Magnetic Fields in Matter, Solution Manual

Magnetic fields in matter Earth dipole (a) Equation (11.15) gives the field at position R along the axis of a dipole as Br = µ0 m/2πR3 , so m= 2πR3 Br 2π (6.4 · 106 m)3 (6.2 · 10−5 T) = = 8.1 · 1022 J/T. µ0 4π · 10−7 kg 2m (697) C (b) Equation (6.53) gives the field at position z on the axis of a current ring as Bz = µ0 Ib2 /2(z 2 + b2 )3/2 . If R is the radius of the earth, then we have z = R and b ≈ R/2, so in terms of R the field is Bz = µ0 I/(53/2 R). Therefore, I= 53/2 RBz 53/2 (6.4 · 106 m)(6.2 · 10−5 T) = = 3.5 · 109 A. µ0 4π · 10−7 kg2m (698) C If we instead treat the current ring as a dipole with moment m = 8.1 · 1022 J/T, then we have m = I(πb2 ) =⇒ I = 4(8.1 · 1022 J/T) m = 2.5 · 109 A, = π(6.4 · 106 m)2 π(R/2)2 (699) which is a so-so approximation to the correct result of 3.5 · 109 A. Disk dipole Let’s divide the disk into rings and then add up the magnetic moments of all the rings. The surface current density at radius r is σv, where v = ωr. This is true because σℓ(v dt) is the amount of charge that crosses a transverse segment with length ℓ in a time dt. So the charge per time per unit transverse length (that is, the surface current density) equals σℓ(v dt)/(ℓ dt) = σv. The current in a given ring with radius r and thickness dr is therefore Ir = (σv)dr = σωr dr. The magnetic moment of this ring is then Ir (πr2 ) = πσωr3 dr. Integrating from r = 0 to r = R gives the total magnetic moment of the disk as πσωR4 /4. θ R dθ Figure 161 Note that this result can be written as ω(πR2 σ)R2 /4 = ωQR2 /4, where Q is the total charge on the disk. If all of this charge were instead located on the rim, then takes a time of 2π/ω, the magnetic moment would be IπR2 = (since one revolution ) Q/(2π/ω) πR2 = ωQR2 /2, which is twice the moment of the disk. Sphere dipole Let the axis of rotation be vertical. Consider a strip located at angle θ, with width dθ, as shown in Fig. 161. The speed of any point in this strip is v = ωr = ω(R sin θ). The effective linear charge density of the strip, dλ, equals σ times the width, so dλ = σ(R dθ). The current dI in the strip is then dI = v dλ = (ωR sin θ)(σR dθ) = ωσR2 sin θ dθ = ωQ sin θ dθ , 4π (700) where we have used σ = Q/4πR2 . Alternatively, you can find dI by multiplying the total charge in the ring, which is σ(2πR sin θ)(R dθ), by the number of revolutions per second, which is ω/2π . The magnetic moment of the ring is dm = (dI)π(R sin θ)2 = ωQR2 sin3 θ dθ . 4 (701) Integrating this over the whole sphere to obtain the total magnetic moment gives (using the integral table in Appendix K or writing sin3 θ as sin θ(1 − cos2 θ)) )  π ( ∫ ωQR2 4 ωQR2 π 3 ωQR2 ωQR2 cos3 θ   = · = sin θ dθ = m= . − cos θ +  3 3 4 4 4 3 0 0 (702) Note that if all of the charge Q were located on the equator, then the current would 2 be I = (ω/2π)Q, and the magnetic moment would be m = IπR = (ωQ/2π)(πR2 ) = ωQR2 /2. The m for a spinning shell is therefore 2/3 of the m for a spinning ring with the same radius and total charge. Also, from Exercise 11.13 the m for a spinning shell is 4/3 times the m for a spinning disk with the same radius and total charge. It makes sense that this factor is larger than 1, because if the shell is projected onto the equatorial disk, the density near the rim is larger than the density at the center. A solenoid as a dipole To estimate roughly the magnetic dipole moment of the solenoid, let us suppose that it is equivalent to a point dipole that would produce, 20 cm away on its axis, a field strength Bz equal to that at the end of the solenoid, namely 1.8 T. This is reasonable because the magnetic field configuration near the end of the solenoid and beyond looks not very different from a dipole field. On this assumption, Eq. (11.15) gives B= µ0 m 2πr3 =⇒ m = 2π (0.2 m)3 (1.8 T) 2πr 3 B = = 7.2 · 104 J/T. µ0 4π · 10−7 kgC2m (703) However, there is actually no need to know this value of m, because we are interested only in a rough estimate of the field at the location of the complaining physicists. All we care about is that the field decreases like 1/r3 ; the exact nature of the radial and tangential components isn’t critical. The physicists are 100 feet away, which is about 30 meters. This is 150 times the 0.2-meter distance at which the field is 18, 000 gauss, so the desired field is smaller by a factor of 1503 . It is therefore roughly equal to 18, 000/1503 ≈ 5 · 10−3 gauss. (If you want to find the two components, you can simply plug the above value of m into Eq. (11.15), with θ = tan−1 (4/3) ≈ 53◦ .) This field is about 100 times smaller than the earth’s field of about 0.5 gauss. If it were perfectly steady it would not be noticed. But if it were frequently switched on and off, it might cause trouble. Dipole in a uniform field Let the magnetic dipole point upward, which means that the uniform field B0 points downward. From Eq. (11.15) the radial component of the dipole field is (µ0 m/2πr3 ) cos θ. And the radial component of the uniform field is −B0 cos θ. So the total radial component on the surface of a sphere with radius r is Br = (µ0 m/2πr3 − B0 ) cos θ. This equals zero everywhere on a sphere with radius r = (µ0 m/2πB0 )1/3 . Since Br = 0 on this sphere, no field lines pass through the sphere, so the internal field lines must look something like those shown in Fig. 162. On the equator, we need to add B0 to the field from the dipole. For the dipole field, we are concerned with the tangential component, which from Eq. (11.15) points downward with magnitude µ0 m/4πr 3 on the equator. Using r = (µ0 m/2πB0 )1/3 , this equals B0 /2. Adding this to the downward uniform field B0 gives a total field of 3B0 /2 pointing downward. If we create the identical field outside the sphere by replacing the dipole by the appropriate distribution of surface current on the sphere, the field inside will be zero. This is true because a field line can’t end on the spherical surface (there are no magnetic monopoles, so magnetic field lines can’t end) or pass through it, so it would have to be ∫ a closed loop inside. But that region is now empty of current, so we must have B · ds = 0 around any path. A closed field-line loop would make this integral be nonzero. So there must be no closed loops, and hence no magnetic field at all. Trapezoid dipole The left and right sides of the trapezoid each produce zero field at P , because they point directly at P so the cross product in the Biot-Savart law is zero. For the top and bottom sides, the angle in the Biot-Savart law is essentially 90◦ for the whole length, so the magnitude of the Biot-Savart contribution is µ0 Iℓ/4πR2 , where ℓ is the length of the particular side and R is the distance to the side. Let d be half the height of the trapezoid; this is essentially equal to a/2, but it will be easier to work with d. The top and bottom sides are distances r − d and r + d from P . If the bottom side has length b, then from similar triangles the top side has length b(r − d)/(r + d). At point P , the top side produces a magnetic field into the page, and the bottom side produces a field out of the page. So the net field into the page at P is ) ( b µ0 I b(r − d)/(r + d) − B = Btop − Bbottom = 2 2 (r − d ) (r + d) 4π ) ( µ0 I ba 1 µ0 I 1 b 2d µ0 I b ≈ − , (704) = = 4π r3 4π r + d r2 − d2 4π r + d r − d r + d where we have used 2d ≈ a, and where we have ignored the d’s in the denominator because they are small compared with r. Since the area of the trapezoid is essentially equal to ba, we have B ≈ µ0 m/4πr3 , in agreement with the Bθ in Eq. (11.15) when θ = 90◦ . Figure 162 Note that the above result is first order in b. You can show that the approximations we made (setting the Biot-Savart angle equal to 90◦ , using 2d ≈ a) involve errors that are second order in b, so we were justified in ignoring them. Field somewhat close to a solenoid Consider a single loop of the solenoid at its middle. From Eq. (11.15), this little dipole creates a field at P equal to µ0 m/4πℓ3 = µ0 (πR2 I)/4πℓ3 . Up to numerical factors, all the other loops create this same field too; the only differences are some trig factors of order 1. So if there are N loops in all, the total field at P behaves like B∼N µ0 (πR2 I) R2 = µ0 (N/ℓ)I 2 . 4ℓ 4πℓ3 (705) Ignoring the factor of 1/4, and recalling that µ0 (N/ℓ)I ≡ µ0 nI is the field B0 inside the solenoid, we obtain B ∼ 0B R2 /ℓ2 , as desired. Using reciprocity Let the dipole m(t) consist of a ring of area a carrying current I2 (t) = I2 cos ωt. So m(t) = (I2 cos ωt)a =⇒ m0 = I2 a. If a current I1 (t) in C1 causes a field B1 (t) at the location of the dipole ring, then the flux through the ring is Φring (t) = B1 (t) · a. The coefficient of mutual inductance is therefore M = Φring /I1 = (B1 · a)/I1 . (We’ve suppressed the t arguments here.) Due to the reciprocity theorem, this is also the M going the other way. The emf in circuit C1 is therefore given by E1 (t) = −M dI2 (t)/dt, where I2 (t) is the current in the dipole ring. Hence the emf in C1 is E1 (t) = = B1 · a d(I2 cos ωt) B1 · a (−ωI2 sin ωt) =− I1 I1 dt ω ω B · (I2 a) sin ωt = B1 · m0 sin ωt ≡ E1 sin ωt, I1 1 I1 − (706) with E1 = (ω/I1 )B1 · m0 , as desired. I1 and B1 are technically functions of time. But since they have the same time dependence, we can take them to be the amplitudes (a vector amplitude in the B1 case). Force between a wire and a loop At an arbitrary point on the wire, the magnetic field from the square-loop dipole has both an upward vertical (z) component and a horizontal component along the wire. But the latter produces no force on the current in the wire, so we care only about the z component. Since the current in the wire in Fig. 6.47 points into the page, the right-hand rule gives the magnetic force on the wire as pointing to the right. Equation (11.14) gives the z component of the dipole field, with m equal to m = I2 ℓ2 . The rightward force on a little piece dx of the wire equals I1 Bz dx. With θ measured away from the vertical axis, dx is given by the usual expression, dx = z dθ/ cos2 θ. (See the reasoning in the paragraph following Eq. (1.37).) Also, the distance r to the little piece is r = z/ cos θ. Integrating over the entire infinite wire, we find the total rightward force on it to be ( ) ∫ ∞ ∫ ∞ µ0 I2 ℓ2 3 cos2 θ − 1 F = I1 Bz dx = I1 dx 3 4π r −∞ −∞ 2 ∫ π/2 3 cos2 θ − 1 µ0 I 1 I 2 ℓ z dθ = 3 2 4π −π/2 (z/ cos θ) cos θ 2 ∫ π/2 µ0 I 1 I 2 ℓ 2 µ0 I 1 I 2 ℓ (3 cos3 θ − cos θ) dθ = = . (707) 2πz 2 4πz 2 −π/2 The integral here equals 2, as you can check with Mathematica or the integral table in Appendix K. This force is consistent with the magnitude of the leftward force on the square loop we found in Eq. (458) in Exercise 6.54, because z ≈ R for large z. Dipoles on a chessboard (a) Equation (11.23) gives the force on a dipole as F = ∇(m · B). The applied force is the negative of this, so the associated work equals the line integral of −∇(m · B). The line integral of the gradient of a quantity is simply the change in that quantity. So the work required to move a dipole to infinity equals the change in −m · B, which is 0 − (−m · B0 ) = m · B0 , because the field is zero at infinity. B0 here is the field (due to all the other dipoles) at the initial location of a dipole on a particular square. This result is consistent with the fact that the energy of a dipole is given by −m · B. Due to the opposite directions of the dipoles on the white and black squares, the B fields from the four nearest neighbors (or two or three, if the dipole is near the edge) point parallel to a given dipole m. So we expect that the sum of all 63 of the m · B0 contributions to the work will be positive. That is, we expect all of the dipoles to be bound. Eq. (11.15) gives the magnetic field due to a dipole, in the plane of the dipole, as µ0 m/4πr3 . The distance r between the various squares is found from the Pythagorean theorem. For a given point labeled by the coordinates (a, b), where a and b each run from 1 to 8, the sum of the m · B0 contributions to the work equals 8 8 (−1)a−i (−1)b−j µ0 m2 ∑ ∑ , (708) − ( ) 3 4πs i=1 j=1 (a − i)2 + (b − j )2 3/2 with the caveat that the term with (i, j) = (a, b) is excluded from the sum. You can check that the negative sign out front makes the overall sign correct. There must be a way to cleanly exclude the (i, j) = (a, b) term in Mathematica, but I can’t figure out what it is. At any rate, this program gets the job done (the values of a and b can be changed): (b) The field from a magnetic dipole takes the same form as the field from an electric dipole. And the force on a dipole does also, because there aren’t any outside currents in the setup; see the discussion following Eq. (11.24). So the magnetic force on each of our magnetic dipoles is conservative, just as the electric force on an electric dipole is. So we can use the same reasoning we used back in Chapters 1 and 2 to say that the total work required to remove all of the dipoles far from 64 2. 68 21 64 2. 22 71 2. 65 01 64 2. 2. 91 28 82 65 63 2. 2. 21 2. 56 40 2. 27 Up to a factor of µ0 m2 /4πs3 , the amounts of work for the various squares on the chessboard are shown in Fig. 163. These entries are repeated in other parts of the board; there are only ten independent entries. The most tightly bound dipoles are the ones on the “bishop’s pawn” and equivalent squares; there are eight such squares. However, these are only negligibly more bound than the other interior dipoles. The binding energies of the 36 interior dipoles differ from one another by less than 1%. 56 {i,1,a-1}, {j,1,8}]+ {i,a+1,8}, {j,1,8}]+ {i,a,a}, {j,1,b-1}]+ {i,a,a}, {j,b+1,8}] 1. 56 a = 3; b = 2; NSum[-(-1)^(a-i)(-1)^(b-j)/((a-i)^2+(b-j)^2)^(3/2), NSum[-(-1)^(a-i)(-1)^(b-j)/((a-i)^2+(b-j)^2)^(3/2), NSum[-(-1)^(a-i)(-1)^(b-j)/((a-i)^2+(b-j)^2)^(3/2), NSum[-(-1)^(a-i)(-1)^(b-j)/((a-i)^2+(b-j)^2)^(3/2), Figure 163 each other equals one half of the sum of all the individual works. The factor of 1/2 gets rid of the double counting in the energy. From Fig. 163 you can show that the total work required is 77.67, in units of µ0 m2 /4πs3 . Potential momentum We quickly see that ∇ × A gives the desired B field in the negative z direction because    ẑ  ŷ B  x̂ ∇ × A =  ∂/∂x ∂/∂y ∂/∂z  = −Bẑ. (709) 2  −x 0  y If B = 0 and v = v0 initially, then L is initially M v0 rẑ. When the field is turned on, Eq. (11.37) gives the increase in the speed as ∆v = qrB/2M . So the change in r × (M v) is M ∆v rẑ = (qr2 B/2)ˆz. But the change in r × (qA) is    x̂ ŷ ẑ   qBr2 qB  qB 2 (x + y 2 )ẑ = − x y 0  = − ẑ. (710) 2 2 2   y −x 0 So the total change in r × (M v + qA) is zero, as desired. Energy of a dipole configuration If you consider the little loops of current that make up the dipoles, you can quickly verify that neither of the dipoles experiences a force along the dotted line (which we’ll call the x axis). Therefore, no work is done in bringing the dipoles toward each other, along the x axis, to the configuration in Fig. 11.39(b). Alternatively, if we look at, say, m1 , then Eq. (11.23) tells us that the force is ∇(m · B) = ∇(mx Bx ) = mx ∇Bx . But ∇Bx has no x component because the Bx due to m2 is identically zero along the x axis. ∇Bx does have a nonzero y component (where y is the direction of m2 ), because ∂Bx /∂y = 0. So there is a force in the y direction. But this force is irrelevant in finding the work done in moving m1 along the x axis. Alternatively again, there is no need to even mention forces. The energy of a dipole is m · B, so if m is always perpendicular to B, as it is in Fig. 11.39(b), then the energy is always zero. So no work is done. The work required to rotate a dipole is the integral of the torque with respect to angle. If θ is defined relative to the x axis as in Fig. 11.39(a), then the magnitude of the torque on m1 is N = |m1 × B2 | = m1 B2 cos θ, because B2 is perpendicular to the x axis. (We could try to keep track of the signs, but it’s easier to just work with the magnitudes and then put in the correct sign by hand at the end.) The integral of this from 0 to θ1 is m1 B2 sin θ1 . From Eq. (11.15), B2 equals µ0 m2 /4πr3 . So the work done by the external torque agency is µ0 m1 m2 sin θ1 /4πr3 . The sign is correct because the work is positive; we are making m1 be more antiparallel to B2 . Now let’s rotate m2 through an angle of π/2 − θ2 to its final position. We can treat m1 as the sum of two separate dipoles: x̂m1 cos θ1 and ŷm1 sin θ1 . At the position of m2 , the former produces a field Bx = µ0 (m1 cos θ1 )/2πr3 , and the latter produces a field By = −µ0 (m1 sin θ1 )/4πr3 . If α is the variable angle of m2 with respect to the y axis, then the torque associated with Bx involves the factor cos α. The integral of this gives a factor of sin α. But since α starts at zero and ends at π/2 − θ2 , this factor becomes cos θ2 . So the work done is −m2 (µ0 m1 cos θ1 /2πr3 ) cos θ2 . We have put in the minus sign because the required work is negative; we are making m2 be more parallel to Bx x̂. Similarly, with the By field, if α is defined in the same way, the torque involves the factor sin α. The integral of this gives a factor of − cos α. But since α starts at zero and ends at π/2 − θ2 , this yields a factor of (1 − sin θ2 ) So the work done is m2 (µ0 m1 sin θ1 /4πr3 )(−1 + sin θ2 ). We have put in a minus sign because the required work is negative; we are making m2 be more parallel to the negative Byŷ field. The total required work done (that is, the potential energy of the system) is the sum of the above three results: ] µ0 m1 m2 [ sin θ1 − 2 cos θ1 cos θ2 + (− sin θ1 + sin θ1 sin θ2 ) W = 4πr3 ( ) µ0 m1 m2 − 2 cos θ1 cos θ2 + sin θ1 sin θ2 , (711) = 4πr3 as desired. If θ1 = 0 and θ2 = 90◦ , then we have the configuration in Fig. 11.39(b), and W is correctly zero. Note that if θ1 √ = θ2 ≡ θ, then the potential energy is positive √ if tan θ > 2 and negative if tan θ < 2. This cutoff angle is the angle for which the field of one dipole is perpendicular to the other dipole at its location. This makes sense, because if the dipoles are bought together in this configuration, then m · B is always zero. So the energy is always zero, and no work is required. Octahedron energy The connecting lines between the 15 pairs of dipoles in the octahedron fall into four groups: • 8 lines are tilted at 45◦ with respect to the dipoles; these are the lines connected to either of the two vertices on the z axis. The associated angles θ1 and θ2 in Exercise 11.23 are both equal to 45◦ . And the distance r equals b. • 4 lines form a square in the xy plane. The θ’s are 90◦ , and r = b. • 2 lines are diagonals of the square in the xy plane. The θ’s are 90◦ , and r = √ • 1 line is vertical along the z axis. The θ’s are 0◦ , and r = 2b. √ 2b. Using the U = (µ0 m1 m2 /4πr3 )(sin θ1 sin θ2 −2 cos θ1 cos θ2 ) result from Exercise 11.23, the potential energy is ( ) ( )) ) 1 ( µ 0 m2 ( 1 1) 1 ( U = −2· + 4 1 − 2 · 0 + 2 3/2 1 − 2 · 0 + 1 3/2 0 − 2 · 1 8 2 2 2 2 4πb3 = 0. (712) Remark: It isn’t obvious why the potential energy should be zero. However, as you can check, it is also zero in the analogous cases of a tetrahedron, a cube with the dipoles aligned parallel to an edge, and a cube with the dipoles aligned parallel to a long diagonal (this case involves some tricky angles). So it is reasonable to conjecture that the potential energy is zero in the case of any platonic solid, for any (common) orientation of the dipoles. Unfortunately, I can’t think of a general proof. Note that our setup with magnetic dipoles is equivalent to one with electric dipoles, because the forces and torques take the same form in electric and magnetic dipoles (since there are no external currents involved; see the discussion following Eq. (11.24)). An extreme case is the limit of an infinite number of infinitesimal electric dipoles lying on the surface of a sphere (so we effectively have two opposite shells of charge near each other). You can show that the potential energy of this system is zero. There must be a general proof for the platonic solids, and perhaps other configurations with sufficient symmetry. . . Rotating a bacterium The dipole moment of one of the crystals is m = M0 V = (4.8 · 105 J/Tm3 )(5 · 10−8 m)3 = 6 · 10−17 J/T. (713) From the discussion in Section 11.6, the work required to rotate the dipole by 90◦ is mB (assuming that it starts aligned with the field). If we have 10 crystals, and if we take the earth’s magnetic field to be 0.5 gauss, then the required work is W = 10mB = 10(6 · 10−17 J/T)(5 · 10−5 T) = 3 · 10−20 J. (714) At room temperature, kT equals 4·10−21 J, so W is roughly 10 times kT . A bacterium therefore maintains close alignment with the earth’s field; the thermal fluctuations have only a small effect. Electric vs. magnetic dipole moments The magnetic and electric forces behave like qvB and qE. From the expressions for the dipole fields in Eq. (10.18) and Eq. (11.15), these forces will be in the ratio of v(µ0 m) to p/ϵ0 . (Since we’re doing things roughly, we’ll ignore any factors of order 1, including trig factors.) Therefore (using ϵ0 µ0 = 1/c2 ), vϵ0 µ0 m (m/c) v Fm = = · . Fe c p p (715) With the given values of p, m, and v, this becomes (10−23 A m2 )/(3 · 108 m/s) 1 Fm · = ≈ 3 · 10−4 . 10−30 C m 100 Fe (716) Using 10−29 C m for p would make the result even smaller. And even if v ≈ c, the ratio will still be small. Diamagnetic susceptibility of water The magnetic susceptibility is given by M = χm B/µ0 . (The M = χm H definition would give essentially the same result, because χm will turn out to be very small; see Exercise 11.38.) The magnetic dipole moment of a given volume V is m = MV = χm BV /µ0 . The force on a dipole is therefore F =m ∂Bz χm Bz V ∂Bz = . µ0 ∂z ∂z (717) From Table 11.1 we have (being careful with the signs; we’ll take upward to be positive for all quantities) Bz = 1.8 tesla, ∂Bz /∂z = −17 tesla/m, and F = 0.22 newtons. The volume taken up by 1 kilogram of water (which is what the data in Table 11.1 are given for, even though an actual sample would of course be much smaller) is 10−3 m3 , so Eq. (717) gives ( m) 4π · 10−7 kg µ0 F C2 (0.22 N) = = −9.0 · 10−6 . (718) χm = −3 3 V (∂B /∂z ) Bz (1.8 T)(10 m )(−17 T/m) z Paramagnetic susceptibility of water (a) We have χ = µ0 Nm2 /kT . There are about 6 · 1023 nuclei in a gram of anything, because the mass of a nucleon is about 1.67 · 10−24 g, while the mass of an electron is negligible in comparison. So a cubic meter of water, which has a mass of 106 grams, contains 6 · 1029 protons. In water, 2 out of 18 nuclei are hydrogen protons, so the number of protons per cubic meter is N = (2/18)(6 · 1029 m−3 ) = 6.7 · 1028 m−3 . The proton magnetic moment is 1/700 of the electron magnetic moment, or (9.3 · 10−24 A m2 )/700 = 1.3 · 10−26 A m2 . At room temperature, kT = 4 · 10−21 J. So χ= (4π · 10−7 µ0 N m2 = kT kg m C2 )(6.7 · 1028 m−3 )(1.3 · 10−26 A m2 )2 = 3.6 · 10−9 . 4 · 10−21 J (719) (b) The magnetization is given by M = χB/µ0 . (The M = χm H definition would give essentially the same result, because the above χ is very small; see Exercise 11.38.) The magnetic moment in a volume V is m = M V . V = 10−3 m3 here, so m= (3.6 · 10−9 )(10−3 m3 )(1.5 T) χV B = = 4.3 · 10−6 A m2 . µ0 4π · 10−7 kg m/C2 (720) You should check that these units work out. (c) If the flask were cubical, the cross section would be 10 cm by 10 cm, or 100 cm2 . So let’s say the area is a = 50 cm2 . The magnetic moment is m = Ia, so I= m 4.3 · 10−6 A m2 = ≈ 9 · 10−4 A, a 5 · 10−3 m2 (721) or 900 microamps. Work on a paramagnetic material If a dipole m is located on the axis of a solenoid, parallel to B, then the magnetic force on it is F = m ∂B/∂z. In terms of the specific susceptibility χ, the magnetic moment of a material of mass m̃ (there are only so many ways to write the letter “m”) is m = χBm̃/µ0 . (You can show that the units of the specific susceptibility χ are m3 /kg.) The force on the mass m̃ is then F = (χB m̃/µ0 ) ∂B/∂z. The work done against this force is ∫ ∫ ∫ ∂B χm̃ χm̃ B dz = − W = − F dz = − d(B 2 ). (722) ∂z µ0 2µ0 If the material is moved along the axis from a point where the field is B1 to a point where the field is B2 , the work done is (χm̃/2µ0 )(B12−B22 ). If B2 is negligible compared with B1 , then W = χB12m̃/2µ0 . So the work per kilogram is χB 21 /2µ0 , as desired. From above, we have χ = µ0 F/(m̃B ∂B/∂z), so we can eliminate χ from the result for W and write ( ) µ0 F FB m̃ m̃ W = χB 2 = B2 = . (723) 2µ0 m̃B ∂B/∂z 2µ0 2 ∂B/∂z From Table 11.1, the data for 1 kg of liquid oxygen are F = 75 N, B = 1.8 T, and ∂B/∂z = 17 T/m. So the work for 1 kg is W = (75 N)(1.8 T) = 4 J. 2(17 T/m) −3 The work for 1 gram is 1/1000 of this, or 4 · 10 (724) J. Greatest force in a solenoid If a dipole m is located on the axis of a solenoid, parallel to B, then the force on it is F = m ∂B/∂z. Per unit volume of the sample, we also have m = χB/µ0 . The force per unit volume is therefore F = (χ/µ0 )B ∂B/∂z. So our goal is to find B and ∂B/∂z and then maximize their product by setting the derivative equal to zero: )2 ( ( ) ∂ ∂B ∂B ∂2B ∂F ∝ 0= + B . (725) =B ∂z ∂z ∂z2 ∂z ∂z Equation (6.56) gives the field in the interior of a finite solenoid as B = (µ0 nI/2)(cos θ1 − cos θ2 ), where the √ angles are shown in Fig. 6.16. In the present case we have θ2 = π and cos θ1 = z/ z 2 + r02 , where z is the distance from the end (with positive z being inside the solenoid). So Eq. (6.56) gives ) ( µ0 nI r02 ∂B µ0 nI z B= =⇒ = 1+ √ 2 2 (z 2 + r02 )3/2 ∂z 2 z + r20 =⇒ Equation (725) then becomes ( 0 = =⇒ 0 = √ =⇒ 3z z 2 + r20 = 1+ √ (√ z2 + z z2 + r20 r02 )( ∂2B µ0 nI −3r02 z = . ∂z2 2 (z 2 + r02)5/2 (726) ) ( )2 r2 0 −3r2 0 z + 2 2 5/2 3/2 2 2 (z + r 0 ) (z + r0 ) ) + z (−3z ) + r20 r02 − 3z 2 . (727) Squaring yields 4 4 + 9z 2 r20 = r40 − 6r02z 2 + ✟ 9z✟ =⇒ 15z 2 = r20 9z✟ ✟ √ =⇒ z = r0 / 15. (728) We have chosen the positive root because the negative root was introduced in the squaring operation and isn’t a solution to Eq. (727). The desired point therefore lies slightly inside the end of the solenoid. Boundary conditions for B From Problem 11.8(a) the external field is the field of a dipole with strength m = (4πR3 /3)M . So from Eq. (11.15) the components are µ0 m cos θ 2µ0 M cos θ = , 3 2πR3 µ µ M sin θ m sin θ 0 0 Bθout = = . 3 4πR3 Brout = (729) From Problem 11.8(b) the internal field is B = (2/3)µ0 M upward (assuming M points upward). The radial component of this is Brin = (2/3)µ0 M cos θ, and the tangential component is Bθin = −(2/3)µ0 M sin θ. The negative sign comes from the fact that the field points toward the top of the sphere, which is the direction of decreasing θ. As predicted, Br is continuous across the surface, and Bθ has a discontinuity of µ0 M sin θ. This gives us the desired result of µ0 Jθ provided that the surface current density Jθ equals M sin θ. And it does indeed, from the reasoning in the example at the end of Section 11.8; the component of M parallel to the surface is M∥ = M sin θ. B at the center of a solid rotating sphere From Problem 11.7 we know that the magnetic field due to a spinning shell with radius r and uniform surface charge density σ is the same as the magnetic field due to a sphere with uniform magnetization Mr = σωr (both inside and outside the shell). And then from Problem 11.8 we know that the internal field of a magnetized sphere is B = 2µ0 Mr /3 = 2µ0 σωr/3. (Alternatively, we could have just invoked the result from Problem 6.11 to obtain this field. But the idea here was to parallel the solution to Problem 11.9.) We can consider the solid spinning sphere to be the superposition of many spinning shells with uniform surface charge density σ = ρ dr. The center of the sphere is inside all of the shells, so we can use the above form of B for every shell. The total field at the center is therefore ∫ R 2µ0 (ρ dr)ωr µ0 ρωR2 B= . (730) = 3 3 0 In terms of the total charge Q = (4πR3 /3)ρ, this result can be written as B = µ0 ωQ/4πR. This field is 5/2 as large as the field at the north pole; see Problem 11.9. Spheres of frozen magnetization (a) From Problem 11.8(a), we know that the external magnetic field of a uniformly magnetized sphere with radius R is the same as the field of a magnetic dipole m located at the center, with magnitude m = (4πR3 /3)M . From Eq. (11.15) the field at points on the axis of the dipole is radial and has magnitude Br = µ0 m/2πR3 . Writing m in terms of M gives ( )( ) 2 kg m J 2 Br = µ0 M = 4π · 10−7 7.5 · 105 = 0.628 T. 3 3 C2 T m3 (731) You should check that the units work out correctly. Note that this result is independent of R. (b) From Eq. (11.15) the field at points on the equator of the sphere is tangential and has magnitude Bθ = µ0 m/4πR3 . This is half of the above Br , which yields 0.314 T. (c) The force acting on each of the two uniformly polarized spheres must be the same as if the other sphere were replaced by a point dipole m at its center, because that leaves the external field unchanged. So we need to find the force between two point dipoles separated by 2R. From Eq. (11.20) this force is F = m ∂Bz /∂z, where ±ẑ are the directions of the dipoles. Since Bz = µ0 m/2πz 3 , we have ∂Bz /∂z = 3µ0 m/2πz 4 (in magnitude). So the attractive force between the dipoles is 3µ0 m2 3µ0 (4πR3 M/3)2 πµ0 R2 M 2 = = 6 32πR4 2π (2R)4 )2 ( ) ( π kg m J −7 2 = 4π · 10 (0.01 m) 7.5 · 105 3 2 Tm 6 C = 37 N ≈ 8.3 pounds. = F (732) The units are easier to see if you write the units of M as C/(m s). Muon deflection Since U = γmc2 (we’ll use U for the energy) we have γ= 1010 eV U = = 50. mc2 2 · 108 eV (733) So β ≡ v/c is essentially equal to 1. The magnitude of the momentum is p = γmv = γmβc = β(γmc2 )/c ≈ U/c. (734) The magnetization is M = N m = (1.5 · 1029 m−3 )(9.3 · 10−24 J/T) = 1.4 · 106 J/(T m3 ). (735) From Eq. (11.55) the bound surface current on each face of the plate is J = M . But we know from Section 6.6 that the magnetic field between two such sheets (with the surface currents pointing in opposite directions) is µ0 J . So the magnetic field inside the plate equals B = µ0 J = µ0 M = (4π · 10−7 )(1.4 · 106 J/m3 T) = 1.76 T. (736) Equivalently, since there are no free currents, we have H = 0, which means B = µ0 M. The transverse force is F⊥ = evB , so the transverse momentum gained is p⊥ = evBt = eB (vt) = eBs, where s is the thickness of the plate. The angle of deflection is therefore (using the definition of an eV) θ = = p⊥ eBs eBsc eBsc Bsc = = = 10 = 10 p U/c U 10 · e(1 V) 10 (1 V) (1.76 T)(0.2 m)(3 · 108 m/s) = 0.011, 1010(1 V) (737) which is a shade more than 0.6◦ . As long as β ≈ 1, the angle of deflection is inversely proportional to the energy U . More generally, it is inversely proportional to βU (assuming the angle is small). Volume integral of near field The product of the near-region volume and the field strength in that volume is proportional to s3 · Q/s2 = sQ = p. But we are assuming that p is held constant. Therefore ∫ E dv over the near region remains constant as s shrinks down. See Problem 10.5 for some quantitative aspects of this setup. In the case of a current loop with magnetic dipole moment m = Iπr2 , the nearregion volume behaves like r3 , and the near field behaves like I/r (using the form of the field from a wire). So the product of the volume and the field strength is r∫3 · I/r = Ir2 = m/π. And we are assuming that m is held constant. Therefore B dv over the near region remains constant as r shrinks down. ∫ The average of, say, E over a macroscopic volume V equals ( E dv)/V . There will be finite contributions to the numerator from the near fields of all the dipoles in the volume, no matter how small they are. So the near fields will necessarily contribute to the average of E. Likewise for B. B Equilibrium orientations First consider the more general case of the isosceles triangle in Fig. 164. If the dipoles m1 and m2 make the same angle θ shown, then the sum of their fields at the top vertex is horizontal. This follows from the form of the dipole field, but it also follows from a symmetry argument: Start with just m1 and B1 , and then rotate the setup 180◦ around the vertical axis to produce the m1′ and B′1 vectors shown in Fig. 165. Then negate m1′ to obtain m2 and B2 . The B1 and B2 fields make the same angles above and below the horizontal, so the total B field points horizontally. B1 B1' m1 θ θ Figure 164 B1 B2 B B2 m'1 m1 m2 m1 m2 (a) Figure 165 In the case of an equilateral triangle, we see that if the dipoles are arranged as shown in Fig. 166(a), then each dipole will point in the direction of the field produced by the other two. Since a dipole has the least energy when it points in the direction of the field it lies in, this is the stable equilibrium configuration that the dipoles will assume. Similarly in the case of other N -gons, we obtain the situations shown in Fig. 166(b,c). The same symmetry argument that we used above can be used to show that each dipole points in the direction of the field due to all the other dipoles. (b) B inside a magnetized sphere Since H ≡ B/µ0 − M, we have B = µ0 (H + M). The two magnetic equations can therefore be written as ∇ · (H + M) = 0 and ∇ × H = 0. (738) These take exactly the same form as the electric equations, ∇ · (E + P/ϵ0 ) = 0 and ∇ × E = 0, (c) (739) with the correspondence being H ←→ E and M ←→ P/ϵ0 . (740) Now, the solution for E in the polarized sphere is completely determined by the polarization P, along with the two equations in Eq. (739). Likewise, the solution Figure 166 m2 for B in the magnetized sphere is completely determined by the magnetization M, along with the two equations in Eq. (738). So to obtain the solution in the latter case, we can simply replace the letters in the solution in the former case, via Eq. (740). The solution E = −P/3ϵ0 for the polarized sphere therefore becomes the solution H = −M/3 for the magnetized sphere. And since B = µ0 (H + M), the desired B field inside the sphere is given by ( ) M 2µ0 M B = µ0 − , (741) +M = 3 3 in agreement with the result from Problem 11.8. Two susceptibilities ′ B/µ0 . And from Eq. (11.72) we have M = χm H. From Eq. (11.52) we have M = χm Equating these two expressions for M, and using H = B/µ0 − M, gives χ′m B/µ0 = χm (B/µ0 − M) = χm (B/µ0 − χ′m B/µ0 ). (742) Canceling the B/µ0 gives ′ ) =⇒ χm = χ′m /(1 − χ′m ), χ′m = χm (1 − χm as desired. In cases where we have χm ≈ χ′m . ′ χm (743) ≪ 1, the denominator here is essentially equal to 1, so Magnetic moment of a rock (a) Ignore the rock for a moment and imagine that there is a current I in the coils. Using the expression in Eq. (6.53) for the field due to a single ring, we find that the field at the location of the rock due to the current in the coils is =⇒ B = B I = µ0 Ir2 2(r2 + z 2 )3/2 1500(4π · 10−7 kg m/C2 )(0.06 m)2 = 0.0225 T/A. ((0.06 m)2 + (0.03 m)2 )3/2 2(1500) (744) Now let’s reintroduce the rock. Its angular frequency is ω = 2π(1740/(60 s)) = 182 s−1 . The vertical component of its dipole moment produces zero net flux through the coils. But the horizontal component does, and this component oscillates sinusoidally. So we effectively have the setup presented in Exercise 11.19, with m0 pointing along the axis of the coils. So m0 is parallel to B1 in the E1 = (ω/I1 )B1 · m0 result from that exercise. The magnetic moment of the rock is therefore given by m0 = 10−3 V E 1 1 = = 2.4 · 10−4 J/T. · · ω B/I 182 s−1 0.0225 T/A (745) (b) The electron magnetic moment is about 10−23 J/T, so the above magnetic moment corresponds to (2.4 · 10−4 )/10−23 = 2.4 · 1019 electrons. Each iron atom can contribute 2 electron spins, so we need 1.2 · 1019 atoms. There are 56 nucleons in an iron atom, so each atom has a mass of about 9 · 10−26 kg. The minimum mass is therefore (1.2 · 1019 )(9 · 10−26 kg) ≈ 10−6 kg, or 0.001 g. Deflecting high-energy particles (a) From the B-H curve in Fig. H ≈ 5000 A/m. In the 20 cm estimate, the curve bcdea has ∫ ∫ H · dl = 11.41(d), we find that B = 1.6 T corresponds to gap, H = (1.6 T)/µ0 = 1.3 · 106 A/m. As a rough a length of about 300 cm. So ∫ H · dl H · dl + gap = = = iron (1.3 · 106 A/m)(0.2 m) + (5000 A/m)(3 m) (260, 000 + 15, 000) A 275, 000 A. (746) We see that, as mentioned in the statement of the exercise, the integral of H is dominated by the gap contribution. From Eq. (11.70) this integral equals N I , where N is the number of turns in the two coils, which is twice the number of turns in each coil. So 275, 000 A is the desired number of ampere turns (a fancy name for the total current passing through the loop). (b) Each coil is roughly rectangular with sides of 300 and 100 cm. So the length of one full turn is about L = 8 m. If the total cross section of copper is 1500 cm2 (in both coils), and if we have N turns (in both coils), then the cross section of each wire is (0.15 m2 )/N . The total resistance in both coils is then R= ρN L ρN (8 m) = N 2 ρ(53 m−1 ). = (0.15 m2 )/N A (747) So the power is (using the above result, N I = 275, 000 A) P = I 2 R = I 2 N 2 ρ(53 m−1 ) = (275, 000 A)2 (2·10−8 ohm-m)(53 m−1 ) = 8·104 J/s. (748) This is independent of N and I individually, because the product N I takes on a given value. (c) Another expression for the power is P = IV . If V = 400 V, then I= Therefore, 8 · 104 J/s P = 200 A. = 400 V V (749) NI 275, 000 A ≈ 1400, (750) = 200 A I 2 or 700 turns per coil. The cross-sectional area of the wire is (1500 cm )/1400 ≈ 1.1 cm2 . If the voltage source were instead, say, 800 V, then the current would be 100 A, so the number of turns would be 2800, and the cross-sectional area would be roughly 0.5 cm2 . N=