ragsdale (zdr82) HW5 ditmire (58335) 1

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1 ragsdale (zdr82) HW5 ditmire (58335) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points The quantity of charge passing through a surface of area 2.6 cm 2 varies with time as q q 1 t 3 + q 2 t + q 3, where q C/s 3, q C/s, q C, and t is in seconds. What is the instantaneous current through the surface at t 1.1 s? Correct answer: A. et : q C/s 3, q C/s, q C, and t 1.1 s. I d q dt 3 q 1 t 2 + q 2 3 ( 5.8 C/s 3) (1.1 s) C/s A. 002 (part 2 of 2) 10.0 points What is the value of the current density at t 1.1 s? Correct answer: A/m 2. et : a 2.6 cm m 2 t 1.1 s. J I A A m A/m 2. and points An incandescent bulb has a resistance of 27 Ω when it is at room temperature (25 C) and 432 Ω when it is hot and delivering light to the room. The temperature coefficient of resistivity of the filament is ( C) 1, where the base resistance R 0 is determined at 0 C. What is the temperature of the bulb when in use? Correct answer: 2900 C. et : 27 Ω, R Ω, and α ( C) 1. Temperature dependence of resistance is R T R 0 [1 + α (T 0 C)] R 0 [1 + α T]. (1) Using Eq. (1), at room temperature; e.g., 27 Ω and T 1 25 C R 0 [1 + α T 1 ] 27 Ω 1 + [0.006 ( C) 1 ] (25 C) Ω. Using Eq. (1), with resistance R 2, R 2 1 R T 2 0 α 432 Ω Ω ( C) C. (2) (3) Or directly substituting Eq. (1) into (2), we have T 2 R 2 1 R 0 α

2 ragsdale (zdr82) HW5 ditmire (58335) 2 R 2 [1 + α T 1 ] 1 α R 2 + αr 2 T 1 α R 2 + R 2 T 1 (4) α 432 Ω 27 Ω [0.006 ( C) 1 ] (27 Ω) Ω 27 Ω (25 C) 2900 C. 004 (part 1 of 4) 10.0 points If the current carried by a metallic conductor is doubled, (a) What happens to the charge carrier density? Assume that the temperature of the metallic conductor remains constant. 1. It is unchanged. correct 2. It quadruples. 3. It doubles. 4. It is quartered. 5. It is halved. J σ E n q v d v d q E m τ. The charge carrier density n is unaffected. This is simply a property of the material, and is related to the mass density and the number of valence electrons available petom. 005 (part 2 of 4) 10.0 points (b) What happens to the current density? 1. It is quartered. 2. It is unchanged. 3. It doubles. correct 4. It quadruples. 5. It is halved. The current density J I only depends A on the current I and the cross-sectional area of the conductor A, so if I doubles and A remains the same, J must also double: J 2 I A 2 I A 2 J. 006 (part 3 of 4) 10.0 points (c) What happens to the electron drift velocity? 1. It is quartered. 2. It quadruples. 3. It is unchanged. 4. It is halved. 5. It doubles. correct The electron drift velocity is given by v d I n q A, where q is the charge of an electron. So if I doubles, v d must also double: v d 2 I n q A 2 I n q A 2 v d. 007 (part 4 of 4) 10.0 points (d) What happens to the average time between collisions? 1. It is unchanged. correct 2. It doubles. 3. It is quartered.

3 ragsdale (zdr82) HW5 ditmire (58335) 3 4. It is halved. 5. It quadruples. The average time τ between collisions is given by τ v d m q E, where m is the mass of an electron and E is the applied electric field (determined by Ohm s aw: J σ E). Substituting for E, τ v d mσ q J. Since v d and J are each multiplied by a factor of two, the factors of two cancel and the other variables remain the same. All the other variables remain the same. Therefore τ remains unchanged points The current in a wire decreases with time according to the relationship I (2.63 ma) e a t where a 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: C. q t t0 I dt I dq dt ( A)e s 1 t dt t0 ( A ) e s 1 t s C. 0 A conductor with cross-sectional area 3 cm 2 carries a current of 10 A. If the concentration of free electrons in the conductor is electrons/m 3, what is the drift velocity of the electrons? Correct answer: mm/s. et : I 10 A, n electrons/m 3, q e C, and A 3 cm 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 10 A electrons/m C mm m2 1 m mm/s. 010 (part 1 of 3) 10.0 points Consider the application of two potential differences across the same metallic ohmic conductor where the conductor is maintained at a constant temperature of 300 K. J 2 V points R R

4 ragsdale (zdr82) HW5 ditmire (58335) 4 Case 1: The potential difference is, the current density is, and the average thermal speed of the free electron within the conductor is. Case 2: The corresponding quantities are V 2, J 2, and v 2. If V 2 2, the corresponding ratio J 2 of current densities is given by 1. J J J 2 2. correct 4. J J J J J J J Using the relation between current density fon ohmic material E ρ J, we find J 2 E 2 ρ ρ E 2 V 2 V (part 2 of 3) 10.0 points The ratio v 2 is approximately 1. v 2 1. correct 2. v v v v v v v v v When two different voltages are applied across the conductor, there will be two different drift velocities. However, we recall that drift velocity is negligible compared with the average thermal velocity of free electrons. We denote the average velocity of free electrons in the absence of applied voltage to be v absent. Thus, the ratio of the magnitude of the average velocities (the vector sum of v absent and average drift velocities) is v 2 v absent + v drift2 v absent + v drift1 v absent v absent (part 3 of 3) 10.0 points Consider the following effects due to the increase of the temperature. A. Variations on the average collision time: (i) the average collision time increases (ii) the average collision time remains the same

5 ragsdale (zdr82) HW5 ditmire (58335) 5 (iii) the average collision time decreases B. Variations on the resistivity: (a) the resistivity increases (b) the resistivity remains the same (c) the resistivity decreases Choose the correct pair of statements which represent the variations as the temperature increases. 1. (ii) and (b) 2. (ii) and (c) 3. (i) and (c) 4. (i) and (a) 5. (i) and (b) 6. (iii) and (c) 7. (iii) and (a) correct 8. (iii) and (b) 9. (ii) and (a) 10. None of them. As temperature increases, thermal motion of particles in the conductor increases, so the collision time τ decreases (i.e., the electrons collide more frequently). Since ρ 1, we find τ that ρ increases. 013 (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 while conductor 2 has a radius r 2 and length l 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 and V 2, and the electric fields within the conductors as and E 2. r 1 E1 I 1 r 2 E2 I 2 V 2 If ρ 2 ρ 1, r 2 2 r 1, l 2 3 and V 2, find the ratio R 2 of the resistances. 1. R R R R correct 5. R R R R R R The relation between resistance and resistivity is given by R ρ l A ρ l π r 2. Then since r 2 2 r 1 and l 2 3, the ratio of the resistances is ρ l 2 R 2 π r 2 2 π r 2 1 ρ l 2 r 2 1 r 2 2 l 2

6 ragsdale (zdr82) HW5 ditmire (58335) 6 (3 ) r 2 1 (2 r 1 ) (part 2 of 3) 10.0 points When the two conductors are attached to a battery of voltage V, determine the ratio E 2 of the electric fields. 1. E correct 2. E E E E E E E E E Fo uniform electric field, E V l. Hence the ratio of the electric fields with the condition V 2 is E 2 V 2 l 2 l (part 3 of 3) 10.0 points Now conside different case where the currents in the two conductors are the same (i.e., I 1 I 2 ). Determine the ratio E 2 of the electric fields. 1. E E E E E E correct 7. E E E E Using J I and E ρ J, we can write A E ρ I A ρ I π r 2. Then the ratio of the electric fields with the condition I 1 I 2 is E 2 ρ 2 I 2 π r 2 2 ρ 1 I 1 π r 2 1 r2 1 r 2 2 r2 1 (2 r 1 ) Alternate Solution. Using Ohm s law, we can write E V l I R l

7 ragsdale (zdr82) HW5 ditmire (58335) 7 Then the ratio of the electric fields with the condition I 1 I 2 is E 2 I 2 R 2 l 2 I 1 R 2 l1 3 l points A wire is made of a material with a resistivity of Ω m. It has length m and diameter mm. What is the resistance of the wire? Correct answer: Ω. et : ρ Ω m, l m, and r mm m. By definition, the resistance of the wire is R ρ l A ρ l ( π r Ω m ) ( m) Ω. π ( m) points A wire with a circular cross section and a resistance R is lengthened to 5.99 times its original length by pulling it through a small hole. After being pulled through the hole, the total volume of the wire is unchanged. Find the resistance of the wire after it is stretched. Correct answer: R. et : l 5.99 l. The volume V of the wire is given by V l A, where l is the length of the wire and A is the cross-sectional area. When the wire is stretched to a length l, the volume will remain the same. The new cross-sectional area is then, l A l A A l A l l A 5.99l A The resistance of a conductor of resistivity ρ is given by so Thus R R ρ l A ρ l A (5.99) 2. R ρ l A, l A l A (5.99l) ( A ) A l 5.99 R (5.99) 2 R R points A resistor is constructed by forming a material of resistivity ρ into a shape of a hollow cylindrical shell of length with inner radius and outer radius r b. Suppose a potential difference is applied between the innend outer surfaces such that the current flows radially outward. r b What is the total resistance R? 1. R ρ ( 2 π ln 1 + r ) b 2. R ρ ( 2 π ln 1 r ) b

8 ragsdale (zdr82) HW5 ditmire (58335) 8 3. R ρ ( 2 π ln 1 r ) a r b 4. R ρ ( 2 π ln 1 + r ) a r b 5. R ρ ( ) 2 π ln ra r b 6. R ρ ( ) 2 π ln rb correct The resistance due to the cylindrical element is the resistivity of the material times the length of the element divided by the crosssectional area (the area seen by the current; i.e., perpendicular to the direction of the current): R ρ l A, To visualize the setup, since the potential difference is applied between the innend the outer surfaces, we may cut the cylinder surface and partially straighten out the side surface. Now the length l here is the distance traveled by the current in crossing the cylindrical element, l r, and the cross sectional area is that of the side of the cylindrical element, A 2 π r. Remember, is the length of the cylinder, while l is the length traveled by the current while crossing the area A. Thus we have R ρ r 2 π r. The total resistance is obtained by letting these cylindrical elements be very thin ( r dr) and integrating from to r b : R ρ 2 π ρ rb d r r 2 π [lnr b ln ] ρ ( ) 2 π ln rb points At 50 C, the resistance of a segment of gold wire is 85 Ω. When the wire is placed in a liquid bath, the resistance increases to 154 Ω. The temperature coefficient is ( C) 1 at 20 C. What is the temperature of the bath? Correct answer: C. et : 85 Ω, R Ω, T 0 20 C, T 1 50 C, and α ( C) 1. Neglecting change in the shape of the wire, we have and where T 0 20 C. Thus R 0 [1 + α(t 1 T 0 )] R α (T 1 T 0 ) R 2 R 0 [1 + α (T 2 T 0 )], R 2 [1 + α (T 2 T 0 )] 1 + α (T 1 T 0 ) R 2 + αr 2 (T 1 T 0 ) + α (T 2 T 0 ) α T 2 R 2 + αr 2 (T 1 T 0 ) + α T 0 so that T 2 R 2 + αr 2 (T 1 T 0 ) + α T 0 α. Since R 2 + α R 2 (T 1 T 0 ) + α T Ω 85 Ω + [0.0034( C) 1 ] (154 Ω) (50 C 20 C) (85 Ω) [0.0034( C) 1 ] (20 C) Ω,

9 ragsdale (zdr82) HW5 ditmire (58335) 9 then T Ω (85 Ω) (0.0034( C) 1 ) C points A current of 1.71 A flows through a 61.9 Ω resistor for 7.7 min. How much heat was generated by the resistor? Correct answer: J. et : i 1.71 A, r 61.9 Ω, t 7.7 min. and Energy is E P t (i 2 r) t (1.71 A) 2 (61.9 Ω) (7.7 min) J. ( ) 60 s min

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