1. Calculate the torque exerted on the square loop shown in Fig.
6.6, due to the circular loop (assume r is much larger than a or b). If
the square loop is free to rotate, what will its equilibrium orientation
be? ... Get solution
2. Starting from the Lorentz force law, in the form of Eq. 5.16, show that the torque on any steady current distribution (not just a square loop) in a uniform field B is m × B. Get solution
3. Find the force of attraction between two magnetic dipoles, m1 and m2, oriented as shown in Fig. 6.7, a distance r apart, (a) using Eq. 6.2, and (b) using Eq. 6.3. ... Eq. 6.2. ... Eq.6.3 ... Get solution
4. Derive Eq. 6.3. [Here’s one way to do it: Assume the dipole is an infinitesimal square, of side _ (if it’s not, chop it up into squares, and apply the argument to each one). Choose axes as shown in Fig. 6.8, and calculate ... along each of the four sides. Expand B in a Taylor series—on the right side, for instance, ... For a more sophisticated method, see Prob. 6.22.] ... Reference prob 6.22 In Prob. 6.4, you calculated the force on a dipole by “brute force.” Here’s a more elegant approach. First write B(r) as a Taylor expansion about the center of the loop: ... Derive Eq. 6.3. [Here’s one way to do it: Assume the dipole is an infinitesimal square, of side _ (if it’s not, chop it up into squares, and apply the argument to each one). Choose axes as shown in Fig. 6.8, and calculate ... along each of the four sides. Expand B in a Taylor series—on the right side, for instance, ... For a more sophisticated method, see Prob. 6.22.] ... Reference prob 6.22 In Prob. 6.4, you calculated the force on a dipole by “brute force.” Here’s a more elegant approach. First write B(r) as a Taylor expansion about the center of the loop: ... Get solution
5. A uniform current density ... fills a slab straddling the yz plane, from x = −a to x = +a. A magnetic dipole ...is situated at the origin. (a) Find the force on the dipole, using Eq. 6.3. (b) Do the same for a dipole pointing in the y direction ... (c) In the electrostatic case, the expressions F = ∇(p · E) and F = (p · ∇)E are equivalent (prove it), but this is not the case for the magnetic analogs (explain why). As an example, calculate (m · ∇)B for the configurations in (a) and (b). Equation 6.3 ... Get solution
6. Of the following materials, which would you expect to be paramagnetic and which diamagnetic: aluminum, copper, copper chloride (CuCl2), carbon, lead, nitrogen (N2), salt (NaCl), sodium, sulfur, water? (Actually, copper is slightly diamagnetic; otherwise they’re all what you’d expect.) Get solution
7. An infinitely long circular cylinder carries a uniform magnetization M parallel to its axis. Find the magnetic field (due to M) inside and outside the cylinder. Get solution
8. A long circular cylinder of radius R carries a magnetization M = ... where k is a constant, s is the distance from the axis, and ... is the usual azimuthal unit vector (Fig. 6.13). Find the magnetic field due toM, for points inside and outside the cylinder. ... Get solution
9. A short circular cylinder of radius a and length L carries a “frozen-in” uniform magnetizationMparallel to its axis. Find the bound current, and sketch the magnetic field of the cylinder. (Make three sketches: one for L ≫ a, one for L≫ a, and one for L ≈ a.) Compare this bar magnet with the bar electret of Prob. 4.11. Get solution
10. An iron rod of length L and square cross section (side a) is given a uniform longitudinal magnetization M, and then bent around into a circle with a narrow gap (width w), as shown in Fig. 6.14. Find the magnetic field at the center of the gap, assuming w ≫ a≫ L. [Hint: treat it as the superposition of a complete torus plus a square loop with reversed current.] ... Get solution
11. In Sect, 6.2.1, we began with the potential of a perfect dipole (Eq. 6.10), whereas in fact we are dealing with physical dipoles. Show, by the method of Sect. 4.2.3, that we nonetheless get the correct macroscopic field. Get solution
12. An infinitely long cylinder, of radius R, carries a “frozen-in” magnetization, parallel to the axis, ... where k is a constant and s is the distance from the axis; there is no free current anywhere. Find the magnetic field inside and outside the cylinder by two different methods: (a) As in Sect. 6.2, locate all the bound currents, and calculate the field they produce. (b) Use Ampère’s law (in the form of Eq. 6.20) to find H, and then get B from Eq. 6.18. (Notice that the second method is much faster, and avoids any explicit reference to the bound currents.) Equation 6.18 ... Equation 6. 20 ... Get solution
13. Suppose the field inside a large piece of magnetic material is B0, so that ... where M is a “frozen-in” magnetization. (a) Now a small spherical cavity is hollowed out of the material (Fig. 6.21). Find the field at the center of the cavity, in terms of B0 and M. Also find H at the center of the cavity, in terms of H0 and M. (b) Do the same for a long needle-shaped cavity running parallel to M. (c) Do the same for a thin wafer-shaped cavity perpendicular to M. Figure6.21 ... Get solution
14. For the bar magnet of Prob. 6.9, make careful sketches of M, B, and H, assuming L is about 2a. Compare Prob. 4.17. Reference 4.17For the bar electret of Prob. 4.11, make three careful sketches: one of P, one of E, and one of D. Assume L is about 2a. [Hint: E lines terminate on charges; D lines terminate on free charges.] Reference prob 4.11 A short cylinder, of radius a and length L, carries a “frozen-in” uniform polarization P, parallel to its axis. Find the bound charge, and sketch the electric field (i) for L ≫ a, (ii) for L ≫ a, and (iii) for L ≈ a. [This is known as a bar electret; it is the electrical analog to a bar magnet. In practice, only very special materials—barium titanate is the most “familiar” example—will hold a permanent electric polarization. That’s why you can’t buy electrets at the toy store.] Reference 6.9 A short circular cylinder of radius a and length L carries a “frozen-in” uniform magnetizationMparallel to its axis. Find the bound current, and sketch the magnetic field of the cylinder. (Make three sketches: one for L ≫ a, one for L≫ a, and one for L ≈ a.) Compare this bar magnet with the bar electret of Prob. 4.11. Get solution
15. If J f = 0 everywhere, the curl of H vanishes (Eq. 6.19), and we can express H as the gradient of a scalar potential W: ... According to Eq. 6.23, then, ... so W obeys Poisson’s equation, with ∇ ·M as the “source.” This opens up all the machinery of Chapter 3. As an example, find the field inside a uniformly magnetized sphere (Ex. 6.1) by separation of variables. [Hint: ∇ ·M = 0 everywhere except at the surface (r = R), so W satisfies Laplace’s equation in the regions r R; use Eq. 3.65, and from Eq. 6.24 figure out the appropriate boundary condition on W.] reference 3.65 ... Reference 6.24 ... Get solution
16. A coaxial cable consists of two very long cylindrical tubes, separated by linear insulating material of magnetic susceptibility χm. A current I flows down the inner conductor and returns along the outer one; in each case, the current distributes itself uniformly over the surface (Fig. 6.24). Find the magnetic field in the region between the tubes. As a check, calculate the magnetization and the bound currents, and confirm that (together, of course, with the free currents) they generate the correct field. ... Get solution
17. A current I flows down a long straight wire of radius a. If the wire is made of linear material (copper, say, or aluminum) with susceptibility χm, and the current is distributed uniformly, what is the magnetic field a distance s from the axis? Find all the bound currents. What is the net bound current flowing down the wire? Get solution
18. A sphere of linear magnetic material is placed in an otherwise uniform magnetic field B0. Find the new field inside the sphere. [Hint: See Prob. 6.15 or Prob. 4.23.] Reference prob 6.15 If J f = 0 everywhere, the curl of H vanishes (Eq. 6.19), and we can express H as the gradient of a scalar potential W: ... According to Eq. 6.23, then, ... Get solution
19. On the basis of the naïve model presented in Sect. 6.1.3, estimate the magnetic susceptibility of a diamagnetic metal such as copper. Compare your answer with the empirical value in Table 6.1, and comment on any discrepancy. Get solution
20. How would you go about demagnetizing a permanent magnet (such as the wrench we have been discussing, at point c in the hysteresis loop)? That is, how could you restore it to its original state, with M = 0 at I = 0? Get solution
21. (a) Show that the energy of a magnetic dipole in a magnetic field B is ... Assume that the magnitude of the dipole moment is fixed, and all you have to do is move it into place and rotate it into its final orientation. The energy required to keep the current flowing is a different problem, which we will confront in Chapter 7.] Compare Eq. 4.6. Figure 6.30 ... (b) Show that the interaction energy of two magnetic dipoles separated by a displacement r is given by ... Compare Eq. 4.7. (c) Express your answer to (b) in terms of the angles θ1 and θ2 in Fig. 6.30, and use the result to find the stable configuration two dipoles would adopt if held a fixed distance apart, but left free to rotate. (d) Suppose you had a large collection of compass needles, mounted on pins at regular intervals along a straight line. How would they point (assuming the earth’s magnetic field can be neglected)? [A rectangular array of compass needles aligns itself spontaneously, and this is sometimes used as a demonstration of “ferromagnetic” behavior on a large scale. It’s a bit of a fraud, however, since the mechanism here is purely classical, and much weaker than the quantum mechanical exchange forces that are actually responsible for ferromagnetism.13] Get solution
22. In Prob. 6.4, you calculated the force on a dipole by “brute force.” Here’s a more elegant approach. First write B(r) as a Taylor expansion about the center of the loop: ... where r0 is the position of the dipole and ∇0 denotes differentiation with respect to r0. Put this into the Lorentz force law (Eq. 5.16) to obtain ... Or, numbering the Cartesian coordinates from 1 to 3: ... where δi j is the Kronecker delta (Prob. 3.52). Get solution
23. A familiar toy consists of donut-shaped permanent magnets (magnetization parallel to the axis), which slide frictionlessly on a vertical rod (Fig. 6.31). Treat the magnets as dipoles, with mass md and dipole moment m. (a) If you put two back-to-back magnets on the rod, the upper one will “float”—the magnetic force upward balancing the gravitational force downward. At what height (z) does it float? (b) If you now add a third magnet (parallel to the bottom one), what is the ratio of the two heights? (Determine the actual number, to three significant digits.) ... Get solution
24. Imagine two charged magnetic dipoles (charge q, dipole moment m), constrained to move on the z axis (same as Problem 6.23(a), but without gravity). Electrically they repel, but magnetically (if both m’s point in the z direction) they attract. (a) Find the equilibrium separation distance. (b) What is the equilibrium separation for two electrons in this orientation. (c) Does there exist, then, a stable bound state of two electrons? Get solution
25. Notice the following parallel: ... Thus, the transcription D→B, E→H, P→ μ0M, ϵ 0 → μ0 turns an electrostatic problem into an analogous magnetostatic one. Use this, together with your knowledge of the electrostatic results, to rederive (a) the magnetic field inside a uniformly magnetized sphere (Eq. 6.16); (b) the magnetic field inside a sphere of linear magnetic material in an otherwise uniform magnetic field (Prob. 6.18); (c) the average magnetic field over a sphere, due to steady currents within the sphere (Eq. 5.93). Get solution
26. Compare Eqs. 2.15, 4.9, and 6.11. Notice that if ρ, P, and M are uniform, the same integral is involved in all three: ... Therefore, if you happen to know the electric field of a uniformly charged object, you can immediately write down the scalar potential of a uniformly polarized object, and the vector potential of a uniformly magnetized object, of the same shape. Use this observation to obtain V inside and outside a uniformly polarized sphere (Ex. 4.2), and A inside and outside a uniformly magnetized sphere (Ex. 6.1). Equation 6.11 ... Get solution
27. At the interface between one linear magnetic material and another, the magnetic field lines bend (Fig. 6.32). Show that tan θ2/ tan θ1 = μ2/μ1, assuming there is no free current at the boundary. Compare Eq. 4.68. Equation 4.68 ... Figure 6.32 ... Get solution
28. A magnetic dipole m is imbedded at the center of a sphere (radius R) of linear magnetic material (permeability μ). Show that the magnetic field inside the sphere (0 r ≤ R) is ... What is the field outside the sphere? Get solution
29. You are asked to referee a grant application, which proposes to determine whether the magnetization of iron is due to “Ampère” dipoles (current loops) or “Gilbert” dipoles (separated magnetic monopoles). The experiment will involve a cylinder of iron (radius R and length L = 10R), uniformly magnetized along the direction of its axis. If the dipoles are Ampère-type, the magnetization is equivalent to a surface bound current ... if they are Gilbert-type, the magnetization is equivalent to surface monopole densities σb = ±M at the two ends. Unfortunately, these two configurations produce identical magnetic fields, at exterior points. However, the interior fields are radically different—in the first case B is in the same general direction as M, whereas in the second it is roughly opposite to M. The applicant proposes to measure this internal field by carving out a small cavity and finding the torque on a tiny compass needle placed inside. Assuming that the obvious technical difficulties can be overcome, and that the question itself is worthy of study, would you advise funding this experiment? If so, what shape cavity would you recommend? If not, what is wrong with the proposal? [Hint: Refer to Probs. 4.11, 4.16, 6.9, and 6.13.] Reference prob 4.11 A short cylinder, of radius a and length L, carries a “frozen-in” uniform polarization P, parallel to its axis. Find the bound charge, and sketch the electric field (i) for L ≫ a, (ii) for L ≫ a, and (iii) for L ≈ a. [This is known as a bar electret; it is the electrical analog to a bar magnet. In practice, only very special materials—barium titanate is the most “familiar” example—will hold a permanent electric polarization. That’s why you can’t buy electrets at the toy store.] Get solution
2. Starting from the Lorentz force law, in the form of Eq. 5.16, show that the torque on any steady current distribution (not just a square loop) in a uniform field B is m × B. Get solution
3. Find the force of attraction between two magnetic dipoles, m1 and m2, oriented as shown in Fig. 6.7, a distance r apart, (a) using Eq. 6.2, and (b) using Eq. 6.3. ... Eq. 6.2. ... Eq.6.3 ... Get solution
4. Derive Eq. 6.3. [Here’s one way to do it: Assume the dipole is an infinitesimal square, of side _ (if it’s not, chop it up into squares, and apply the argument to each one). Choose axes as shown in Fig. 6.8, and calculate ... along each of the four sides. Expand B in a Taylor series—on the right side, for instance, ... For a more sophisticated method, see Prob. 6.22.] ... Reference prob 6.22 In Prob. 6.4, you calculated the force on a dipole by “brute force.” Here’s a more elegant approach. First write B(r) as a Taylor expansion about the center of the loop: ... Derive Eq. 6.3. [Here’s one way to do it: Assume the dipole is an infinitesimal square, of side _ (if it’s not, chop it up into squares, and apply the argument to each one). Choose axes as shown in Fig. 6.8, and calculate ... along each of the four sides. Expand B in a Taylor series—on the right side, for instance, ... For a more sophisticated method, see Prob. 6.22.] ... Reference prob 6.22 In Prob. 6.4, you calculated the force on a dipole by “brute force.” Here’s a more elegant approach. First write B(r) as a Taylor expansion about the center of the loop: ... Get solution
5. A uniform current density ... fills a slab straddling the yz plane, from x = −a to x = +a. A magnetic dipole ...is situated at the origin. (a) Find the force on the dipole, using Eq. 6.3. (b) Do the same for a dipole pointing in the y direction ... (c) In the electrostatic case, the expressions F = ∇(p · E) and F = (p · ∇)E are equivalent (prove it), but this is not the case for the magnetic analogs (explain why). As an example, calculate (m · ∇)B for the configurations in (a) and (b). Equation 6.3 ... Get solution
6. Of the following materials, which would you expect to be paramagnetic and which diamagnetic: aluminum, copper, copper chloride (CuCl2), carbon, lead, nitrogen (N2), salt (NaCl), sodium, sulfur, water? (Actually, copper is slightly diamagnetic; otherwise they’re all what you’d expect.) Get solution
7. An infinitely long circular cylinder carries a uniform magnetization M parallel to its axis. Find the magnetic field (due to M) inside and outside the cylinder. Get solution
8. A long circular cylinder of radius R carries a magnetization M = ... where k is a constant, s is the distance from the axis, and ... is the usual azimuthal unit vector (Fig. 6.13). Find the magnetic field due toM, for points inside and outside the cylinder. ... Get solution
9. A short circular cylinder of radius a and length L carries a “frozen-in” uniform magnetizationMparallel to its axis. Find the bound current, and sketch the magnetic field of the cylinder. (Make three sketches: one for L ≫ a, one for L≫ a, and one for L ≈ a.) Compare this bar magnet with the bar electret of Prob. 4.11. Get solution
10. An iron rod of length L and square cross section (side a) is given a uniform longitudinal magnetization M, and then bent around into a circle with a narrow gap (width w), as shown in Fig. 6.14. Find the magnetic field at the center of the gap, assuming w ≫ a≫ L. [Hint: treat it as the superposition of a complete torus plus a square loop with reversed current.] ... Get solution
11. In Sect, 6.2.1, we began with the potential of a perfect dipole (Eq. 6.10), whereas in fact we are dealing with physical dipoles. Show, by the method of Sect. 4.2.3, that we nonetheless get the correct macroscopic field. Get solution
12. An infinitely long cylinder, of radius R, carries a “frozen-in” magnetization, parallel to the axis, ... where k is a constant and s is the distance from the axis; there is no free current anywhere. Find the magnetic field inside and outside the cylinder by two different methods: (a) As in Sect. 6.2, locate all the bound currents, and calculate the field they produce. (b) Use Ampère’s law (in the form of Eq. 6.20) to find H, and then get B from Eq. 6.18. (Notice that the second method is much faster, and avoids any explicit reference to the bound currents.) Equation 6.18 ... Equation 6. 20 ... Get solution
13. Suppose the field inside a large piece of magnetic material is B0, so that ... where M is a “frozen-in” magnetization. (a) Now a small spherical cavity is hollowed out of the material (Fig. 6.21). Find the field at the center of the cavity, in terms of B0 and M. Also find H at the center of the cavity, in terms of H0 and M. (b) Do the same for a long needle-shaped cavity running parallel to M. (c) Do the same for a thin wafer-shaped cavity perpendicular to M. Figure6.21 ... Get solution
14. For the bar magnet of Prob. 6.9, make careful sketches of M, B, and H, assuming L is about 2a. Compare Prob. 4.17. Reference 4.17For the bar electret of Prob. 4.11, make three careful sketches: one of P, one of E, and one of D. Assume L is about 2a. [Hint: E lines terminate on charges; D lines terminate on free charges.] Reference prob 4.11 A short cylinder, of radius a and length L, carries a “frozen-in” uniform polarization P, parallel to its axis. Find the bound charge, and sketch the electric field (i) for L ≫ a, (ii) for L ≫ a, and (iii) for L ≈ a. [This is known as a bar electret; it is the electrical analog to a bar magnet. In practice, only very special materials—barium titanate is the most “familiar” example—will hold a permanent electric polarization. That’s why you can’t buy electrets at the toy store.] Reference 6.9 A short circular cylinder of radius a and length L carries a “frozen-in” uniform magnetizationMparallel to its axis. Find the bound current, and sketch the magnetic field of the cylinder. (Make three sketches: one for L ≫ a, one for L≫ a, and one for L ≈ a.) Compare this bar magnet with the bar electret of Prob. 4.11. Get solution
15. If J f = 0 everywhere, the curl of H vanishes (Eq. 6.19), and we can express H as the gradient of a scalar potential W: ... According to Eq. 6.23, then, ... so W obeys Poisson’s equation, with ∇ ·M as the “source.” This opens up all the machinery of Chapter 3. As an example, find the field inside a uniformly magnetized sphere (Ex. 6.1) by separation of variables. [Hint: ∇ ·M = 0 everywhere except at the surface (r = R), so W satisfies Laplace’s equation in the regions r R; use Eq. 3.65, and from Eq. 6.24 figure out the appropriate boundary condition on W.] reference 3.65 ... Reference 6.24 ... Get solution
16. A coaxial cable consists of two very long cylindrical tubes, separated by linear insulating material of magnetic susceptibility χm. A current I flows down the inner conductor and returns along the outer one; in each case, the current distributes itself uniformly over the surface (Fig. 6.24). Find the magnetic field in the region between the tubes. As a check, calculate the magnetization and the bound currents, and confirm that (together, of course, with the free currents) they generate the correct field. ... Get solution
17. A current I flows down a long straight wire of radius a. If the wire is made of linear material (copper, say, or aluminum) with susceptibility χm, and the current is distributed uniformly, what is the magnetic field a distance s from the axis? Find all the bound currents. What is the net bound current flowing down the wire? Get solution
18. A sphere of linear magnetic material is placed in an otherwise uniform magnetic field B0. Find the new field inside the sphere. [Hint: See Prob. 6.15 or Prob. 4.23.] Reference prob 6.15 If J f = 0 everywhere, the curl of H vanishes (Eq. 6.19), and we can express H as the gradient of a scalar potential W: ... According to Eq. 6.23, then, ... Get solution
19. On the basis of the naïve model presented in Sect. 6.1.3, estimate the magnetic susceptibility of a diamagnetic metal such as copper. Compare your answer with the empirical value in Table 6.1, and comment on any discrepancy. Get solution
20. How would you go about demagnetizing a permanent magnet (such as the wrench we have been discussing, at point c in the hysteresis loop)? That is, how could you restore it to its original state, with M = 0 at I = 0? Get solution
21. (a) Show that the energy of a magnetic dipole in a magnetic field B is ... Assume that the magnitude of the dipole moment is fixed, and all you have to do is move it into place and rotate it into its final orientation. The energy required to keep the current flowing is a different problem, which we will confront in Chapter 7.] Compare Eq. 4.6. Figure 6.30 ... (b) Show that the interaction energy of two magnetic dipoles separated by a displacement r is given by ... Compare Eq. 4.7. (c) Express your answer to (b) in terms of the angles θ1 and θ2 in Fig. 6.30, and use the result to find the stable configuration two dipoles would adopt if held a fixed distance apart, but left free to rotate. (d) Suppose you had a large collection of compass needles, mounted on pins at regular intervals along a straight line. How would they point (assuming the earth’s magnetic field can be neglected)? [A rectangular array of compass needles aligns itself spontaneously, and this is sometimes used as a demonstration of “ferromagnetic” behavior on a large scale. It’s a bit of a fraud, however, since the mechanism here is purely classical, and much weaker than the quantum mechanical exchange forces that are actually responsible for ferromagnetism.13] Get solution
22. In Prob. 6.4, you calculated the force on a dipole by “brute force.” Here’s a more elegant approach. First write B(r) as a Taylor expansion about the center of the loop: ... where r0 is the position of the dipole and ∇0 denotes differentiation with respect to r0. Put this into the Lorentz force law (Eq. 5.16) to obtain ... Or, numbering the Cartesian coordinates from 1 to 3: ... where δi j is the Kronecker delta (Prob. 3.52). Get solution
23. A familiar toy consists of donut-shaped permanent magnets (magnetization parallel to the axis), which slide frictionlessly on a vertical rod (Fig. 6.31). Treat the magnets as dipoles, with mass md and dipole moment m. (a) If you put two back-to-back magnets on the rod, the upper one will “float”—the magnetic force upward balancing the gravitational force downward. At what height (z) does it float? (b) If you now add a third magnet (parallel to the bottom one), what is the ratio of the two heights? (Determine the actual number, to three significant digits.) ... Get solution
24. Imagine two charged magnetic dipoles (charge q, dipole moment m), constrained to move on the z axis (same as Problem 6.23(a), but without gravity). Electrically they repel, but magnetically (if both m’s point in the z direction) they attract. (a) Find the equilibrium separation distance. (b) What is the equilibrium separation for two electrons in this orientation. (c) Does there exist, then, a stable bound state of two electrons? Get solution
25. Notice the following parallel: ... Thus, the transcription D→B, E→H, P→ μ0M, ϵ 0 → μ0 turns an electrostatic problem into an analogous magnetostatic one. Use this, together with your knowledge of the electrostatic results, to rederive (a) the magnetic field inside a uniformly magnetized sphere (Eq. 6.16); (b) the magnetic field inside a sphere of linear magnetic material in an otherwise uniform magnetic field (Prob. 6.18); (c) the average magnetic field over a sphere, due to steady currents within the sphere (Eq. 5.93). Get solution
26. Compare Eqs. 2.15, 4.9, and 6.11. Notice that if ρ, P, and M are uniform, the same integral is involved in all three: ... Therefore, if you happen to know the electric field of a uniformly charged object, you can immediately write down the scalar potential of a uniformly polarized object, and the vector potential of a uniformly magnetized object, of the same shape. Use this observation to obtain V inside and outside a uniformly polarized sphere (Ex. 4.2), and A inside and outside a uniformly magnetized sphere (Ex. 6.1). Equation 6.11 ... Get solution
27. At the interface between one linear magnetic material and another, the magnetic field lines bend (Fig. 6.32). Show that tan θ2/ tan θ1 = μ2/μ1, assuming there is no free current at the boundary. Compare Eq. 4.68. Equation 4.68 ... Figure 6.32 ... Get solution
28. A magnetic dipole m is imbedded at the center of a sphere (radius R) of linear magnetic material (permeability μ). Show that the magnetic field inside the sphere (0 r ≤ R) is ... What is the field outside the sphere? Get solution
29. You are asked to referee a grant application, which proposes to determine whether the magnetization of iron is due to “Ampère” dipoles (current loops) or “Gilbert” dipoles (separated magnetic monopoles). The experiment will involve a cylinder of iron (radius R and length L = 10R), uniformly magnetized along the direction of its axis. If the dipoles are Ampère-type, the magnetization is equivalent to a surface bound current ... if they are Gilbert-type, the magnetization is equivalent to surface monopole densities σb = ±M at the two ends. Unfortunately, these two configurations produce identical magnetic fields, at exterior points. However, the interior fields are radically different—in the first case B is in the same general direction as M, whereas in the second it is roughly opposite to M. The applicant proposes to measure this internal field by carving out a small cavity and finding the torque on a tiny compass needle placed inside. Assuming that the obvious technical difficulties can be overcome, and that the question itself is worthy of study, would you advise funding this experiment? If so, what shape cavity would you recommend? If not, what is wrong with the proposal? [Hint: Refer to Probs. 4.11, 4.16, 6.9, and 6.13.] Reference prob 4.11 A short cylinder, of radius a and length L, carries a “frozen-in” uniform polarization P, parallel to its axis. Find the bound charge, and sketch the electric field (i) for L ≫ a, (ii) for L ≫ a, and (iii) for L ≈ a. [This is known as a bar electret; it is the electrical analog to a bar magnet. In practice, only very special materials—barium titanate is the most “familiar” example—will hold a permanent electric polarization. That’s why you can’t buy electrets at the toy store.] Get solution
