Version 001 – Electromagnetism – tubman – (IBII20142015) This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Holt SF 22C 02 001 10.0 points The current in an AC circuit is measured with an ammeter, which gives a reading of 8.5 A. Calculate the maximum AC current. 1 003 (part 2 of 2) 10.0 points Find the maximum potential difference output of the source. Correct answer: 249 V. Explanation: The maximum potential difference is Correct answer: 12.0208 A. Explanation: Vmax = 249 V . Let : Irms = 8.5 A . The rms current√is 2 Imax Irms = Imax = √ , 2 2 so the maximum AC current is √ Imax = Irms 2 √ = (8.5 A) 2 AP B 1998 MC 45 004 10.0 points A metal spring (wrapped loosely around a cardboard doughnut as a toroid) has its ends attached so that it forms a circle. It is placed in a uniform magnetic field, as shown. = 12.0208 A . Holt SF 22Rev 44 002 (part 1 of 2) 10.0 points The alternating potential difference of a generator is represented by the equation E = (249 V) sin(359 rad/s) t , where E is in volts and t is in seconds. Find the frequency of the potential difference of the source. Correct answer: 57.1366 Hz. Explanation: Let : E = Vmax sin ω t = (249 V) sin(359 rad/s) t , ω = 2πf = 359 rad/s . The frequency of the source is ω f= 2π 359 rad/s = 2π = 57.1366 Hz . B B B B Which of the following will NOT cause a current to be induced in the spring? and 1. Moving the spring parallel to the magnetic field correct 2. Rotating the spring about a diameter 3. Moving the spring in and out of the magnetic field 4. Changing the magnitude of the magnetic field Version 001 – Electromagnetism – tubman – (IBII20142015) 5. Increasing the diameter of the circle by stretching the spring 2. clockwise. correct Explanation: There will be an induced current if there is an induced emf around the spring. By Faraday’s law, we know that 3. counter-clockwise. E =− AP EM 1993 FR 2 B 005 (part 1 of 3) 10.0 points A rectangular loop of copper wire of resistance R has width a and length b. The loop is stationary in a uniform magnetic field. The magnetic field B at time t = 0 seconds is directed into the page as shown below. The uniform magnetic field varies with time t according to the relationship B = B0 cos ω t, where ω and B0 are positive constants and B is positive when the field is directed into the page. B a b n turns B Explanation: d ΦB , dt where φB is the magnetic flux through the loop formed by the spring. To obtain a non-zero emf, we must change the magnetic flux. This can be accomplished by either changing the magnitude of the magnetic field or changing the area of the loop perpendicular to the magnetic field. Of the five choices, only moving the spring parallel to the magnetic field will NOT change the magnetic flux, thus will NOT cause a current to be induced in the spring. B 2 π π When ω t = , B = B0 cos = 0; i.e., 2 2 the field has been decreasing, and is about to change direction. The induced current will be in a direction to oppose this change; i.e., clockwise. 006 (part 2 of 3) 10.0 points What is the expression for the magnitude of the induced current in the loop as a function of time in terms of a, b, B0 , ω, R, t, and fundamental constants. 1. I = R ω B0 | sin ω t| ab 2. I = a b ω B0 R | sin ω t| a b ω B0 | sin ω t| correct R a b B0 4. I = | sin ω t| R R B0 5. I = | sin ω t| ab a b B0 | sin ω t| 6. I = ωR Explanation: Calculating the flux, 3. I = B The direction ofπthe induced current in the loop when ω t = , after the magnetic field 2 begins to oscillate is 1. undetermined, since the current is zero. Φ = a b B0 cos ω t . Calculating the emf, dΦ (negative sign not required) dt = a b ω B0 sin ω t . E =− Version 001 – Electromagnetism – tubman – (IBII20142015) Using Ohm’s Law, the magnitude of the current is I ωt 5. I= π 2 0 |E| a b ω B0 = | sin ω t| . R R 007 (part 3 of 3) 10.0 points Select a graph of the induced current I vs ω t, taking clockwise current to be positive. 3 π 3π 2 2π 5π 2 3π I ωt 6. π 2 0 I π 3π 2 2π 5π 2 3π ωt 1. 0 π 2 π 3π 2 2π 5π 2 3π Explanation: The graph is a sine wave with period 2 π . I I ωt ωt 2. 0 π 2 π 3π 2 2π 5π 2 ωt 0 π 2 π 3π 2 2π 5π 2 π 2 π 3π 2 2π 5π 2 3π 3π I 3. 0 3π Faraday Equation 008 10.0 points Suppose you are looking into the end of a long cylindrical tube in which there is a uniform magnetic field pointing away from you. What is the direction of the induced electric field if the magnitude of the magnetic field is decreased with time? 1. radially inward toward the axis of the tube I 2. counterclockwise ωt 4. 0 correct π 2 π 3π 2 2π 5π 2 3π 3. toward you 4. clockwise correct 5. radially outward from the axis of the tube Version 001 – Electromagnetism – tubman – (IBII20142015) 4 = 312.429 N . 6. away from you Explanation: If the magnetic field decreases with time, the electric field will be generated in order to oppose the change. The right hand rule can be applied to find that the direction of electric field is in the clockwise direction. Applied Force on a Bar 02 009 (part 1 of 2) 10.0 points In the arrangement shown in the figure, the resistor is 7 Ω and a 9 T magnetic field is directed out of the paper. The separation between the rails is 3 m . An applied force moves the bar to the right at a constant speed of 3 m/s . To maintain the motion of the bar, a force must be applied on the bar to balance the magnetic force F = FB = 312.429 N . 010 (part 2 of 2) 10.0 points At what rate is energy dissipated in the resistor? Correct answer: 937.286 W. Explanation: The power dissipated in the resistor is P = I 2 R = (11.5714 A)2 (7 Ω) 9T m≪1 g 7Ω 3m I = 937.286 W . 3 m/s 9T Calculate the applied force required to move the bar to the right at a constant speed of 3 m/s. Assume the bar and rails have negligible resistance and friction. Neglect the mass of the bar. AP B 1998 MC 48 011 10.0 points A single circular loop of wire in the plane of the page is perpendicular to a uniform mag~ directed into the page, as shown. netic field B B Correct answer: 312.429 N. Explanation: The motional emf induced in the circuit is E = B ℓ v = (9 T) (3 m) (3 m/s) = 81 V . B If the magnitude of the magnetic field is decreasing, then the induced current in the wire loop is From Ohm’s law, the current flowing through the resistor is E 81 V I= = = 11.5714 A , R 7Ω so the magnitude of the force exerted on the bar due to the magnetic field is 2. counterclockwise around the loop. FB = I ℓ B = (11.5714 A)(3 m)(9 T) 4. directed upward out of the paper. 1. clockwise around the loop. correct 3. zero. (No current is induced.) Version 001 – Electromagnetism – tubman – (IBII20142015) Explanation: Using Lenz’s law, the induced current must be directed to counter the change of the magnetic flux through the loop; i.e., the magnetic field generated by the induced current is pointing downward into the page. Thus, using right hand rule, the induced current must be clockwise along the loop. Sliding Metal Rod 03 012 (part 1 of 2) 10.0 points The resistance of the rectangular current loop is R, and the metal rod is sliding to the left. The length of the rod is d, while the width of the rails is ℓ. a and b are the contact points where the rod touches the rails, and d > ℓ . a B d m v B ℓ b What is the magnitude of the induced current around the loop? 1. |Iind | = 2. |Iind | = 3. |Iind | = 4. |Iind | = 5. |Iind | = 6. |Iind | = 7. |Iind | = B d2 v R B d v2 R Bdv R2 Bℓv correct R Bℓv R2 Bdv R B d v2 R2 B ℓ v2 R2 B ℓ2 v 9. |Iind | = R B ℓ v2 10. |Iind | = R Explanation: Note: The part of the rod which extends past the rails does not have a bearing on the answers. Lenz’s law states that the induced current appears such that it opposes the change in the magnetic flux. In this case the magnetic flux through the rectangular loop is decreasing (since the area of the loop is decreasing) with the direction of the flux into the page, so that the induced magnetic field must point into the page in order to keep the flux through the loop constant. This corresponds to an induced current which flows in a counter-clockwise direction. Hence, if you look at the potential drop across the resistor, then you can see that the potential at a is greater than the potential at b, and the direction of the induced current is down through the metal rod. The rate of change of the area of the rectangular loop is 8. |Iind | = 5. directed downward into the paper. R 5 dA dx =ℓ . dt dt Then from Faraday’s law, the magnitude of the induced emf is given by E = Bℓv. From Ohm’s law, E = I R so the magnitude of the induced current is Iind = E Bℓv = . R R 013 (part 2 of 2) 10.0 points The magnitude of the force exerted on the metal rod by the magnetic field, and the direction of the current through the metal bar are given respectively by B 2 ℓ2 v 2 ~ , current flow a → b 1. kF k = R Version 001 – Electromagnetism – tubman – (IBII20142015) 2 ~ k = B d ℓ v , current flow a → b 2. kF R 2 2 ~ k = B d v , current flow a → b 3. kF R 2 2 ~ k = B ℓ v , current flow b → a 4. kF R 2 2 ~ k = B d v , current flow b → a 5. kF R 2 2 2 ~ k = B ℓ v , current flow b → a 6. kF R 2 B ℓ2 v ~ 7. kF k = , current flow a → b corR rect 2 2 2 ~ k = B d v , current flow a → b 8. kF R 2 ~ k = B d ℓ v , current flow b → a 9. kF R 2 B d2 v 2 ~ 10. kF k = , current flow b → a R Explanation: The magnitude of the magnetic force exerted on the metal rod is given by F = I ℓB, where i is the induced current found in part 2. ~ = 0 outside the Note: ℓ was used since B rectangle. Substituting in the expression for the induced current yields F = B 2 ℓ2 v . R The direction of the induced current is down (a → b) through the metal rod as explained in Part 1. Flat Coil of Wire 014 10.0 points A flat coil of wire consisting of 14 turns, each with an area of 150 cm2 , is positioned perpendicularly to a uniform magnetic field that increases its magnitude at a constant rate from 1.4 T to 14 T in 4.7 s. If the coil has a total 6 resistance of 0.2 Ω, what is the magnitude of the induced current? Correct answer: 2.81489 A. Explanation: d ΦB Zdt ~ · dA ~ ΦB = N B E=− = NBA N A (B2 − B1 ) |E| = t |E| I= R N A (B2 − B1 ) = Rt = 2.81489 A . keywords:
© Copyright 2024