PHY 232 Exam 3 October 15, 2012 Exam 3 November 19, 2012 Instructor: Timothy Martin Student Information Name and section: UK Student ID: Seat #: Instructions Answer the questions in the space provided. On the long form questions, show your work! Answers with 0 work shown will not be graded. Points will be given based on correctness and coherency if it is difficult to follow your work, you will likely receive less partial credit. If you run out of room for an answer, continue on the back of the page. You are not allowed use of a calculator or any other electronic device during this test. Question Points Score 1 25 2 25 3 20 4 30 Total: 100
PHY 232 Exam 3, Page 2 of 7 October 15, 2012 1. Multiple choice Multiple choice questions (a) (3 points) Two long parallel wires are oriented vertically, both carrying current upward. The forces on the left and right wire respectively will be: A. Both to the right. B. To the right and to the left. C. To the left and to the right. D. Both to the left. Solution: B. Use the RHR like currents attract. (b) (3 points) Which of the following types of magnetic behavior persists even in the absence of a magnetic field? A. Ferromagnetism B. Paramagnetism C. Diamagnetism D. None of the above Solution: A. (c) (3 points) A loop of current sits in a uniform magnetic field. The resulting torque causes it to start to rotate. Imagine if we suddenly reversed the direction of the current, while keeping everything else the same. What effect would this sudden change have on the magnitude of the torque? A. Increase. B. Decrease. C. Stay the same. D. Not enough information is given. Solution: C. At that moment, the direction of the torque change, but the magnitude will be the same. (d) (3 points) A square loop of wire of side-length a is travelling to the right with speed v. It enters a region of magnetic field directed into the page. (Perpendicular to the plane of the loop.) The magnitude of the generated emf as it moves between regions is A. a 2 vb B. avb C. vb/a D. B/av Solution: B. As derived several times in class. Note that it s also the only option with the correct units. Page 2
PHY 232 Exam 3, Page 3 of 7 October 15, 2012 (e) (3 points) If a solenoid with current flowing through it is compressed like a spring to half its original length, what happens to the magnetic field inside? A. It remains constant B. It is halved. C. It doubles. D. It changes direction Solution: C. The magnetic field in a solenoid is proportional to N/L, so halving the length will double the field. (f) (3 points) A simple circuit consists of an inductor, a resistor, and a battery. At t =0, the current is I =0. As we watch the circuit, which of the following is true? A. The voltage across the resistor will asymptotically approach 0. B. The magnetic field in the inductor will asymptotically approach some maximum value. C. The current in the circuit will oscillate, but will at first increase in magnitude. D. The energy in the inductor will oscillate, but will at first decrease in magnitude. Solution: B. This LR circuit will asymptotically approach a maximum current I = E/R. The magnetic field in the inductor is proportional to the current. (Just like in a solenoid.) (g) (3 points) A circular loop of current is centered around the z-axis. We can use the Biot-Savart law to calculate the magnetic field at some point P along the z-axis, by integrating over the loop. As we perform this integral, which is true of the individual contribution of an infinitesimal segment of the loop? A. Only the magnitude of the contribution db will vary over the length of the wire. B. Only the direction of the contribution will vary over the length of the wire. C. Both magnitude and direction vary. D. Both magnitude and direction are constant. Solution: B. Each infinitesimal bit of current is the same distance from the point, with the relative angle between ˆr and ds likewise constant. But the direction will vary. (h) (3 points) As the current in a 4Hinductor increases, we measure a constant potential difference of 12 V across it. If the initial current was I = 10A, what is the current 1slater? A. 13 A B. 19 A C. 26 A D. 58 A Solution: A. We find that di/dt =3A/ s. That means that in one second, the current will have increased by 3A. (i) (1 point) Hopefully, you remembered to write your name, section, and ID on the cover page. A. Definitely. B. Absolutely. Page 3
PHY 232 Exam 3, Page 4 of 7 October 15, 2012 2. (20 points) Three long wires are arranged along the x-axis as shown below. The middle wire is at the origin, the other two at x = ±a. All three currents have the same magnitude. (I 1 = I 2 = I 3.) (a) (15 points) Find the direction and magnitude of the magnetic field at some point P located along the x-axis at x =3a. (b) (10 points) Find the direction and magnitude of the magnetic force on an electron as it passes through point P. The electron has velocity v = v 0 î and charge q = e. Solution: For (a), simply use that the magnetic field magnitude will be given by B = µ0i 2πr. Use the RHR to get the directions: at point P, wires 1 and 3 will produce a magnetic field directly downwards, and wire 2 directly upwards. So B tot =( B 1 + B 2 B 3 )ĵ = µ 0I 1 2π 2a + 1 3a 1 ĵ 4a The result is that B tot = 5µ 0I 24πaĵ For (b), just use that F = qv B: F B =( e)(v 0 î) B Substituting the answer from (a), we get F B = ev 0 5µ 0 I 24πa ˆk Page 4
PHY 232 Exam 3, Page 5 of 7 October 15, 2012 3. A long straight wire of radius R carries a current I. Assume that the current is uniformly distributed throughout the wire, and flows along the z-axis into the page. As is conventional, the +y direction is up and the +x direction is to the right. (a) (15 points) What is the magnitude of the magnetic field generated halfway between the center and edge of the wire along the y axis, at y = R/2? (About at the indicated point in the diagram.) (b) (5 points) If we replaced the wire with a hollow wire carrying the same total current, with inner radius R, what would the field at the indicated point be then? 3 4 Solution: This problem is best solved with Ampere s Law. Imagine a circular surface of the appropriate radius. For (a), one quarter of the total current will be inside a circle of radius r = R/2 (ratio of areas!), so B(2π R 2 )=µ I 0 4 B = µ 0I 4πR For (b), no current will sit inside the Amperian surface, so the magnetic field will be 0. Page 5
PHY 232 Exam 3, Page 6 of 7 October 15, 2012 4. A square wire loop is rotating in a constant magnetic field B at an angular speed ω, which will induce a current. At t =0the magnetic field is parallel to the plane of the loop, as shown in the diagram. In the diagram, we are looking down at the top of the loop and it is rotating counter-clockwise. It has side-length a and total resistance R. (a) (5 points) At t =0, the current will be at maximum amplitude what will be the direction of this current through the top part of the loop? (You can indicate the direction as just up or down, based on the illustration.) (b) (15 points) What will the magnitude of the current be, as a function of time? (Be careful to check that the answer at t =0makes sense!) (c) (10 points) Calculate the magnitude of the magnetic force acting on just the top part of the loop as a function of time. Solution: For (a), the current will flow up the page initially. The flux will be increasing through the loop it will be to the right of the loop, so Lenz s Law tells us the direction of the current. For (b), first find the flux as a function of time: Φ B = BAcos(θ BA )=Ba 2 sin(ωt) It s in the final expression sin so that the phase works out you could also have written something like cos(π/2 ωt), but it s easier to work with just sin. The emf is the time derivative of the flux, and I = E/R I(t) = ωba2 R cos(ωt) For (c), we use F = I L B. L is the sidelength a, and we can see that the angle between L and B will be θ = ωt. So Plugging in the answer from (b) we d get F B = IaB sin(ωt) F B = ωb2 a 3 R cos(ωt)sin(ωt) Page 6
PHY 232 Exam 3, Page 7 of 7 October 15, 2012 Formula sheet Magnetic force on a moving particle Magnetic force on a length of wire F B = qv B F B = I L B Faraday s Law Biot-Savart Law E = dφ B dt db = µ 0 Ids ˆr 4π r 2 Magnetic field magnitude around a long current carrying wire B = µ 0I 2πr Magnetic field inside a solenoid B = µ 0 ni Ampere s Law B ds = µ 0 I enc An inductor in a circuit obeys The energy stored in the inductor is L = E/ di dt U = 1 2 LI2 The energy density of a magnetic field is u B = 1 2µ 0 B 2 Page 7