Problem Set 6: Solutions

Transcription

1 University of Aabama Department of Physics and Astronomy PH 102 / LeCair Summer II 2010 Probem Set 6: Soutions 1. A conducting rectanguar oop of mass M, resistance R, and dimensions w by fas from rest into a magnetic fied B, as shown at right. At some point before the top edge of the oop reaches the magnetic fied, the oop attains a constant termina veocity v T. Show that the termina veocity is: v T = MgR B 2 w 2 NB termina veocity is reached when the net acceeration is zero. See the schematic figure beow. w Bin v First, et us anayze the situation quaitativey. As the oop fas into the region of magnetic fied, more of its area is exposed to the fied, which increases the tota fux through the oop. This increase in magnetic fux wi cause an induced potentia difference around the oop, via Faraday s aw, which wi create a current that tries to counteract this change in magnetic fux. Since the fux is increasing, the induced current in the oop wi try to act against the existing fied to reduce the change in fux, which means the current wi circuate countercockwise to create a fied out of the page. Once there is a current fowing in the oop, each current-carrying segment wi fee a magnetic force. The eft and right segments of the oop wi have equa and opposite forces, eading to no net effect, but the current fowing (to the right) in the bottom segment wi ead to a force F B =BIw upward. Again, this is consistent with Faraday s (and Lenz s) aw - any magnetic force on the oop must act in such a way to reduce the rate at which the fux changes, which in this case ceary means

2 sowing down the oop. The upward force on the oop wi serve to counteract the gravitationa force, which is utimatey responsibe for the fux change in this case anyway. The faster the oop fas, the arger the upward force it experiences, and at some point the magnetic force wi baance the gravitationa force perfecty, eading to no net acceeration, and hence constant veocity. This is the termina veocity. Of course, once the whoe oop is inside the magnetic fied, the fux is again constant, and the oop just starts to fa normay again. i Quantitativey, we must first find the induced votage around the oop, which wi give us the current. The current wi give us the force, which wi finay give us the acceeration. As the oop fas into the magnetic fied, at some instant t we wi say that a ength x of the oop has moved into the fied, out of the tota ength. At this time, the tota fux through the oop is then: Φ B = B A = BA = Bwx From the fux, we can easiy find the induced votage from Faraday s aw. V = Φ B t = Bw x t = Bwv Here we made use of the fact that the rate at which the ength of the oop exposed to the magnetic fied changes is simpy the instantaneous veocity, x/ t = v. Once we have the induced votage, given the resistance of the oop R, we know the current via Ohm s aw: I = V R = Bwv R From Lenz s aw we know the current circuates countercockwise. In the right-most segment of the oop, the current is fowing up, and the magnetic fied into the page. The right-hand rue then dictates that the force on this current-carrying segment must be to the eft. The eft-most segment of the oop has a force equa in magnitude, since the current I, the ength of wire, and the magnetic fied are the same, but the force is in the opposite direction. Thus, taken together, the eft and right segments of the oop contribute no net force. The bottom segment, however, experiences an upward force, since the current is to the right. For a constant magnetic fied and constant current (true at east instantaneousy), the force is easiy found: F B = BIw i We woud sti have eddy currents, which woud provide some retarding force, but for thin wires eddy current forces are probaby going to be negigibe. This is basicay what we demonstrated with our conducting penduums swinging through a magnetic fied. The penduums that had ony thin segments of conductor (it ooked ike a fork) experienced very itte damping compared to a pain fat pate.

3 We can substitute our expression for I above: F B = BIw = B2 w 2 v R At the termina veocity v T, this upward force wi exacty baance the downward gravitationa force: B 2 w 2 v T F = mg R = v T = mgr B 2 w 2 = 0 2. An ocean current fows at a speed of 1.4 m/s in a region where the vertica component of the earth s magnetic fied is T. The resistivity of seawater in that region is about ρ=0.25 Ω m. On the assumption that there is no other horizonta component of E other than the motiona term v B, find the horizonta current density J in A/m 2? NB reca the genera version of Ohm s aw, viz. E = ρj. If you carried a botte of seawater through the earth s fied at this speed, woud such a current be fowing in it? Ions in the seawater in motion experience a magnetic force qvb, which wi separate positive and negative ions. This resuts in an eectric force qe. In equiibrium, the two wi baance, giving E =vb. Using Ohm s aw, J =E/ρ=vB/ρ. More formay, et the ocean current be aong the ŷ direction and the magnetic fied aong the ẑ direction. The eectric fied in the frame of the moving ions is then E = v B = vb ˆx (1) The current density is then given by Ohm s aw: J = 1 ρ E = vb ρ ˆx A/m 2 (2) 3. Consider an MRI (magnetic resonance imaging) magnet that produces a magnetic fied B =1.5 T at a current of I = 140 A. Assume the magnet is a soenoid with a radius of 0.30 m and a ength of 2.0 m. (a) What is the number of turns of the soenoid? (b) What is its inductance? (c) How much energy is stored in this magnet? (d) If a the energy in part (b) were converted to the kinetic energy of a car (m=1000 kg), what woud the speed of the car be?

4 We know the fied B, current I, and ength. For a soenoid as such with N turns of wire, we must have: ( ) N B = µ o I (3) N = B 17, 000 µ o I (4) The inductance for a soenoid of cross sectiona area A = πr 2 (given the radius r) can be found in your textbook: L = µ on 2 A = µ oπr 2 N 2 51 H (5) The stored energy is U = 1 2 LI J. If this energy were converted to kinetic energy 1 2 mv2 of a car with mass 1000 kg, we have U = 1 2 mv2 = 1 2 LI2, and the speed woud be v = 2U/m = I L/m 32 m/s. 4. Very arge magnetic fieds can be produced using a procedure caed fux compression. A metaic cyindrica tube of radius R is paced coaxiay in a ong soenoid of somewhat arger radius. The space between the tube and the soenoid is fied with a highy exposive materia. When the exposive is set off, it coapses the tube to a cyinder of radius r <R. If the coapse happens very rapidy, induced current in the tube maintains the magnetic fux neary constant inside the tube, even though the area shrinks. If the initia magnetic fied in the soenoid is 2.50 T, and R/r =12.0, what is the maximum fied that can be reached? The basic idea here is that the fux through the tube is the same before and after the exposion. Since after the exposion the cross-sectiona area is severey reduced, the fied must be much arger in order to make the fux the same. expaining things in more detai: Here s a bit from the Wikipedia about fux compression, Magneto-exposive generators use a technique caed magnetic fux compression, which wi be described in detai ater. The technique is made possibe when the time scaes over which the device operates are sufficienty brief that resistive current oss is negigibe, and the magnetic fux on any surface surrounded by a conductor (copper wire, for exampe) remains constant, even though the size and shape of the surface may change. This fux conservation can be demonstrated from Maxwe s equations. The most intuitive expanation of this conservation of encosed fux foows from the principe that any change in an eectromagnetic system provokes an effect in order to oppose the change. For this reason, reducing the area of the surface encosed by a conductor, which woud reduce the magnetic fux, resuts in the induction of current in the eectrica conductor, which tends to return the encosed fux to its origina vaue. In magneto-exposive generators, this phenomenon is obtained by various techniques which depend on powerfu exposives. The compression process aows the chemica energy of the exposives to be (partiay) transformed into the energy of an intense

5 magnetic fied surrounded by a correspondingy arge eectric current. Wikipedia, Fux Compression So: a we need to do is cacuate the fux before and after the exposion, set them equa to each other, and sove for the fied after the exposion. Quantities with i subscripts refer to before the exposion, those with f after the exposion, and a symbos have their usua meanings. Φ B,i = B i A i = B i πr 2 Φ B,f = B f A f = B f πr 2 Φ B,i = Φ B,f = B i πr 2 = B f πr 2 ( ) R 2 B f = B i = (12.0) T = 360 T r 5. In a mass spectrometer, a beam of ions is first made to pass through a veocity seector with perpendicuar E and B fieds. Here, the eectric fied E is to the right, between parae charged pates, and the magnetic fied B in the same region is into the page. The seected ions are then made to enter a region of different magnetic fied B, where they move in arcs of circes. The radii of these circes depend on the masses of the ions. Assume that each ion has a singe charge e. Show that in terms of the given fied vaues and the impact distance the mass of the ion is m = ebb 2E Bin E Bin ion

6 Standard mass spectrometer probem from the text/notes/ecture, just be sure to use the new fied B in the region with no eectric fied, and note that =2r.