Week 10 - Electromagnetic Induction

Transcription

1 Week 10 - Electromagnetic Induction October 28, 2012 Exercise 10.1: Discussion Questions a) Two circular loops lie side by side in the same plane. One is connected to a source that supplies an increasing current; the other is a simple closed ring. Is the induced current in the ring in the same direction as the current in the loop is connected to the source, or opposite? What if the current in the first loop is decreasing? Explain. The best ting here is to draw a sketch of the situation. Let s for the sake of the argument suppose that the current is going clockwise in the first coil. As the current is increasing in the first coil, the magnetic field is increasing trough the second coil from below. By Lenz law the current will therefore flow in a direction which opposes this increasing magnetic field. If it has to oppose a magnetic field coming from below it has to be going in the same direction as the current in the first coil. But if the current in the first coil is decreasing the magnetic field coming from below trough the second coil is decreasing in strength. Lenz law tells us the current will again flow to oppose this decrease in strength and thus the current in the second coil will now have to flow the opposite way of the first coil. b) A sheet of copper is placed between the poles of an electromagnet with the magnetic field perpendicular to the sheet. When the sheet is pulled out, a considerable force is required, and the force required increases with speed. Explain. As the copper sheet is placed between the poles of the electromagnet there is a considerable magnetic flux trough it. When this sheet is now pulled out of the field, this flux decreases and by Lenz law there will be currents in the sheet that flows in such a way that opposes the decrease in flux. These current will now feel a magnetic force proportional to the speed at which you pull the sheets out of the magnetic field. This is the reason for the force felt. Draw the situation to ensure yourself that this force is indeed in the direction opposite to the motion of the sheet. Note that the same phenomenon would not happen if you had a material with a very high resistance. Current would then not be able to flow and there would not be a corresponding magnetic force. c) A student asserted that if a permanent magnet is dropped down a vertical copper pipe, it eventually reaches a terminal velocity even if there is no air resistance. Why should this be? Or should it? 1

2 One explanation comes about when viewing the pipe as a series of vertically stacked circular coils of copper wire. When the permanent magnet falls toward one such coil it s magnetic flux is increasing, so it sets up an eddy current creating a magnetic field which opposes that of the permanent magnet. The resulting magnetic forces on the permanent magnet from this field is indeed one which opposes it s motion. And this field will be larger the faster the magnet falls which indicates that there comes a velocity for which the force is equal to gravitational force. Exercise 10.2: Search Coils and Credit Cards A search coil consists of N loops, each with area A, connected trough a circuit of resistance R to an ammeter. The search coil is first placed such that the area vector A is parallel with the magnetic field lines in a certain region. Then it is turned in a short time t such that A is now perpendicular to the magnetic field lines and there is no flux trough the search coil. a) Derive the equation relating the total charge Q flowing trough the search coil during the time t to the magnetic field magnitude B. Q = NBA R. (1) b) In a credit card reader, the magnetic strip on the back of the credit card is rapidly swiped past a coil within the reader. Explain, using the same ideas that underlie the operation a search coil, how the reader can decode the information stored in the pattern of magnetization on the strip. When the different regions of different magnetization swipe past this coil they induce a current in it. This current is dependent on the magnetic field strength from different parts of the magnetized pattern, but it should also be dependent on how fast you run your card trough the reader. c) Is it necessary that the credit card be swiped trough the reader at exactly the right speed? Why or why not? From the answer it (b) it seems like it should depend on exactly that, since the current is proportional to the velocity of the card. However it might be the shape of the signal produced that matters. Alternatively, from the relation in (a), the total charge is independent of the time period of the swipe. So card readers can read the total charge and give an identification from that. Exercise 10.3: Make a Generator? You are shipwrecked on a deserted tropical island. You have some electrical devices that you could operate using a generator, but you have no magnets. The earth s magnetic field at your location is horizontal and has magnitude T, and you decide to try to use this field for a generator by rotating a large circular coil of wire at a high rate. You need to produce a peak emf of 9.0 V and 2

3 estimate that you can rotate the coil at 30 rpm by turning a crank handle. You also decide that to have an acceptable coil resistance, the maximum number of turns the coil can have is a) What area must the coil have? A = ε p NπB = 9.0 π m2 = 17.9 m 2. (2) b) If the coil is circular, what is the maximum translational speed of a point on the coil as it rotates? Do you think this device is feasible? Explain. v = ωr = π 2.4 m/s = 7.5 m/s. (3) Exercise 10.4: emf in a Bullet At the equator, the earth s magnetic field has a value of T and is approximately horizontal, pointing toward the north. A solider fighting a war in the region recalls his electromagnetism and starts to wonder weather it is created an emf in his bullets while in flight. The bullets have a length of l = 1.0 cm and a diameter of d = 0.4 cm and have a speed of v = 300 m/s while airborne. a) If the solider shots toward the east, is there an emf? If so what is it s value and which side (top, bottom, front or back) is at the higher potential? V = vbd = V = 96 µv. (4) b) What is the emf the solider shots toward the south? If the bullet is shot toward the south, there will be no magnetic force on the bullet. This is because the velocity and the magnetic field is anti-parallel. c) What is the emf between the front and the back of the bullet for any horizontal velocity? For any horizontal velocity, the magnetic force will always be perpendicular to it and the magnetic field. Since both the magnetic field and the velocity lies in the horizontal plane the force must be pointing up or down. Thus there can not be an emf between the two sides of the bullet. 3

4 Figure 1 Exercise 10.5: Terminal Speed A conducting rod with length L, mass m and resistance R moves without friction on metal rails as shown in figure 1. A uniform magnetic field B is directed into the plane of the figure. The rod starts from rest and is acted on my a constant force F directed to the right. The rails are infinitely long and have negigble resistance. a) Sketch the speed of the rod as a function of time. Here you might solve the equation of motion for the speed or use your intuition. Figure 2 b) Find an expression for the terminal speed (the speed when the acceleration of the rod is zero). Explain why this speed is prorportional to the resistance R? v = F R (BL) 2 (5) 4

5 Exercise 10.6: A Persistent Current (There is currently no answers or solutions included for this exercise.) A circular wire loop of radius a and resistance R initially has a magnetic flux through it due to an external magnetic field. The external field then decreases to zero. A current is induced in the loop while the exernal field is changing; however, this current does not stop at the instant that the external field stops changing. The reason is that the current itself generates a magnetic field, whic gives rise to a flux trough the loop. If the current changes, the flux trough the loop changes as well, and an induced emf appears in the loop to oppose the change. a) The magnetic field at the center of the loop of radius a is given by B = µ 0 i/2a. If we use the crude approximation that the field has this same value at all points within the loop, what is the flux of this field trough the loop? b) By using Faraday s law, and the relationship ε = ir, show that ater the external field has stopped changing, the current in the loop obes the differential equation ( ) di 2R dt = i. (6) πµ 0 a c) If the current has the value i 0 at t = 0, the instant that the external field stops changing, solve the equation in part (b) to find i as a function of time for t > 0. d) If the loop has radius a = 50 cm and resistance R = 0.10 Ω, how long after the external field stops changing will the current be equal to i 0? e) In the examples in the book, this effect was ignored. Explain why this is a good approximation. 5