Em research
Upcoming SlideShare
Loading in...5

Like this? Share it with your network


Em research






Total Views
Views on SlideShare
Embed Views



0 Embeds 0

No embeds



Upload Details

Uploaded via as Adobe PDF

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
Post Comment
Edit your comment

Em research Document Transcript

  • 1. Research-inspired problems for electricity and magnetism Beth Parksa͒ Physics Department, Colgate University, Hamilton, New York 13346 ͑Received 17 August 2005; accepted 16 December 2005͒ I present several problems related to current research problems and techniques that are suitable for introductory and intermediate electricity and magnetism. © 2006 American Association of Physics Teachers. ͓DOI: 10.1119/1.2166369͔ I. INTRODUCTION Electricity and magnetism is a well-established field, so it is difficult to inspire students with problems from current research topics. One reason that the subject is still an impor- tant part of the curriculum is the frequency with which we need to solve basic problems in electricity and magnetism as part of our research. Incorporating some of these real-life problems into courses can help motivate students to learn electricity and magnetism. In this article I will share several of these problems, sketches of their solutions, and some background informa- tion. Some of the problems are suitable for an introductory course in electricity and magnetism, and others belong in a second course. I have not included problems for use in an advanced course because at that level the research applica- tions are much more numerous. II. INTERACTIONS BETWEEN ELECTRON SPINS The magnetic dipole-dipole interaction is almost always negligible between electrons in solids in comparison to the much larger magnetic couplings caused by Coulomb ex- change interactions. However, in the layered material Sr2Cu3O4Cl2, the symmetry of the crystal lattice completely frustrates the exchange interactions, and a weak ferromag- netism is caused by magnetic dipole-dipole interactions. Figure 1 shows the structure of the Cu3O4 planes. The black circles are copper ions and the white circles are oxygen ions. Copper ions have one unpaired electron, giving the ion a net spin of 1 2. ͑Do not confuse this spin with the nuclear spin, which produces a much smaller magnetic moment that is irrelevant to this problem.͒ The spins of the copper ions at the corners of the squares are strongly antiferromagnetically coupled to their neighbors and align as shown. ͑This align- ment is caused by Coulomb exchange interactions, as is the case for almost every magnetic material.͒ The distance be- tween the antiferromagnetically coupled spins is a =0.386 nm. The planes are separated from each other by 0.626 nm. Therefore, the density of the center copper atoms is one per 0.187 nm3 . Problem 1. The copper ions at the centers of the squares ͑labeled II͒ interact ferromagnetically with each copper ion at the corners ͑labeled I͒. ͑Do not confuse the labels I and II with the charge of the ions, which are all +2.͒ Discuss why these interactions do not result in magnetic ordering of the center coppers. ͑Hint: Along what direction would the copper II spin point due to interactions with each of the corner cop- per atoms individually?͒ Problem 2. Because the stronger exchange interactions are ineffective, consider the effect of dipole-dipole interactions. In what direction does this interaction try to align the spins on the two copper ͑II͒ sites shown? Consider only interac- tions with the four nearest neighbors, because all other inter- actions will be much weaker. It would help to sketch the field lines due to the four neighbors. Problem 3. What is the strength of the interaction; that is, what is the energy difference between the highest and lowest energy alignments of the electrons at the centers of the squares? Compare this energy with kBT at 40 K. Problem 4. What would the magnetization M of the mate- rial be at 40 K? ͑Just below 40 K, the material undergoes a phase transition to a different magnetic structure.͒ Recall from statistical mechanics that if the alignment energy ⌬E is much less than kBT, the magnetization is reduced from its maximum value by the ratio ⌬E/6kBT.1 Solutions: Because the ferromagnetic interactions cancel by symmetry, the remaining interaction is the dipole-dipole interaction, which tends to orient the spins in these two cen- ter copper ͑II͒ ions along the −y-direction as defined in Fig. 1. If there were copper ions at the centers of the other two squares, they would be oriented along the +y-direction, and no bulk ferromagnetic moment would be measured. How- ever, because there are copper ions only in the centers of alternate squares, there is a net ferromagnetic moment. The alignment energy is E=m·B, where B is the magnetic field at the center due to the four corner ions. The field due to one of the corner electrons at the location of the center electron is B1 = ␮0m 4␲r3͑2 cos ␪rˆ + sin ␪␪ˆ͒, ͑1͒ where r=a/ͱ2 and ␪=␲/4. Only the component of B1 that lies along the direction of magnetization will contribute; the other component will cancel with the B-field from another copper ion. The result for the total B-field from all four ions is B=3ͱ2␮0m/␲a3 . Therefore the alignment energy is ⌬E =mup·B−mdown·B=2ϫ3ͱ2m2 ␮0/␲a3 =5.1ϫ10−24 J, be- cause the magnetic moment of an electron is ␮B=9.27 ϫ10−24 J/T ͑gϷ2͒. At 40 K, kBT=5.5ϫ10−22 J, so the ther- mal energy is much greater than this alignment energy. In this case, the magnetization will be M = nm⌬E 6kBT = ͱ2m3 n␮0 kBT␲a3 , ͑2͒ where n is the density of spins, 5.4ϫ1027 m−3 . Equation ͑2͒ gives a magnetization of 77 J/T-m3 . I have presented only the simplest explanation of the mag- netic behavior that is consistent with experimental observa- tions; more complicated possibilities also exist. More infor- mation about this material can be found in Ref. 2. 351 351Am. J. Phys. 74 ͑4͒, April 2006 http://aapt.org/ajp © 2006 American Association of Physics Teachers
  • 2. III. SQUID MAGNETOMETER Measurements of the magnetic moment as a function of applied field are essential to the characterization of supercon- ductors, ferromagnets, and antiferromagnets such as Sr2Cu3O4Cl2. A SQUID magnetometer is one of the most sensitive ways to measure the magnetic moment of a small sample. A SQUID ͑superconducting quantum interference device͒ converts a magnetic flux through its pick-up loop into a voltage that can be easily measured. When the sample is held in a magnetic field inside the magnetometer, the sample develops a dipole moment that produces a flux through the SQUID pickup loop. A charac- teristic voltage signal is produced that is proportional to the flux. ͑This phenomenon is very different from induction in a normal pickup loop in which the induced voltage is propor- tional to the rate of change of the flux.͒ The flux through the loop is a function of the distance of the sample from the loop, so as the sample is moved through the pickup loop, the voltage changes as a function of position. The magnetic mo- ment of the sample is calculated by fitting the observed variation in the voltage signal to the predicted dipole shape. Your assignment is to calculate this shape. Problem 5. The geometry is shown in Fig. 2. Assume that the pick-up loop is 2.5 cm in diameter ͑R=1.25 cm͒. The sample moves vertically in a total path of 4 cm. The voltage from the SQUID, which is proportional to the magnetic flux through the loop, is measured as a function of the position of the sample. The sample is much smaller than the loop, so you can approximate it as a perfect dipole. Calculate the shape of the signal observed. Problem 6. The magnetic moment of a 15 mm3 sample of Sr2Cu3O4Cl2 at 40 K is easily detectable. Compare the flux through the pick-up loop due to this sample to the flux of the Earth’s field. Solutions: This problem can be solved numerically or ana- lytically. In the most straightforward analytical approach, the field of the dipole is expressed in cylindrical coordinates. This expression is found most easily by starting with the vector potential of a dipole in cylindrical coordinates and taking its curl to find B. ͑It is only necessary to consider the z-component of B.͒ The result is Bz = ␮0m 4␲ 2z2 − s2 ͑s2 + z2 ͒5/2 , ͑3͒ where s is the distance from the z axis to the point at which the B-field is measured and z is the displacement of the sample from the center of the loop along the z axis. Equation ͑3͒ for Bz can be integrated over the area of the pick-up loop to find the flux as a function of the distance z between the sample and the pick-up loop: ⌽͑z͒ = ␮0m 2 R2 ͑z2 + R2 ͒3/2 , ͑4͒ where R is the radius of the pick-up loop. The flux as a function of sample position displays a characteristic peak at z=0. The problem is a bit simpler if we realize that because the divergence of B is zero, the flux is the same through the loop whether we choose the flat surface or the surface of the sphere centered on the sample. In the latter case, we can use the expression for the radial part of the B-field in spherical coordinates: Br = ␮0m 4␲r32 cos ␪. ͑5͒ Here r2 =R2 +z2 . This result for Br can easily be integrated over the area of the loop, letting ␪ range from 0 to arctan͑R/z͒, which yields the same result as before. We found that the magnetization for Sr2Cu3O4Cl2 should be 77 J/T-m3 . By using the volume given, we find that m =1.2ϫ10−6 J/T, and the maximum flux through a loop with R=0.0125 m is ␮0m/2R=7.5ϫ10−13 T-m2 . The Earth’s field is roughly 5ϫ10−5 T, so its flux through the loop is 2.5ϫ10−8 T-m2 . This difference tells us that SQUID magne- tometers must be designed very carefully to be sensitive to small changes in a large background flux. Advanced students might be asked what would be ob- served if, for some reason, the sample became magnetically polarized perpendicular to the applied magnetic field. Their first answer will ͑hopefully͒ be that the SQUID signal would be zero due to symmetry. However, you might remind them that symmetry is never perfect in real experiments; the sample will not be exactly centered in the pick-up loop. They can then numerically calculate the SQUID signal for this geometry. The resulting antisymmetric SQUID signal was the first indication that I had that something very strange was going on in the magnetism of Sr2Cu3O4Cl2; more details are in Ref. 3. IV. SUPERCONDUCTING MAGNETS Problem 7. Superconducting magnets are fairly common laboratory tools, and their strong magnetic fields sometimes have unintended effects on other experiments in the physics Fig. 1. Cu–O plane of Sr2Cu3O4Cl2. The black circles are copper ions and the white circles are oxygen ions. The two different Cu sites are labeled by Roman numerals; these labels should not be confused with the charge of the copper ion, which is +2 for both sites. Fig. 2. The SQUID produces a voltage that is proportional to the magnetic flux through the pick-up loop as the sample moves through the loop. 352 352Am. J. Phys., Vol. 74, No. 4, April 2006 Beth Parks
  • 3. building. If a superconducting magnet is roughly a solenoid 8 in. long and 3 in. in diameter with a 1 in. bore and a field of 6 T in the center, how far away from the solenoid along its axis must you be to have the magnetic field be compa- rable to that of the Earth? You may consider the current density to be uniform and in the azimuthal direction inside the solenoid. Solution: This geometry is the superposition of loops of current. Each current loop creates a field along its axis, B͑z͒ = ␮0I 2 r2 ͑r2 + z2 ͒3/2 , ͑6͒ where r is the radius of the loop and z is the distance from its center. Students should be able to set I=Jdrdz and integrate to find the field. I have not given the current density J, so students need to determine what current density would create this field in the center. They can find it by forcing the B-field to be 6 T in the center: Bcenter = 6T = ͵−L/2 L/2 ͵R1 R2 ␮0J 2 r2 ͑r2 + z2 ͒3/2 dr dz, ͑7͒ where L=8 in. is the length of the magnet, R1=0.5 in. is the inner radius, and R2=1.5 in. is the outer radius. The integral is evaluated more easily if z is integrated first. For the di- mensions given, the current density is J=1.97ϫ108 A/m2 . ͑Ask the students to compare this density to what could safely go through a copper wire.͒ Using this current density, they can find the B-field as a function of distance d from the magnet: Bd = ͵−L/2 L/2 ͵R1 R2 ␮0J 2 r2 dr dz ͑r2 + ͑d + z͒2 ͒3/2 . ͑8͒ This integral can be done, but the result is not pretty, and the equation cannot be solved for d analytically. However, at this point the student should think like a physicist. We are not looking for the exact distance; rather, we’re trying to deter- mine how far away we must go before the B-field of the magnet doesn’t matter. At this distance, it is likely that d ӷr,z. In this case, the integrand can be simplified: Bd Ϸ ͵−L/2 L/2 ͵R1 R2 ␮0J 2 r2 dr dz d3 = ␮0JL 2 R2 3 − R1 3 3d3 . ͑9͒ The result, d=2.1 m, is consistent with our initial assumption that dӷR1,R2,L and is accurate to within 0.15%. In the spirit of making reasonable approximations, a stu- dent could come up with an answer much more easily by assuming that the solenoid can be approximated by a current loop with a diameter of R=2 in. The ratio of the field at the center of the loop to the field a distance d away is ͑R2 +d2 ͒3/2 /R3 . This approach gives the result that the field will equal that of the Earth at a distance of 1.3 m, which is not as close as our previous result, but is good enough to be useful in a lab. V. MAGNETS AND OPTICAL TABLES Problem 8. In magneto-optical experiments it is necessary to place a superconducting magnet on an optical table. These tables are typically made of ferromagnetic steel. They can be made of noninductive stainless steel, but because this mate- rial is more expensive, it is avoided when possible. I recently purchased a superconducting magnet with a field of 7 T at the center of its bore. I planned to locate it on a noninductive table that was rigidly joined to an ordinary steel table. The magnet’s specifications showed that the field should be 0.01 T at a distance of 93 cm along the axis, which is roughly the distance from the magnet center to the nearest point on the steel table. According to my calcula- tions, the force between the magnet and the cryostat should be less than 500 N, or 20 lb in this location, which is much less than the weight of the magnet. However, when I con- sulted with the engineers at the cryostat company, I was told that the force between the magnet and a 2000 lb steel table would be 3.4ϫ106 lb. How did each of us arrive at our an- swers and who was right? Solution: An upper limit to the force is found by assuming that the table is an infinite plane of infinitely permeable ma- terial. This assumption makes the problem analogous to an electric dipole held above an infinite conductor: the force between them is the same as between a dipole and its mirror image. The main remaining difficulty is to find the size of the dipole. We can use the fact that the field is 0.01 T at a dis- tance of 0.93 m. This distance is certainly far enough that the dipole approximation should be valid. We find m=4.0 ϫ104 J/T. Then the force on the magnet ͑or on the table͒ is equal to the force on this dipole separated by ͑2ϫ0.93͒ m from an identical dipole. The B-field along the axis is B͑r͒ =␮0m/2␲r3 , so the force is F=ٌ͑m·B͒=m·ٌB=m‫ץ‬B/‫ץ‬r. This force is 81 N, or 18 lb. This value is a high upper bound on the actual force, because the table is not infinitely perme- able and does not occupy the entire half-plane. The approach probably taken by the magnet company was to assume that the table reaches its saturation magnetization due to the influence of the magnet. According to the CRC Handbook, steel has a saturation magnetization of around 20 000 G, which is reached at an applied field of about 100 Oe=0.01 T/␮0, roughly equal to the field at the closest point to the cryostat. Because H is negligible, the magneti- zation M is approximately B/␮0=1.6ϫ106 A/m. We can find the magnetic moment m using the mass ͑2000 lb͒ and the density ͑7800 kg/m3 ͒: m=1.9ϫ105 J/T. The force on this dipole is calculated in the same way as before, but at a distance of 0.93 m. The result of 6000 lb is still much lower than the magnet company’s estimate. I’m not sure exactly what happened in that calculation—perhaps the engineers actually used the maximum magnetic suscep- tibility of steel to overestimate the resulting magnetic moment—but I am happy to use their result in class as an example of how important it is to ensure that your answers are reasonable. If students see this second approach, they may have a difficult time deciding why it is incorrect. The error is in the assumption that the steel is fully magnetized. One way to understand it is to realize that although a small piece of steel might become fully magnetized when placed in a region with H=0.01 T/␮0, a large piece of steel will not be fully mag- netized in the same H-field. This point can be understood in terms of demagnetizing fields, but I prefer to consider an energy argument. The energy gain due to aligning with the field is M·B. The energy cost is the extra field energy cre- ated, and because the field is proportional to M, the energy is proportional to M2 . For a small M, the alignment energy is 353 353Am. J. Phys., Vol. 74, No. 4, April 2006 Beth Parks
  • 4. larger, but as M increases, the cost of creating the field in- creases. So for a sample as large as an optical table, the cost of full alignment is prohibitive. VI. CARBON NANOTUBES Carbon nanotubes are almost one-dimensional carbon structures with electrical properties that can range from semi- conducting to nearly ballistic conductance. They are objects of intense research, due both to fundamental physical interest in their one-dimensional properties and to their possible technological applications as nanoscale semiconductors and conductors. Problem 9. To measure its properties, the density of elec- trons on the nanotube is tuned by applying a voltage differ- ence between the nanotube and a ground plane, which can be approximated as an infinite conducting plane. ͑The nanotube actually rests on an insulating layer that separates it from the ground plane.͒ The density of electrons is determined by the capacitance between the nanotube and the ground plane. Cal- culate this capacitance per unit length of the nanotube. Solution: This problem can be mapped onto a fairly stan- dard problem: finding the potential due to two infinite paral- lel lines carrying opposite charge densities. It can be shown that for parallel line charges, the equipotential lines are cylinders.1 This demonstration is straightforward but not easy and should probably be given before this problem is assigned. Alternatively, the result could be provided to the students. Once that problem is solved, the difficulty of mapping the problem should not be underestimated; it was nontrivial for some of my best students. The first step is to recognize that the nanotube surface will be ͑approximately͒ an equipoten- tial, and therefore the line charges can be replaced by nano- tubes that are placed on the equipotential surfaces. ͑By using Gauss’s law, we see that the charge on the nanotube must be the same as the charge on the line.͒ The second step is to realize that the problem of two parallel nanotubes with op- posite potentials is equivalent ͑due to image charges͒ to the problem of a nanotube suspended above a conducting sur- face. Finally, it is necessary to figure out how to manipulate the results from the parallel line problem to obtain a capacitance. For lines with charge density ␭ separated by distance 2a, Griffiths1 finds that the surface with potential V0 is a circle with radius r = a cschͩ2␲⑀0V0 ␭ ͪ, ͑10͒ and its center a distance above the plane is given by y = a cothͩ2␲⑀0V0 ␭ ͪ. ͑11͒ Because a is irrelevant to the nanotube problem ͑the cylinder of constant voltage is not centered on the line charge͒, the first step is to take the ratio of these two equations to elimi- nate a. Then we can solve to find the capacitance per unit length: C/L = ␭/V0 = 2␲⑀0 cosh−1 ͑y/r͒ . ͑12͒ Typically the nanotube might have a diameter of 1.4 nm and lie on an insulating silicon oxide layer of thickness 500 nm, yielding a capacitance per unit length of 8.5 ϫ10−12 C/m. A more exact treatment of this problem would also include the dielectric constant of the insulating layer, but that correction is beyond the scope of most intermediate- level courses. Another correction is the quantum capacitance of the nanotube, that is, the additional energy an electron must have in order for it to find an unoccupied electronic state in the nanotube. This quantum capacitance turns out to be comparable in size to the geometrical capacitance calcu- lated here.4 a͒ Electronic address: meparks@mail.colgate.edu 1 David J. Griffiths, Introduction to Electrodynamics ͑Prentice Hall, Engle- wood Cliffs, NJ, 1999͒, 3rd ed., The magnetization can be found from the alignment energy using a straightforward extension of Problem 4.40 on pp. 200–201. The potential of parallel wires is considered in Problem 2.47 on p. 107. 2 F. C. Chou, Amnon Aharony, R. J. Birgeneau, O. Entin-Wohlman, M. Greven, A. B. Harris, M. A. Kastner, Y. J. Kim, D. S. Kleinberg, Y. S. Lee, and Q. Zhu, “Ferromagnetic moment and spin rotation transitions in tetragonal antiferromagnetic Sr2Cu3O4Cl2,” Phys. Rev. Lett. 78͑3͒, 535– 538 ͑1997͒. 3 Beth Parks, M. A. Kastner, Y. J. Kim, A. B. Harris, F. C. Chou, O. Entin-Wohlman, and Amnon Aharony, “Magnetization measurements of antiferromagnetic domains in Sr2Cu3O4Cl2,” Phys. Rev. B 63͑13͒, 134433-1–10 ͑2001͒. 4 Peter J. Burke, “An rf circuit model for carbon nanotubes,” IEEE Trans. Nanotechnol. 2͑1͒, 55–58 ͑2003͒. 354 354Am. J. Phys., Vol. 74, No. 4, April 2006 Beth Parks