7.0

1.50

qvB = mv²r → r = mvqB

F = q·v × B = q(v x̂) × (−B ẑ) = −qvB (x̂ × ẑ) = −qvB(−ŷ) = +qvB ŷ

t_½ = πrv

t_½ = πv · mvqB = πmqB

t_½ = 7·10⁻⁹ s → T = 14·10⁻⁹ s

t_½ = πmqB → B = πmq·t_½ B = π · 1.67·10⁻²⁷1.6·10⁻¹⁹ · 7·10⁻⁹ = 5.246·10⁻²⁷1.12·10⁻²⁷ = 4.7 T

Δv = (2.4 − 2.0)·10⁵ = 4·10⁴ m/s , Δt = 3·10⁻⁹ s a = ΔvΔt = 4·10⁴3·10⁻⁹ = 1.33·10¹³ m/s²

qE = ma → E = m·aq E = 1.67·10⁻²⁷ · 1.33·10¹³1.6·10⁻¹⁹ = 2.23·10⁻¹⁴1.6·10⁻¹⁹ = 1.4·10⁵ V/m

d_C = v₁ + v₂2 · Δt = 2.2·10⁵ · 3·10⁻⁹ = 6.6·10⁻⁴ m

F = q·v·B F₂F₁ = v₂v₁

ΔE_k = q·E·d_C = 1.6·10⁻¹⁹ · 1.4·10⁵ · 6.6·10⁻⁴ = 1.47·10⁻¹⁷ J E_k,1 = ½mv₁² = 3.34·10⁻¹⁷ J → E_k,3 = 3.34·10⁻¹⁷ + 2 · 1.47·10⁻¹⁷ = 6.28·10⁻¹⁷ J v₃ = √2E_k,3m = 2.74·10⁵ m/s F₂F₁ = 2.742.00 ≈ 1.37

N = 2500 | L = 25 cm | ℓ_ab = 2.8 cm | g = 10 m/s²
K = 1 → I² = 16.0 | 2 → 25.0 | 3 → 42.3 | 4 → 56.3 | 5 → 72.3 (A²)

4.00 A

1

5.03

5.63

5.00

16.0

0.56

B_{סילונית} = μ₀·N·IL , בכיוון ציר הסילונית משמאל לימין

F = I·ℓ × B

B = μ₀·N·IL

F_B = B·I·ℓ_ab·sin90° = B·I·ℓ_ab

F_B = μ₀·N·IL · I · ℓ_ab = μ₀·N·ℓ_ab·I²L

F_B·D = K·m₀·g·D → F_B = K·m₀·g

μ₀·N·ℓ_ab·I²L = K·m₀·g

I² = m₀·g·Lμ₀·N·ℓ_ab · K

m = Σ K·I²Σ K² = 779.655 = 14.2 A²

m = m₀·g·Lμ₀·N·ℓ_ab → m₀ = m·μ₀·N·ℓ_abg·L

L = 25 cm = 0.25 m , ℓ_ab = 2.8 cm = 0.028 m

m₀ = 14.2 · 4π·10⁻⁷ · 2500 · 0.02810 · 0.25 = 14.2 · 8.80·10⁻⁵2.5 = 5.0·10⁻⁴ kg

F = I·ℓ × B → (−I·ℓ) × (−B) = +(I·ℓ × B)

F_B = μ₀·N·ℓ_ab·I²L , I² > 0 תמיד

R = 1.2 Ω | S = 100 cm² = 0.01 m² | הכיוון החיובי של הזרם: מ-L ל-M

2.0 ms

2.0

−0.20

−100

1.00

0.833

L(0,0) → M(1,0) → K(0,1) → L A = 12[x_L(y_M−y_K) + x_M(y_K−y_L) + x_K(y_L−y_M)] = +12

ε = −S · ΔBΔt , I = εR , S = 0.01 m²

ΔBΔt = −0.5 − 05·10⁻³ = −100 T/s ε = −0.01 · (−100) = +1.0 V → I = 1.01.2 = 0.833 A

ΔBΔt = 0 → ε = 0 → I = 0

ΔBΔt = 0 − 0.515·10⁻³ = −33.3 T/s ε = −0.01 · (−33.3) = +0.333 V → I = 0.3331.2 = 0.278 A

15 ms < 20 ms < 30 ms → ε = 0.333 V , I = 0.278 A

P = ε²R = 0.333²1.2 = 0.11111.2 = 9.26·10⁻² W P = ε·I = 0.333 · 0.278 = 0.0926 W

ε = B·v·ℓ_גבול

ℓ_גבול(t) = v·t → ε(t) = B·v²·t

Φ = B · S_{שבתוך השדה}

Φ_א = Φ_ב = B·S_עיגול → ΔΦ_אΔt = ΔΦ_בΔt → ε_א = ε_ב

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