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- [SOLUTIONS] Mastering Physics HW27.pdf

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HW27 Due: 11:59pm on Tuesday, November 10, 2009 Note: To understand how points are awarded, read your instructor's Grading Policy. [Return to Standard Assignment View] Motional EMF in a Conducting Rod In the figure, a conducting rod with length = 34.0 moves in a magnetic field of magnitude 0.370 directed into the plane of the figure. The rod moves with speed = 6.00 in the direction shown. Part A When the charges in the rod are in equilibrium, which point, a or b, has an excess of positive charge? Hint A.1 Finding the force on charges moving in a magnetic field The force on a charge moving with velocity in a magnetic field is given by . One can use the right-hand rule to determine the direction in which the charges in the rod are forced. ANSWER: a b Correct Part B In what direction does the electric field then point? Hint B.1Where does the electric field come from? Hint not displayed ANSWER: from a toward b from b toward a Correct Part C When the charges in the rod are in equilibrium, what is the magnitude of the electric field within the rod? Hint C.1 How to approach the problem Use the fact that at equilibrium, the electric and magnetic forces on a test charge in the rod must be equal and opposite to each other. Hint C.2 Formula for the force on a charge moving in a magnetic field The force on a charge moving with velocity in a magnetic field is given by . Hint C.3 Formula for the force on a charge in an electric field The force on a charge in an electric field is . Hint C.4 Units Recall that the SI unit of a magnetic field is the tesla, velocity is measured in meters per second, and electric field in volts per meter. Express your answer in volts per meter to at least three significant figures. ANSWER: = 2.22 Correct Part D Which point, a or b, is at higher potential? Hint D.1 Formula for the potential difference The formula for the potential difference between two points a and b in a region of space containing an electric field is , where the line integral can be taken along any path from b to a. In particular, the straight line connecting b to a (the ends of the rod), i.e., the path along the length of the rod, is a good choice for this problem. ANSWER: a b Correct Some work must have been done in order to create this potential difference, i.e., to separate the charges in the rod. This work was done by the initial force required to pull the rod in opposition to the force on it due to the interaction of the transient current in it with the magnetic field. Note that once the charges are in equilibrium, no force is required to keep the rod moving with constant velocity. Part E What is the magnitude of the potential difference between the ends of the rod? Hint E.1Formula for the potential difference The formula for the potential difference between two points a and b in a region of space containing an electric field is , where the line integral can be taken along any path from b to a. In particular, the straight line connecting b to a (the ends of the rod), i.e., the path along the length of the rod, is a good choice for this problem. Hint E.2Calculating the integral Note that the electric field in the rod is a constant (the rod is a conductor), and so the integral should be relatively straightforward. Express your answer in volts to at least three significant figures. ANSWER: = 0.755 Correct Part F What is the magnitude of the motional emf induced in the rod? Hint F.1 Definition of motional emf Hint not displayed Express your answer in volts to at least three significant figures. ANSWER: = 0.755 Correct In the Hints for Part C of "Rail Gun" take the CCW direction around the circuit for determining the sign of the induced EMF. Please note that, in spite of the blurb which will follow a correct answer to Part C, you do not have to understand anything about differential equations to solve Part D. Just follow the hint for Part D. GBA Rail Gun This problem explores how a current-carrying wire can be accelerated by a magnetic field. You will use the ideas of magnetic flux and the EMF due to change of flux through a loop. Note that there is an involved follow-up part that will be shown once you have found the answer to Part B. Part A A conducting rod is free to slide on two parallel rails with negligible friction. At the right end of the rails, a voltage source of strength in series with a resistor of resistance makes a closed circuit together with the rails and the rod. The rails and the rod are taken to be perfect conductors. The rails extend to infinity on the left. The arrangement is shown in the figure. There is a uniform magnetic field of magnitude , pervading all space, perpendicular to the plane of rod and rails. The rod is released from rest, and it is observed that it accelerates to the left. In what direction does the magnetic field point? Hint A.1 The force on a conducting rod due to a magnetic field There is a force on the rod because a current is flowing through it. Hence charges are moving perpendicular to the magnetic field. The rod experiences a force , which is given by . Hint A.2 The direction of the magnetic field Use the right-hand rule with the cross product: Take your right hand and point with your index finger in the direction of the rail's motion, and point your middle finger in the direction of . Your thumb will then point in the direction of the magnetic field. ANSWER: into the plane of the figure out of the plane of the figure Correct Part B Assuming that the rails have no resistance, what is the most accurate qualitative description of the motion of the rod? Hint B.1Lenz's law An EMF is induced in the circuit due to the change in magnetic flux through it. But will this EMF increase the current through the loop or decrease it by opposing the voltage coming from the source? Lenz's law states that induced currents will always flow in such a way that they oppose the change in flux that caused them. Hint B.2Appyling Lenz's law to this problem Let us apply Lenz's law here. We want to find the direction of the induced current. The change in flux that induced this current is caused by the motion of the rod, which in turn is caused by the current flowing around the circuit due to the voltage source. The induced current works against that source current, and reduces it. Alternatively, one could say that the induced EMF opposes the voltage from the source. Hint B.3The velocity of the rod The higher the velocity of the rod, the higher the induced EMF, and the lower the current flowing through the loop. But the force accelerating the rod is proportional to the current. Hence the acceleration goes down as the velocity goes up. The velocity cannot increase beyond the point at which the induced EMF is equal and opposite to . ANSWER: The rod will accelerate but the magnitude of the acceleration will decrease with time; the velocity of the rod will approach but never exceed a certain terminal velocity. Under these idealized conditions the rod will experience constant acceleration and the velocity of the rod will increase indefinitely. The rod will accelerate indefinitely with acceleration proportional to its (increasing) velocity. Correct Part C What is the acceleration of the rod? Take to be the mass of the rod. Hint C.1 Find the induced EMF To determine the current, we need the EMF induced by the change in magnetic flux through the current loop of area . Faraday's law states that . The motion of the rod will change according to . What is the EMF ? Express in terms of the velocity , the separation of the rails , and the magnetic field . ANSWER: = Correct Hint C.2 Find the current in the rod We can find the current through the rail by using Kirchhoff's rule for a closed circuit: , where is the induced EMF. What is the current ? Express your answer in terms of and other given quantities. ANSWER: ANSWER: = Correct Hint C.3 Find the acceleration of the rod To find the acceleration , we need the force on the rod. We have already determined its direction. The magnitude of this force is given in Part A.i. Write down an expression for , using Newton's second law. Expres your answer in terms of mass , current , , and . ANSWER: = Correct Express your answer as a function of , , the velocity of the rod , , , and the mass of the rod . ANSWER: = Correct Making the substitution , you obtain the dfferential equation , which you can solve to find the velocity of the rod as a function of time: . To achieve a high acceleration, which is necessary for a useful gun, a magnetic field of large magnitude and a high voltage are advantageous. Part D What is the terminal velocity reached by the rod? Hint D.1 Find an expression for the terminal velocity Hint not displayed ANSWER: = Correct A larger magnetic field increases the acceleration of the rod, but lowers the terminal velocity: a trade-off for rail gun engineers! A Primitive DC Generator The conducting rod ab in the figure has a length of 50 and makes contact with metal rails ca and db. The apparatus is in a uniform magnetic field of strength 0.75 , perpendicular to the plane of the figure. Part A Find the magnitude of the emf induced in the rod when it is being pulled toward the right with a constant speed of 7.8 . ANSWER: 2.93 Correct V Part B In what direction does the current flow in the rod? ANSWER: towards the bottom of the screen towards the top of the screen Correct Part C Assume all of the resistance of the circuit abdc is between locations c and d. If that resistance has a value of 1.8 , what is the induced current in the circuit as long as the rod is moving to the right with a constant speed of 7.8 ? ANSWER: 1.63 Correct A Part D What is the magnitude of the magnetic force exerted on the moving bar while it is moving to the right with a constant speed of 7.8 ? ANSWER: 0.609 Correct N Part E What is the direction of the magnetic force described in Part D? ANSWER: to the left to the right towards the bottom of the screen out of the screen towards the top of the screen into the screen Correct Part F Ignoring any possible friction between the rods and the rails, find the magnitude of the force required to keep the rod moving to the right with a constant speed of 7.8 . ANSWER: 0.609 Correct N Part G What is the angle between the force described in Part F and the velocity vector for the moving bar? ANSWER: 0.00×10 0 Correct degrees Part H At what rate is mechanical work being done by the person or machine which is pulling the bar to the right at a constant speed of 7.8 ? ANSWER: 4.75 Correct W Part I At what rate is thermal energy being produced in the resistance between points c and d in the circuit? ANSWER: 4.75 Correct W Part J Compare the rate at which mechanical work is done by the force with the rate at which thermal energy is developed in the circuit. ANSWER: equal nonequal Correct Most of "Introduction to Faraday's Law" is excellent, but Part (E) is arguable. MP prefers answer 2; please use that answer. However, it is possible to envision situations (involving relative velocities between circuits, or between a circuit and a magnet) in which non-electrostatic E fields are being induced by a constant current. I am assuming here that "constant current" means constant in time; if "constant" is taken to mean constant in time AND space, then MP is explicitly correct. One could also argue that MP's answer is technically correct because, in the blurb which follows a correct answer to Part (E), MP is careful to state that the E field in the integral is due to the moving charges and not due to any changing magnetic flux because of the way in which those charges are moving. The main idea is that, for E fields due to charges, the integral of Edotdl over a closed loop is zero; for E fields due to changing magnetic flux, the same integral is not zero. However, please ignore the parenthetical part of the last sentence in the blurb. Whether or not the moving charges pass through the Faraday loop makes absolutely no difference in any way. GBA Introduction to Faraday's Law Learning Goal: To understand the terms in Faraday's law for magnetic induction of electric fields, and contrast these fields with those produced by static charges. Faraday's law describes how electric fields and electromotive forces are generated from changing magnetic fields. It relates the line integral of the electric field around a closed loop to the change in the total magnetic field integral across a surface bounded by that loop: , where is the line integral of the electric field, and the magnetic flux is given by , where is the angle between the magnetic field and the local normal to the surface bounded by the closed loop. Direction: The line integral and surface integral reverse their signs if the reference direction of or is reversed. The right-hand rule applies here: If the thumb of your right hand points along , then the fingers point along . You are free to take the loop anywhere you choose, although usually it makes sense to choose it to lie along the path of the circuit you are considering. Part A Consider the direction of the electric field in the figure. Assume that the magnetic field points upward, as shown. Under what circumstances is the direction of the electric field shown in the figure correct? Hint A.1 How to approach the problem Hint not displayed ANSWER: always if increases with time if decreases with time depending on whether your right thumb is pointing up or down Correct Part B Now consider the magnetic flux through a surface bounded by the loop. Which of the following statements about this surface must be true if you want to use Faraday's law to relate the magnetic flux to the line integral of the electric field around the loop? ANSWER: The surface must be the circular disk in the middle of the loop. The surface must be perpendicular to the magnetic field at each point. The surface can be any surface whose edge is the loop. The surface can be any surface whose edge is the loop as long as no magnetic field line passes through it more than once. Correct You are free to take any surface bounded by the loop as the surface over which to evaluate the integral. The result will always be the same, owing to the continuity of magnetic field lines (they never start or end anywhere, since there are no magnetic charges). It is important to understand the vast differences between electric fields produced by changing magnetic fields via Faraday's law and the more familiar electric fields produced by charges via Coulomb's law. Here are some short questions that illustrate these differences. Part C When can an electric field be measured at any point from the force on a stationary test charge at that point? Hint C.1 Force on a stationary charge Hint not displayed ANSWER: only if the field is generated by the coulomb field of static charges only if the field is generated by a changing magnetic field no matter how the field is generated Correct In fact, this operation defines an electric field. Similarly, if the test charge is moving, it will measure magnetic fields. Part D When can an electric field that does not vary in time arise? ANSWER: only if the field is generated by a coulomb field of static charges only if the field is generated by a changing magnetic field in either of the above two cases Electric fields never vary in time; otherwise, a charge could gain energy from the field. Correct Part E When will the integral around any closed loop of the projection of the electric field along that loop be zero? ANSWER: only if the field is generated by the coulomb field of static charges only if the field is generated by the coulomb field of static charges or a constant current only if the field is generated by a changing magnetic field however the field is generated The loop integral is always zero; otherwise, a charge moving around the loop would gain energy. Correct The electric field generated by a static charge or a constant current always has zero loop integral. A constant current is a continuous line of evenly-spaced charges moving with constant velocity. An electric field generated by any other configuration of moving charges (moving through the loop) would have a non-zero loop integral. Here is a simple quantitative problem that uses Faraday's law. Part F A cylindrical iron rod of infinite length with cross-sectional area is oriented with its axis of symmetry coincident with the z axis of a cylindrical coordinate system as shown in the figure. It has a magnetic field inside that varies according to . Find the theta component of the electric field at distance from the z axis, where is larger than the radius of the rod. Hint F.1 Selecting the loop Hint not displayed Hint F.2 Find the magnetic flux Hint not displayed Hint F.3 Finding the EMF from Faraday's law Hint not displayed Hint F.4 Help from symmetry Hint not displayed Hint F.5 Find the EMF in terms of Hint not displayed Express your answer in terms of , , , , and any needed constants such as , , and . ANSWER: = Correct For Part E of 29.30, remember that the EMF is the work per unit charge done on a hypothetical positive charge moving from one end of the split ring to the other (moving in the direction of the induced E field). GBA Exercise 29.30 The magnetic field at all points within the colored circle of the figure has an initial magnitude of 0.800 . (The circle could represent approximately the space inside a long, thin solenoid.) The magnetic field is directed into the plane of the diagram and is decreasing at the rate of 0.0300 . Part A What is the magnitude of the induced electric field at any point on the circular conducting ring with radius = 0.100 ? ANSWER: = 1.50×10 ?3 Correct Part B What is the direction of this field at any point on the circular conducting ring? ANSWER: clockwise counterclockwise Correct Part C What is the current in the ring if its resistance is 4.40 ? ANSWER: = 2.14×10 ?4 Correct Part D What is the emf between points a and b on the ring? ANSWER: = 4.71×10 ?4 Correct Part E If the ring is cut at some point and the ends are separated slightly, what will be the emf between the ends? ANSWER: = 9.42×10 ?4 Correct In Part A of 29.30, you should have already done the calculation with Faraday's Law that you will need for 29.27. Problem 29.27 should thus be easy; you only have to calculate dB/dt and put in the numbers for r. GBA Problem 29.27 A long, thin solenoid has 850 turns per meter and radius 3.00 . The current in the solenoid is increasing at a uniform rate of 57.0 . Part A What is the magnitude of the induced electric field at a point near the center of the solenoid? ANSWER: 0 Correct V/m Part B What is the magnitude of the induced electric field at a point 0.550 from the axis of the solenoid? ANSWER: 1.67×10 ?4 V/m Correct V/m Part C What is the magnitude of the induced electric field at a point 1.00 from the axis of the solenoid? ANSWER: 3.04×10 ?4 Correct V/m Problem 29.45 contains lots of superfluous information; read the problem carefully. You will find that the question is easy once you have correctly decided which information is useful and which can be ignored. GBA Problem 29.45 A circular coil of wire has a radius of 0.520 , 15.0 turns, and a total resistance of 1.56 . The coil lies in the xy-plane. The coil is in a uniform magnetic field that is in the -z-direction, which is directed away from you as you view the coil. The magnitude B of the field depends on time as follows: increases at a constant rate from 0 at to 0.850 at = 0.480 ; is constant at 0.850 from = 0.480 to = 1.08 ; decreases at a constant rate from 0.850 at = 1.08 to 0 at = 2.04 . Part A What is the maximum induced electric field magnitude in the coil during the to 2.04 time interval? ANSWER: 0.460 Correct Score Summary: Your score on this assignment is 99.5%. You received 54.71 out of a possible total of 55 points. clockwork MasteringPhysics: Assignment Print View

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Author: Anonymous

Textbook: University Physics Vol 2 (Chapters 21-37) (12th Edition) (Chapters 21-37 v. 2)

Created: 2010-01-23

Updated: 2010-01-23

File Size: 9 page(s)

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Views: 17598

Textbook: University Physics Vol 2 (Chapters 21-37) (12th Edition) (Chapters 21-37 v. 2)

Created: 2010-01-23

Updated: 2010-01-23

File Size: 9 page(s)

Keywords: flash card flashcards digital flashcards note sharing notes textbook wiki college dorm class classroom exam homework test quiz university college education learn student teachers tutors share, study blue studyblue studyblu

Views: 17598

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