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PHY303L HW5

# PHY303L HW5 - thwing(jrt2329 Homework 05 Yao(59110 This...

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thwing (jrt2329) – Homework 05 – Yao – (59110) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The current in a wire decreases with time according to the relationship I = (1 . 54 mA) × e at where a = 0 . 13328 s 1 . Determine the total charge that passes through the wire from t = 0 to the time the current has diminished to zero. Correct answer: 0 . 0115546 C. Explanation: I = dq dt q = integraldisplay t t =0 I dt = integraldisplay t =0 (0 . 00154 A) e 0 . 13328 s 1 t dt = (0 . 00154 A ) e 0 . 13328 s 1 t - 0 . 13328 s 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 0 = 0 . 0115546 C . 002 10.0 points If 5 × 10 21 electrons pass through a 20 Ω resis- tor in 14 min, what is the potential difference across the resistor? Correct answer: 19 . 0714 V. Explanation: Let : n = 5 × 10 21 , Δ t = 14 min , R = 20 Ω , and q e = 1 . 602 × 10 19 C . The current is defined as I = Δ Q Δ t . Since each electron carries a q e , we have I = n q e Δ t = ( 5 × 10 21 ) (1 . 602 × 10 19 C) 14 min parenleftbigg 1 min 60 s parenrightbigg = 0 . 953571 A . Then from Ohm’s Law, V = I R = (0 . 953571 A) (20 Ω) = 19 . 0714 V . 003 10.0 points A conductor with cross-sectional area 4 cm 2 carries a current of 9 A. If the concentration of free electrons in the conductor is 8 × 10 28 electrons / m 3 , what is the drift velocity of the electrons? Correct answer: 0 . 00175542 mm / s. Explanation: Let : I = 9 A , n = 8 × 10 28 electrons / m 3 , q e = 1 . 60218 × 10 19 C , and A = 4 cm 2 = 0 . 0004 m 2 . The current in a conductor is given by I = n q v d A , where n is the number of charge carriers per unit volume, q is the charge per carrier, v d is the drift velocity of the carriers and A is the cross section of the conduction. Solving for v d , we have v d = I n q A = 9 A 8 × 10 28 electrons / m 3 × 1 1 . 60218 × 10 19 C × 1 0 . 0004 m 2 × 1000 mm 1 m = 0 . 00175542 mm / s .

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thwing (jrt2329) – Homework 05 – Yao – (59110) 2 004 (part 1 of 3) 10.0 points Consider two cylindrical conductors made of the same ohmic material. Conductor 1 has a radius r 1 and length 1 while conductor 2 has a radius r 2 and length 2 . Denote: The currents of the two conductors as I 1 and I 2 , the potential differences between the two ends of the conductors as V 1 and V 2 , and the electric fields within the conductors as E 1 and E 2 . V 1 vector E 1 I 1 1 r 1 V 2 vector E 2 I 2 2 r 2 If ρ 2 = ρ 1 , r 2 = 2 r 1 , ℓ 2 = 3 1 and V 2 = V 1 , find the ratio R 2 R 1 of the resistances. 1. R 2 R 1 = 1 4 2. R 2 R 1 = 3 3. R 2 R 1 = 2 3 4. R 2 R 1 = 2 5. R 2 R 1 = 4 6. R 2 R 1 = 3 4 correct 7. R 2 R 1 = 1 2 8. R 2 R 1 = 4 3 9. R 2 R 1 = 3 2 10. R 2 R 1 = 1 3 Explanation: The relation between resistance and resis- tivity is given by R = ρ ℓ A = ρ ℓ π r 2 . Then since r 2 = 2 r 1 and 2 = 3 1 , the ratio of the resistances is R 2 R 1 = ρ ℓ 2 π r 2 2 π r 2 1 ρ ℓ 1 = 2 r 2 1 1 r 2 2 = (3 1 ) r 2 1 1 (2 r 1 ) 2 = 3 4 .
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