Version 001 HW 22 EM Induction C&J sizemore (21301jtsizemore) 1

Version 001 HW 22 EM Induction C&J sizemore (21301jtsizemore) 1 This print-out should have 35 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Serway CP 20 21 001 (part 1 of 2) 10.0 points An automobile has a vertical radio antenna 0.72 m long. The automobile travels at 79.5 km/h on a horizontal road where Earth s magnetic field is 34.8 µt directed toward the north and downward at an angle of 43.1 below the horizontal. Specify the direction that the automobile should move in order to generate the maximum motional emf in the antenna, with the top of the antenna positive relative to the bottom. 1. east correct 2. south 3. north 4. Unable to determine 5. west Since the magnetic field of the Earth points north-ward, the automobile should move east or west in order to create a maximum induced emf. The force exerted on the positive charge inside the antenna due to the moving automobile is F q ( v ) So, for the positive charge on top of the antenna, the automobile should move east! 002 (part 2 of 2) 10.0 points Find the magnitude of this induced emf. Correct answer: 0.000404013 V. Let : l 0.72 m, v 79.5 km/h 22.0833 m/s, 34.8 µt 3.48 10 5 T, and θ 43.1. The motional emf is E lv cosθ (3.48 10 5 T)(0.72 m)(22.0833 m/s) cos43.1 0.000404013 V. Serway CP 20 23 003 (part 1 of 2) 10.0 points A coil is suspended around an axis which is colinear with the axis of a bar magnet. The coil is connected to a resistor with ends labeled a and b. The bar magnet moves from right to left with North and South poles labeled as in the figure. Use Lenz s law to answer the following question concerning the direction of induced currents and magnetic fields. S R a b v What is the direction of the induced magnetic field in the coil when the bar magnet is moving from right to left? 1. right to left ( induced ) correct 2. the induced field is zero tesla 3. left to right ( induced ) The induced magnetic field depends on whether the flux is increasing or decreasing. The magnetic flux through the coil is from left to right. When the magnet moves from right to left, the magnetic flux through the coils increases. The induced current in the coil must produce an induced magnetic field from right N

Version 001 HW 22 EM Induction C&J sizemore (21301jtsizemore) 2 to left ( induced ) to resist any change of magnetic flux in the coil (Lenz s Law). 004 (part 2 of 2) 10.0 points What is the direction of the induced current in resistor R when the bar magnet is moving from right to left? 1. the induced current is zero amperes 2. from a through R to b (I ) 3. from b through R to a ( I) correct The helical coil when viewed from the bar magnet winds around the solenoid from terminal b counter-clockwise. As the induced field is right to left ( induced ), the induced current must flow clockwise and therefore it goes from b through R to a ( I). Serway CP 20 37 005 10.0 points A 1.98 H inductor carries a steady current of 0.304 A. When the switch in the circuit is thrown open, the current disappears in 11.5 ms. What is the average induced emf in the inductor during this time? Correct answer: 52.3409 V. Let : L 1.98 H, I 0.304 A, and t 11.5 ms. From Faraday s law, the self-induced emf is proportional to the time rate of change of the current, with the proportionality constant L is the inductance of the inductor. Thus the self inductance is ε L I (1.98 H) 0.304 A 0.0115 s 52.3409 V. Serway CP 20 27 006 10.0 points A copper bar has a constant velocity in the plane of the paper and perpendicular to a magnetic field pointed out of the plane of the paper. If the top of the bar becomes negative relative to the bottom of the bar, what is the direction of the velocity v of the bar? 1. from top to bottom ( ) 2. from left to right ( ) correct 3. from bottom to top ( ) 4. from right to left ( ) Positive charges will move in the direction of the magnetic force, while negative charges move in the opposite direction. To produce the indicated charge separation, the positive charges in the conductor experience downward magnetic forces while the negative charges in the conductor experience upward magnetic forces leaving the charge separation shown in the figure. Using the right-hand rule with F q v to produce this force on positive charges, the velocity v must be directed from left to right ( ).

Version 001 HW 22 EM Induction C&J sizemore (21301jtsizemore) 3 F v Serway CP 20 53 007 (part 1 of 4) 10.0 points An 820-turn wire coil of resistance 22 Ω is placed on top of a 12500-turn, 5.83 cmlong-solenoid, as shown in the figure. The solenoid is connected to a 15 Ω resistor, and 54.5 V battery. oth coil and solenoid have cross-sectional areas of 0.0001 m 2. The permeability of free space is 4π 10 7 Tm/A. negligible length 5.83 cm 12500 turns 820 turns coil 15 Ω 54.5 V Switch Figure: The coil is tightly wraped and of negligible length and is touching the top end of the tightly wrapped solenoid(these features are not obvious from looking at the figure). How long does it take the solenoid current to reach 62.9% of its maximum value? Correct answer: 0.0222631 s. F Let : N s 12500turns, l s 5.83 cm 0.0583 m, R 15 Ω, A 0.0001 m 2, and µ 0 4π 10 7 Tm/A. The primary circuit, containing the battery, theresistorrandthesolenoidisanrlcircuit with S µ 0N S I l S so that Φ S A µ 0N s IA l s. The inductance of the solenoid is L N sφ I µ 0N 2 S A l S (4π 107 Tm/A)(12500) 2 0.0583 m (0.0001 m 2 ) 0.336792 H. The time constant is τ L, and the time for R the current to reach 0.629 times its maximum value is defined by ( I I 0 R I 0 1e t/τ) e t/τ (1R) ( ) t 1 τ ln 1R ( ) 1 t τ ln 1R L ( ) 1 R ln 1R 0.336792 H ( ) 1 ln 15 Ω 10.629 22.2631 ms 0.0222631 s. 008 (part 2 of 4) 10.0 points

Version 001 HW 22 EM Induction C&J sizemore (21301jtsizemore) 4 Assume the magnetic field produced by the solenoid at the location of the coil is one-half as strong as the field at the center of the solenoid. Find the average back emf caused by the self-inductance of the solenoid during this interval. Correct answer: 34.5725 V. The current is Let : R I I 0 0.629 and E 54.5 V. and the average emf is I(t) 0.629I max 0.629 E R E sol L I L I(t)0 t0 L(0.629E) Rt (0.336792 H)(0.629)(54.5 V) (15 Ω)(0.0222631 s) 34.5725 V E sol 34.5725 V. 009 (part 3 of 4) 10.0 points Find the average rate of change in magnetic flux through each turn of the coil during this time interval. Correct answer: 1.3829 mv. When the current in the solenoid is I (0.629)I max 2.28537 A, the field in the solenoid is µ 0 n s I µ 0 ( Ns l s ) I and the flux is Φ A µ 0N s I A l s. Thus the average back emf is Φ Φ0 1 2 t0 µ 0N s IA 2l s t (12500)(2.28537 A)(0.0001 m2 ) (2)(0.0583 m)(0.0222631 s) 1.3829 mv. (1.25664 10 6 V s/am) 103 mv V 010 (part 4 of 4) 10.0 points Determine the magnitude of the average induced current in the coil. Correct answer: 51.5445 ma. Let : N c 820 and R coil 22 Ω. The rate of change of flux in the 820 turn coilisthesameasthatinthesolenoid. Therefore, the average induced emf in the coil is E coil N C Φ. Therefore, the magnitude of the average induced current is I coil E coil R coil N c Φ R coil 820 22 Ω (0.0013829 V) 103 ma A 51.5445 ma.

Version 001 HW 22 EM Induction C&J sizemore (21301jtsizemore) 5 Serway CP 20 27 D 011 10.0 points A indicates the magnetic field pointing up from the surface of the paper and a indicates the magnetic field pointing down into the surface of the paper. A copper bar has a constant velocity v in theplane ofthepaper and isperpendicular to a magnetic field as shown in the figures. How are charges distributed on the bar? 1. v 2. v 3. v Study the case where the magnetic field is pointed out of the plane of the paper and the bar is moving from left to right ( ). To produce the indicated charge separation, the positive charges in the conductor experience downward magnetic forces while the negative charges in the conductor experience upward magnetic forces leaving the charge separation shown in the figure. F Using the right-hand rule F q v, the top will be negative and the bottom will be positive. The other three choices are incorrect since they do not abide by the right-hand rule. Serway CP 20 25 012 10.0 points Consider a loop of wire with a resistor. In the same plane, (with switch S closed) a long wire has a current I flowing from right to left. R v F I S 4. v cor- If switch S is closed for a long period of time, and then suddenly opened, what is the direction of the induced current through R as the current I rapidly decreases? rect 1. The current flows left to right through R. correct 2. There is no current flow. Positive charges will move in the direction of the magnetic force, while negative charges move in the opposite direction. 3. The current flows right to left through R.

Version 001 HW 22 EM Induction C&J sizemore (21301jtsizemore) 6 The induced magnetic field depends on whether the flux is increasing or decreasing, since Lenz s law will resist the change in the magnetic flux. Due to the right-hand rule, the magnetic field due to the current is directed into the plane of the paper through the loop with a resistor in it. When the current decreases, the magnetic flux also decreases. The induced current in the loop must produce a magnetic field to maintain a constant magnetic flux in the loop. Consequently, the current in the loop is clockwise. The current flows left to right through R. Serway CP 20 44 013 (part 1 of 4) 10.0 points A 57 mh inductor, a 9.8 Ω resistor, and a 1.6 V battery are connected in series. The switch is closed at t 0. Find the voltage drop across the resistor at t 0. Correct answer: 0 V. and The time constant is τ L R, so I(t) I 0 ( 1e t/τ) E R (0) RI(0) 0 V. 014 (part 2 of 4) 10.0 points Find the voltage drop across the resistor after one time constant has passed. Correct answer: 1.01139 V. At t τ, Let : E 1.6 V. E R RI(τ) RI 0 ( 1e 1 ) E e1 e (1.6 V) e1 e 1.01139 V. 015 (part 3 of 4) 10.0 points Find the voltage drop across the inductor at t 0. Correct answer: 1.6 V. Since E R E L E, E L (0) E E R (0) 1.6 V. 016 (part 4 of 4) 10.0 points Find the voltage drop across the inductor after one time constant has elapsed. Correct answer: 0.588607 V. E L (τ) E E R (τ) 1.6 V1.01139 V 0.588607 V. Serway CP 20 29 017 (part 1 of 3) 10.0 points Two coils are suspended around a central axis as shown in the figure below. One coil is connected to a resistor with ends labeled a and b. The other coil is connected to a battery E. The coils are moving relative to each other as indicated by the velocity vectors v. Use Lenz s law to answer the following question concerning the direction of induced currents and magnetic fields.

Version 001 HW 22 EM Induction C&J sizemore (21301jtsizemore) 7 v induced E E a R b v What is the direction of the magnetic field in the coil with the battery attached? 1. left to right ( primary ) 2. The primary field is zero. 3. right to left ( primary ) correct The helical coil with the battery attached (when viewed from either end) is wound clockwise (as you go into the coil). Using the right-hand rule, the magnetic flux through the coil with the battery attached has a magnetic field direction right to left. 018 (part 2 of 3) 10.0 points What is the direction of the induced magnetic field in the coil with the resistor attached when thethe coilsare movingaway from each other? 1. right to left ( induced ) correct 2. left to right ( induced ) 3. The induced field is zero. Note: The induced magnetic field depends on whether the flux is increasing or decreasing. primary Themagneticfluxthroughthecoilwiththe battery attached is from right to left. When thecoilsaremovingawayfromeachother,the magnetic flux from the coil with the battery attached decreases its intensity in the coil with the attached resistor. The induced current in the coil must produce an induced magnetic field from right to left ( induced ) to resist any change of magnetic flux in the coil (Lenz s Law). 019 (part 3 of 3) 10.0 points What is the direction of the induced current in resistor R when the coils are moving away from each other? 1. from b through R to a ( I) correct 2. from a through R to b (I ) 3. The induced current is zero. The helical coil with the resistor attached (when viewed from either end) is wound counter-clockwise (as you go into the coil). Since the induced field is right to left ( induced ) the induced current in the coil flows clockwise when viewing the coil from the right-hand side. Therefore the current flows from b through R to a ( I). Serway CP 20 09 020 10.0 points A 31-turn circular coil of wire has a diameter of 1.36 m. It is placed with its axis along the direction of Earth s magnetic field of 55.4 µt, a R b

Version 001 HW 22 EM Induction C&J sizemore (21301jtsizemore) 8 and then in 0.15 s it is flipped 51. What is the strength of the average emf generated in the coil? Correct answer: 6.1652 mv. Let : n 31 turn, d 1.36 m, b 55.4 µt 5.54 10 5 T, 0.15 s, and φ 51. The initial and final angles between the normal to the plane of the circular loop and the magnetic field are θ i 0 θ f 0 51 51. Note: Since the cosine function is symmetric about 0 it does not make any difference how you define the positive direction of the initial angle. Also, since the magnitude of the average emf is to be determined, the direction of the flip-flop does not matter. From Faraday s Law for solenoid, The average emf is E N Φ < E > N Φ ( A) N N πr2 (cosθ f cosθ i ) ( 1.36 m (5.54 10 5 T)π 2 (31) (0.15 s) [cos51 cos0 ] 103 mv V 6.1652 mv. Serway CP 20 55 02. ) 2 021 10.0 points A 13 Ω square loop, whose dimensions are 4.2 m 4.2 m, is placed in a uniform 0.13 T magnetic field that is directed perpendicular to the plane of the loop (see figure). The loop, which is hinged at each vertex, is pulled as shown until the separation between points C and D is d 2.6 m. The process takes 0.095 s. C F 4.2 m D 4.2 m What is the average current generated in the loop? Correct answer: 0.763817 A. asic Concept: Faraday s Law Ohm s Law d E dφ dt I V R. Let : R 13 Ω, a 4.2 m, d 2.6 m, and 0.095 s. Solution: From Faraday s law, the average induced emf is E avg Φ A. The change in area for the square with sides a is A A f A i,. F

Version 001 HW 22 EM Induction C&J sizemore (21301jtsizemore) 9 where A i a 2 17.64 m 2. To find A f, we note that the final shape is made up of four right triangles of hypotenuse a with one leg being d. The area of each of 2 these triangles is 1 4 A f ( ) 1 d 2 2 a 2 ( ) 2 d. 2 Thus ( ) 2 d A f d a 2 2 ( 2.6 m (2.6 m) (4.2 m) 2 2 10.3837 m 2. We then find the average emf to be E avg [A ia f ] ) 2 (0.13 T)[(17.64 m2 )(10.3837 m 2 )] 0.095 s 9.92962 V. Applying Ohm s law, the average current is I avg E avg R (9.92962 V) (13 Ω) 0.763817 A. Serway CP 20 49 022 (part 1 of 2) 10.0 points A 35 V battery is connected in series with a resistanceof4.6ωandaninductanceof5.2h. Find the energy stored in the inductor when the current reaches its maximum value. Correct answer: 150.52 J. so Let : V 35 V, R 4.6 Ω, and L 5.2 H. The maximum current is I max V R, U 1 2 LI max 2 LV 2 2R 2 (5.2 H)(35 V)2 2(4.6 Ω) 2 150.52 J. 023 (part 2 of 2) 10.0 points Find the energy stored in the inductor one time constant after the switch is closed. Correct answer: 60.1442 J. Let : V 35 V, R 4.6 Ω, and L 5.2 H. At t τ, ( I 1 1 ) ( ) e1 V I max e e R, and W 1 2 LI2 L(e1)2 V 2 2e 2 R 2 (5.2 H)(e1)2 (35 V) 2 2e 2 (4.6 Ω) 2 60.1442 J. Serway CP 20 47 024 10.0 points How much energy is stored in a 95.4 mh inductor at the instant when the current is 3.2 A?

Version 001 HW 22 EM Induction C&J sizemore (21301jtsizemore) 10 Correct answer: 0.488448 J. Let : L 95.4 mh 0.0954 H and I 3.2 A. The energy is Serway CP 20 62 026 10.0 points An aluminum ring of radius 10 cm and resistance 0.0005 Ω is placed around the center of a long air-core solenoid with 1400 turns/m and a smaller radius 3 cm. U 1 2 LI2 1 (0.0954 H)(3.2 A)2 2 0.488448 J. Serway CP 20 13 025 10.0 points The plane of a rectangular coil, 4.5 cm by 8.2 cm, is perpendicular to the direction of a uniform magnetic field. Ifthecoilhas65turnsandatotalresistance of 9.6 Ω, at what rate must the magnitude of change to induce a current of 0.14 A in the windings of the coil? Correct answer: 5.6035 T/s. Given : x 4.5 cm 0.045 m, y 8.2 cm 0.082 m, N 65 turns, r 9.6 Ω, and I 0.14 A. The induced emf is E IR N Φ N (A) N A, so IR N (xy) (0.14 A)(9.6 Ω) 65(0.045 m)(0.082 m) 5.6035 T/s. S If the current in the solenoid is increasing at a constant rate of 280 A/s, what is the induced current in the ring? Assume that the solenoid produces a negligible magnetic field outside its cross-sectional area. Correct answer: 27856. The onlyplacethat thereisanonzero magnetic field is through the inner solenoid. Using the formula derived from Ampére s law, we know that the field inside the inner solenoid is µ 0 ni. Therefore, I ind 1 R V ind 1 dφ R dt 1 d R da dt 1 d R dt µ 0nIπri 2 mu 0nπr 2 i R di dt (1.25664 106 T m/a)π(3 cm) 2 0.0005 Ω (1400 turns/m)(280 A/s) 27856

Version 001 HW 22 EM Induction C&J sizemore (21301jtsizemore) 11 Serway CP 20 32 027 (part 1 of 3) 10.0 points A motor has coils with a resistance of 15 Ω and operates from a voltage of 240 V. When the motor is operating at its maximum speed, the back emf is 150 V. Findthecurrentinthecoilswhenthemotor is first turned on. Correct answer: 16 A. Let : R 15 Ω, E 240 V, and E back 150 V. Immediately after the switch is closed, the motor coils are still stationary and the back emf is zero. Thus I max E R 240 V 15 Ω 16 A. 028 (part 2 of 3) 10.0 points Find the current in the coils when the motor has reached maximum speed. Correct answer: 6 A. At the maximum speed the current is reduced to I E E back R 240 V150 V 15 Ω 6 A. 029 (part 3 of 3) 10.0 points If the current in the motor is 7 A at some instant, what is the back emf at that time? Correct answer: 135 V. Let : I 7 A. For the back emf, so E E back IR, E back E IR 240 V(7 A)(15 Ω) 135 V. Serway CP 20 06 030 10.0 points A solenoid 2.37 cm in diameter and 10.1 cm long has 371 turns and carries a current of 17.5 A. Find the magnetic flux through the circular cross-sectional area of the solenoid. Correct answer: 3.56358 10 5 T m 2. Let : µ 0 1.25664 10 6 T m/a, r 1.185 cm 0.01185 m, l 10.1 cm 0.101 m, N s 371 turns, and I 17.5 A. The magnetic field strength is µ 0 ni µ 0N s I l s and since θ 0, the magnetic flux is Φ A cosθ µ 0N s I(πr 2 ) l s (1.25664 10 6 T m/a) (371 turns)(17.5 A)π(0.01185 m)2 0.101 m 3.56358 10 5 T m 2. keywords: Serway CP 20 40

Version 001 HW 22 EM Induction C&J sizemore (21301jtsizemore) 12 031 10.0 points An emf of 37.7 mv is induced in a 524-turn coil when the current is changing at a rate of 9.8 A/s. What is the magnetic flux through each turnofthecoilataninstant whenthecurrent is 4.6 A? Correct answer: 3.37708 10 5 T m 2. Let : E 37.7 mv 0.0377 V, N 524, I 4.6 A, and I 9.8 A/s. From Faraday s Law E N Φ E L I L E I L I 0.0377 V 9.8 A/s 0.00384694 H At the instant where the current is 4.6 A, the flux through each turn is Φ LI N keywords: (0.00384694 H)(4.6 A) 524 3.37708 10 5 T m 2. Serway CP 20 11 032 10.0 points A strong electromagnet produces a uniform field of 1.35 T over a cross-sectional area of 0.308 m 2. A coil having 224 turns with a total resistance of 23.1 Ω is placed around the electromagnet, and the current in the electromagnetisturnedoff,reachingzeroin24.2ms. What current is induced in the coil? Correct answer: 166.612 A. so Given : i 1.35 T, A 0.308 m 2, N 224 turns, r 23.1 Ω, and t 24.2 ms 0.0242 s. The change in flux is Φ 0Φ i i A, E N Φ I E R N ia R N ia (224)(1.35 T)(0.308 m2 ) (23.1 Ω)(0.0242 s) 166.612 A.r Serway CP 20 35 033 (part 1 of 3) 10.0 points In a model AC generator, a 754-turn rectangular coil, 4.7 cm by 21 cm, rotates at 109 rev/min in a uniform magnetic field of 0.92 T. What is the maximum emf induced in the coil? Correct answer: 78.1505 V. Let : N 754, x 4.7 cm 0.047 m, y 21 cm 0.21 m, ω 109 rev/min, and 0.92 T. The area is A xy (0.047 m)(0.21 m) 0.00987 m 2,

Version 001 HW 22 EM Induction C&J sizemore (21301jtsizemore) 13 the angular velocity is 2π rad ω (109 rev/min) rev 11.4145 rad/s, and the maximum emf is 1 min 60 s E max N Aω (754)(0.00987 m 2 )(0.92 T) (11.4145 rad/s) 78.1505 V. 034 (part 2 of 3) 10.0 points What is the instantaneous value of the emf in the coil at t π s? Assume that the emf is 31 zero at t 0. Correct answer: 71.5472 V. Let : t π 31 s. E E max sinωt (78.1505 V) [ ( π )] sin (11.4145 rad/s) 31 s 71.5472 V. 035 (part 3 of 3) 10.0 points What is the smallest value of t for which the emf will have its maximum value? Correct answer: 0.137615 s. The emf will be a maximum at t T 4 2π 4ω π 2ω π 2(11.4145 rad/s) 0.137615 s. keywords: