The Magnetic Field Part 1, Solution Manual

The magnetic field (a) Motion in a B field p2 |dr| |dp| = |p| |r| =⇒ v dt qvB dt = p R =⇒ R = p γmv = . qB qB (b) (418) If B is uniform, then Eq. (417) actually proves that the particle travels in a circle, because it gives the radius of curvature at any point as R = γmv/qB. Since v is constant (because the magnetic force is always perpendicular to the velocity), we see that R is constant, which means that the path is a circle. Second solution: We can also calculate R in a more mathematical way. The Lorentz-force law F = qv × B combined with Newton’s third law F = dp/dt gives d(γmv) = qv × B =⇒ dt dv q = v × B. dt γm (419) Note that we are in fact allowed to take the γ outside the derivative because we know that the speed v is constant. Assume that B is uniform. Let the motion be in the x-y plane, with the magnetic field pointing in the z direction. Then v = (vx , vy , 0) and B = (0, 0, B). So v × B = B(vy , −vx , 0). The x and y components of Eq. (419) can then be written as qB dvx = vy dt γm r1 (417) The time to complete one revolution is 2πR 2π γmv 2πγm t= = = . v qB qB v and dvy qB =− vx . γm dt p1 r2 First solution: The magnitude of the magnetic force is F = qvB, so the magnitude of the change in p during a short time dt is dp = F dt = qvB dt. The momentum itself is p = γmv. Fig. 106(a) shows the r and p vectors at two nearby times. In Fig. 106(b) the angle θ is the same in the two triangles, because each p is perpendicular to the corresponding r. So from similar triangles we have (420) dp p2 θ p1 r2 dr θ r1 Figure 106 Taking the derivative of the first of these equations, and then substituting in the value of dvy /dt from the second, gives ( ) qB 2 d2 v x =− vx . dt2 γm (421) This is a simple-harmonic-oscillator type equation, for which the general solution takes the form, qB vx (t) = A cos(ωt + ϕ), where ω = . (422) γm The first of the equations in Eq. (420) then quickly gives vy (t) = −A sin(ωt + ϕ). A and ϕ are arbitrary constants, determined by the initial conditions. However, if the momentum p = γmv is given, then vx and vy must each have an amplitude of p/γm. Hence A = p/γm. The period is 2π/ω = 2πγm/qB, in agreement with the result in part (a). Integrating vx (t) and vy (t) to find x and y gives (up to arbitrary additive constants, which only affect the position of the center of the circle) ( ) ) A( x(t), y(t) = sin(ωt + ϕ), cos(ωt + ϕ) . ω (423) This describes a circle with radius R = A/ω = (p/γm)/(qB/γm) = p/qB, in agreement with the result in part (a). Proton in space The speed is essentially c, so from Exercise 6.29 the radius of curvature is R= γmv 107 (1.67 · 10−27 kg)(3 · 108 m/s) p = = = 1.0 · 1017 m. qB eB (1.6 · 10−19 C)(3 · 10−10 T) (424) (It was fine to set the factor of v in the numerator equal to c, but we of course can’t set v = c inside the γ factor!) The time to complete one revolution is C B 2I t= I A I (425) Field from three wires BC BA 2πR 2π(1.0 · 1017 m) = = 2.1 · 109 s ≈ 70 years. v 3 · 108 m/s At point P1 at the center of the square, the magnetic fields due √ √ to wires A and C in Fig. 107 cancel. The field due to B is µ0 (2I)/2π(d/ 2) = 2 µ0 I/πd, directed diagonally down to the left, as shown. BB BB BA BC Figure 107 At point P2 √ at the lower right-hand corner, the field due to B is half of the field at P1 , so it is µ0 I/ 2 πd, directed diagonally down to the left, as shown. The field due to A is µ0 I/2πd, directed upward. The √ field due to C is µ0 I/2πd, directed rightward. The sum of these two fields is µ0 I/ 2 πd, directed diagonally up to the right. The vector sum of all three fields is therefore zero at P2 . Oersted’s experiment The compass needle initially points in the direction of the earth’s magnetic field, which has strength 0.2 gauss (in the horizontal direction). In Oersted’s experiment, the wire runs parallel to the initial orientation of the needle. If the needle ends up rotated by 45◦ , the magnetic field from the wire must be 0.2 gauss in the perpendicular direction. In other words, we have a current-carrying wire that produces a magnetic field of 2 · 10−5 T at a distance of about 2 cm. Therefore, B= µ0 I 2πr =⇒ I = 2π (0.02 m)(2 · 10−5 T) 2πrB = = 2 A. µ0 4π · 10−7 kgC2m (426) Force between wires The magnetic field due to one of the wires in Fig. 5.1(b), at the location of the other, is ( m) 4π · 10−7 kg µ0 I C2 (20 A) = B= = 8 · 10−5 T. (427) 2πr 2π(0.05 m) The force per unit length on each wire is then IB = (20 A)(8·10−5 T) = 1.6· 10−3 N/m, and it is repulsive. Torque on a loop From Eq. (6.17) the force on a small piece of the loop is dF = I dl × B. The ẑ component of B produces a force in the plane of the loop, which contributes nothing to the torque around the x axis (or the y axis) and can therefore be ignored. (As an exercise, you can also show that it produces no torque around the z axis.) So for the present purposes we have B = ŷBy . Now, dl × ŷBy = ẑ dl By sin θ, where θ is the angle that dl makes with the y axis. But dl sin θ is simply the dx span of the little interval dl . Hence dFz = IBy dx. The torque about the x axis is then dNx = y dFz = IBy y dx. We haven’t been keeping track of the signs, but this result for dNx does indeed have the correct sign according to the right-hand-rule convention. ∫ We must integrate dNx around the entire loop to find the total torque. But loop y dx equals the area a of the loop. This is true because in Fig. 108 the product y dx for segment A equals the area of the thin rectangle up to A, whereas the product y dx for segment B equals the negative (since dx is negative) of the area of the thin rectangle up to B. The sum of these signed areas is the area of the rectangle from B to A. Adding up all such rectangles gives the area a of the whole loop. This works even if B is below the x axis; y is now negative, so the areas add, which again yields the entire area of the thin rectangle. We therefore have Nx = IaBy . Physically, this Nx component arises because the relevant force on the loop points in the +ẑ direction for the top part (the part with larger values of y), and in the −ẑ direction for the bottom part. The loop will therefore tend to spin around an axis pointing in the x direction. This physical reasoning makes it fairly clear that the loop won’t tend to spin around an axis pointing in the y direction (assuming B has no x component). Mathematically, dNy for our loop equals ∫ x dFz = IBy x dx. So Ny involves the integral x dx around the loop, which you can quickly show is zero; the upper and lower parts of the loop now exactly cancel instead of only partially canceling. The total torque therefore points only in the x direction, and it equals N = x̂IaBy . If we define the magnetic moment as m = Ia = Ia(−ẑ), then we can write N in the general form of N = m × B because m × B = (−Iaẑ) × (ŷBy + ẑBz ) = −IaBy ẑ × ŷ + 0 = x̂IaBy , (428) as desired. The minus sign in the relation a = −aˆ z comes from the given orientation of the current, along with the right-hand convention. Since N = m × B is a vector relation, its validity can’t depend on our specific choice of coordinate system. So if it y A B dx Figure 108 x it true for a particular choice of axes (as we just demonstrated), then it must be true for any choice. Let’s now look at the net force on the loop. If B is uniform over the loop, then the net force is (∫ ) ∫ ∫ dl × B. (429) dF = I dl × B = I loop ∫ But dl = 0 because we have a closed loop. The net force is therefore zero. Note that this result holds even if the loop isn’t planar. Determining c If we follow the wires around in Fig. 6.42, we see that the voltage across both capacitors is equal to E0 cos 2πf t. Let’s first determine the magnetic force between the rings. Neglecting the inductance (the subject of Chapter 7) and resistance of the two rings and leads, the charge on capacitor C2 at any time t takes the form, Q2 = E0 (cos 2πf t)C2 . The current through C2 is then I = dQ2 /dt = −2πf E0 C2 sin 2πf t, with positive defined as flowing into the left terminal of C2 , or equivalently out of the right terminal and into the rings. The two rings are in series, so this current flows through each. If h ≪ b we are justified in computing the magnetic force between the rings as if they were parallel straight wires, with force per unit length µ0 I 2 /2πh. The length is 2πb, so the magnetic force pulling the upper ring down (note that the currents are in the same direction, so the force is attractive) is Fm = µ0 (2πf E0 C2 )2 b µ0 I 2 (2πb) = sin2 2πf t. 2πh h (430) The time average of sin2 2πf t is 1/2, so the average attractive magnetic force between the rings is 2µ0 π 2 f 2 E02 C22b Fm = . (431) h Now let’s determine the electric force between the capacitor plates. As mentioned above, the potential difference between the plates is E0 cos 2πf t. The field between the plates is therefore E = (E0 cos 2πf t)/s. The downward electric force on the upper plate can be obtained in various ways. From Section 3.7 the force is the energy density times the area, which gives Fe = (ϵ0 E 2 /2)(πa2 ). Alternatively, the force is the charge times the average of the fields on either side of the plate, namely E/2 (see Section 1.14), which gives (σπa2 )(E/2). (Equivalently, the force is the charge times the field from the other plate.) These two expressions agree because σ = Eϵ0 . Using the above expression for E, and noting that the time average of cos2 2πf t is 1/2, we find the average attractive electric force between the rings to be Fe = )2 ( ϵ0 πa2 E0 cos 2πf t s 2 =⇒ F e = ϵ0 πa2 E02 . 4s2 (432) We can eliminate s in favor of the capacitance C1 , which is given by Eq. (3.15) as C1 = ϵ0 πa2 /s. So 1/s2 = C12/ϵ02 π 2 a4 , and the electric force becomes Fe = E02C12 . 4πϵ0 a2 (433) When the forces are balanced (which might be brought about by varying C2 ), we have Fe = Fm =⇒ =⇒ =⇒ E 02 C12 2µ0 π 2 f 2 E02C22b = 4πϵ0 a2 h 1 8π 3 a2 bf 2 C22 = hC12 µ0 ϵ 0 ) ( )1/2 ( C2 1 b f, √ = (2π)3/2 a h µ0 ϵ 0 C1 (434) as desired. Note that a constant voltage wouldn’t be useful here, because the capacitors would quickly reach their maximum charge, which would mean the current would be zero. The alternating voltage allows there to be (except at discrete moments during each cycle) both a nonzero charge on the capacitor C1 and a nonzero current in the rings. Let’s see what some reasonable numbers give. If a = 0.1 m, b/h = 10, and C2 /C1 = √ 106 , we find that the righthand side equals 1/ µ0 ϵ0 = c = 3 · 108 m/s when f = 60.2 s−1 . So given all the other parameters, we can determine c by sweeping though frequency values until we find the f (which is 60.2 s−1 in the present case) that makes things balance. If the current rings consist of N turns each, this magnifies the magnetic force by a factor N 2 (the magnetic field is N times as large, and there are N times as many loops that it acts on), which decreases the necessary value of f (or C2 /C1 , etc.) by a factor N . Field at different radii The radius is 2 cm, so 1/4 of the cross-sectional area, and hence current (so 2000 A), is enclosed within r = 1 cm. The current enclosed in both the r = 2 cm and r = 3 cm cases is 8000 A. So we have ) ( 4π · 10−7 kgCm (2000 A) µ0 I 1 2 = 0.04 T, B1 = = 2π (0.01 m) 2πr1 ) ( 4π · 10−7 kgCm (8000 A) µ0 I 2 2 = 0.08 T, B2 = = 2π (0.02 m) 2πr2 ) ( −7 kg m 4π · 10 µ0 I 3 C2 (8000 A) = 0.0533 T. (435) B3 = = 2π (0.03 m) 2πr3 These fields are 400, 800, and 533 gauss, respectively. Off-center hole The given setup is equivalent to the superposition of a complete solid rod with current flowing into the page plus a smaller rod (where the hole is) with current flowing out of the page. If the two current densities are equal and opposite, then there will be zero current in the hole, in agreement with the given setup. Given the ratio of the areas of the two circular cross sections, currents of 1200 A into the page and 300 A out of the page will yield the given 900 A into the page. The large rod produces zero field on its axis, so the desired field is due entirely to the smaller rod with 300 A coming out of the page. The magnitude of the field is ) ( µ0 I 4π · 10−7 kgC2m (300 A) B= = = 0.003 T, (436) 2π (0.02 m) 2πr or 30 gauss, and it points to the left. A more remarkable fact (see Exercise 6.38) is that the field is 30 gauss pointing to the left not only at P but everywhere inside the cylindrical hole. Uniform field in off-center hole Inside a solid cylinder (with no cavity), the current enclosed within a radius r is I(r2 /R2 ), because area is proportional to length squared. So Ampere’s law gives B(2πr) = µ0 Ir2 /R2 =⇒ B = µ0 Ir/2πR2 . Assuming that the z axis (and hence current) points out of the page, the magnetic field points in the tangential direction, θ̂, which can be written as ẑ × r̂. If we combine the magnitude r with the unit vector r̂ to make the full vector r, the complete vector form of the field can be written as B = (µ0 I/2πR2 )ẑ × r. And since I/πR2 equals the current density J, we can write this as B = (µ0 J/2)ẑ × r. r2 r1 a R Figure 109 The given setup can be considered to be the superposition of the given solid rod with radius R and another smaller rod with current flowing in the opposite direction with the same current density J. The current densities will then cancel in the region of the smaller rod, creating the desired cavity. Consider a point inside the cavity, at position r1 with respect to the center of the given rod, and at position r2 with respect to the center of the cavity, as shown in Fig. 109. Using the above expression for B, the field at this point is the sum of the fields from the two rods we are superposing. Since the currents flow in opposite directions, the net field is µ0 J µ0 J µ0 J µ0 J ẑ × r1 − ẑ × r2 = ẑ × (r1 − r2 ) = ẑ × a. 2 2 2 2 (437) All of the quantities in this expression are constants, so the field is uniform, as desired. It is perpendicular to a and proportional to |a|. If a = 0, the field is zero, as expected, because the final object is a thick annular ring. The field doesn’t depend on the radius of either rod, as long as one rod is contained inside the other. Constant magnitude of B If Ir is the current inside radius r, then Ampere’s law gives B · 2πr = µ0 Ir =⇒ B = µ0 I r . 2πr (438) If we want B to be independent of r, then we need Ir to be proportional to r. Ir is found by integrating the current density J(r): ∫ ∫ r Ir = J da = (439) J(r′ ) · (2πr′ dr′ ). 0 It is easiest to guess and check the form of J(r′ ). If J(r′ ) is proportional to 1/r′ , then it takes the form of J(r′ ) = α/r′ , so ∫ r Ir = (440) (α/r′ )(2πr′ dr′ ) = 2παr, 0 as desired. The field is then B= µ0 (2παr) µ0 I r = = µ0 α. 2πr 2πr (441) The above “1/r” result for the current density is the same result that holds for the charge density in the case of the electric field due to a charged cylinder or sphere. In both of these cases the electric field is independent of r if the density ρ is proportional to 1/r. Note that even though the current density diverges at r = 0, the actual current does not. There is a finite amount of current in any cross section with radius r, and it is given (by construction) by Ir = 2παr. Any ring (at any radius) with thickness dr contains the same amount of current, dI = 2πα dr. We can also solve this exercise by using the differential form of Ampere’s law, ∇ × B = µ0 J. Since B points tangentially and has a uniform value, it can be written as B = B0θ̂. Equation F.2 in Appendix F then gives ∇×B= 1 ∂(rB0 ) B0 ẑ = ẑ. r r ∂r (442) Setting this equal to µ0 Jẑ gives J = B0 /(µ0 r), consistent with the 1/r dependence we found above. The factor of B0 /µ0 here equals the α from above. The pinch effect If the conduction electrons are forced closer to the axis, there will be uncompensated negative charge near the axis. This will generate an inward radial electric field E that pushes outward on the electrons, preventing further constriction when the outward electric force balances the inward magnetic force, that is, when eE = evB =⇒ E = vB. The magnetic field at radius r is Br = µ0 Ir /2πr, where Ir is the current contained within radius r. Assuming no redistribution of the charge, Ir is given by Ir = πr2 J , where J = nev is the current density (n is the number of electrons per unit volume, and v is the drift velocity). The B field is therefore Br = µ0 (πr2 nev)/2πr = µ0 rnev/2. Suppose that the cloud of electrons at radius r is squeezed inward by a small distance ∆r. The cylinder of radius r will now contain, per unit length, an excess of negative charge in the amount of ∆λ = (ne)(2πr ∆r); this is the volume charge density times the cross-sectional area. This causes an inward electric field equal to Er = ∆λ/2πϵ0 r = ne ∆r/ϵ0 . The condition for equilibrium in then (using µ0 ϵ0 = 1/c2 ) Er = vBr =⇒ ne ∆r µ0 rnev =v 2 ϵ0 =⇒ ∆r µ0 ϵ 0 v 2 v2 = = 2. 2c r 2 (443) In solid conductors we always have v/c ≪ 1. In metal conduction, v/c is seldom much greater than 10−10 , so (∆r)/r ≈ 10−20 is too small to detect. In highly ionized gases, however, the “pinch effect,” as it is called, can be not only detectable but important. If the effect were large enough to measure, a Hall probe in the spirit of Fig. 6.33 could be used, with one lead connected to the axis (by drilling a thin tube in the rod), and the other lead connected to the surface of the rod. If v ≈ 10−3 m/s and B ≈ 1 T, the resulting E ≈ 10−3 V/m would be large enough to generate a measurable voltage difference. Integral of A, flux of B Using Stokes’ theorem, along with B = ∇ × A, we have ∫ ∫ ∫ B · da = Φ, A · ds = ∇ × A · da = C S (444) S as desired. This relation is similar to Ampere’s law because the differential form of that law, µ0 J = ∇ × B, takes the same form as the above B = ∇ × A relation. Finding the vector potential Since B = ∇ × A, we want ∂Az ∂Ay − = 0, ∂y ∂z ∂Ax ∂Az − = 0, ∂z ∂x ∂Ay ∂Ax − = B0 . ∂y ∂x (445) From inspection, a few choices for A that satisfy these equations are A = (0, B0 x, 0), or (−B0 y, 0, 0), or (−B0 y/2, B0 x/2, 0). In general, any vector of the form (−ay, bx, 0) works if a + b = B0 . And even more generally, adding on any vector with zero curl also works. Vector potential inside a wire Since area is proportional to r2 , the current contained within a radius r is Ir = Ir2 /r20 . The magnitude of the magnetic field at radius r is then B(r) = µ0 I r µ0 Ir , = 2πr 2πr20 (446) and it points in the positive θ̂ direction. The θ̂ vector equals (−y/r, x/r, 0) because this vector has length 1 and has zero dot product with the radial vector (x, y, 0). So the Cartesian components of B are µ0 Iy y Bx = − B = − , r 2πr20 and By = µ0 Ix x B= . r 2πr20 (447) The magnetic field associated with the potential A = A0ẑ(x2 + y 2 ) is B = ∇ × A = x̂ ∂Az ∂A − ŷ z = 2A0 yx̂ − 2A0 xŷ. ∂x ∂y (448) This agrees with the B in Eq. (447) if A0 = −µ0 I/4πr02 . Alternatively, in cylindrical coordinates we have A = A0ẑr2 . From Eq. (F.2) in ˆ Appendix F the associated magnetic field is B = ∇ × A = −(∂Az /∂r)θ̂ = −2A0 rθ. Comparing this with the B in Eq. (446), which points in the positive θ̂ direction, we find A0 = −µ0 I/4πr02 , as above. Since A0 is negative, A points in the direction opposite to the current (which points in the positive ẑ direction). You might be wondering how this can be, in view of the fact that Eq. (6.44) seems to say that A points in the same direction as J. The answer is that we can add an arbitrary constant to the A in Eq. (6.44), and it will still yield the same value of B = ∇ × A. Adding on a sufficiently large vector pointing in the negative ẑ direction will make A point opposite to J. Line integral along the axis The magnetic field on the axis is Bz = µ0 Ib2 /2(b2 + z 2 )3/2 , so the given line integral is (using the integral table in Appendix K) ∞ ∫ ∫ ∞  µ0 Ib2 ∞ µ0 Ib2 2 µ0 Ib2 dz z  = Bz dz = = = µ0 I,  2 2 2 1/2 2 2 3/2 2 2 b (b + z ) 2 b2 −∞ −∞ (b + z ) −∞ (449) as desired. If you want, you can derive this integral with a trig substitution, z = b tan θ. 137 r To see why the integral along the axis should indeed be equal to µ0 I, consider the closed path shown in Fig. 110, which involves a semicircle touching the points z = ±r. Assume that r ≫ b. Along the z axis, Bz behaves like 1/z 3 for z ≫ b. And |B| also behaves like 1/r3 along the (large) semicircle. Accepting that this is true (see below), then since the length of the semicircle is proportional to r, the line integral along the semicircle is at least as small (in order of magnitude) as r/r3 = 1/r2 , which goes to zero as r → ∞. We can therefore ignore the return semicircular path. So the line integral along the whole loop (which encloses a current I) equals the line integral along the z axis, in the r → ∞ limit. r Figure 110 Let’s now argue why |B| behaves like 1/r3 for large r. Consider the point at the “side” of the semicircle in Fig. 110. In order of magnitude, the field at this point, due to the ring, is the same as the field due to a square with side b. But the field due to the square has contributions from two opposite sides (the sides perpendicular to the r̂ vector) that nearly cancel, because the current moves in opposite directions along these sides. The Biot-Savart law says that each side gives a contribution of order 1/r2 . Taking the difference of these contributions is essentially the same as taking a derivative, and the derivative of 1/r2 is proportional to 1/r3 , as desired. Additionally, the two sides parallel to the r̂ vector also happen to produce a contribution of order 1/r3 ; see Problem 6.14. At points in between the axis and the “side” point on the semicircle, there will be various angles that come into play. But these simply bring in factors of order 1 that morph the 1/z 3 result on the axis to the 1/r 3 result at the side point, so they don’t change the general 1/r3 result. dl Consider a small piece of the wire at angle θ, subtending an angle dθ, as shown in Fig.111. If r is the distance from a given point P to the small piece, then Fig. 112 shows that the length of the piece is dl = r dθ/ cos θ. But r equals b/ cos θ, so dl = b dθ/ cos2 θ. (This can also be obtained by taking the differential of l = b tan θ.) In the Biot-Savart law, the cross product between dl and r̂ brings in a factor of sin ϕ, which is the same as cos θ. If the current points rightward, then we have (with ẑ pointing out of the page) B = = θ b φ Field from an infinite wire ∫ µ0 I dl × r̂ = r2 4π ∫ I Figure 111 oP r dθ t θ π/2 (b dθ/ cos2 ) cos θ ẑ (b/ cos θ)2 −π/2  π/2 ∫ π/2  µ0 I µ I µ I cos θ dθ = ẑ 0 sin θ = ẑ 0 , ẑ 4πb 2πb 4πb −π/2 −π/2 µ0 I 4π P dθ r rdθ dl = ____ cosθ (450) Figure 112 B which agrees with the standard result obtained more much quickly via Ampere’s law. Field from a wire frame (a) In Fig. 113 the contributions to the field at P from the diagonally opposite edges AB and EF cancel, as do those from CD and GH. The pair BC and F G produces a field that points in the y direction at P , as does the pair HA and DE. So the total magnetic field at P points in the positive y direction. (b) Imagine superposing on the given setup the current in the two horizontal squares shown in Fig. 114. The currents along six of the edges cancel, and we end up with the desired square loop of current. But the field at P due to the two squares in Fig. 114 is zero, due to a symmetry argument. (Imagine rotating the setup by 180◦ around either the x or y axis. The setup is unchanged, so the magnetic field must point along both the x and y axes. The zero vector is the only vector with A z C G D P x y H E F Figure 113 B A z C D G F P x y H E Figure 114 this property.) Or you can just note that diagonally opposite edges combine to give zero field at P , as we saw in part (a). Therefore, since we added on zero field at P , the field at P in the original setup must be the same as the field at P in the case of the single square loop. Field at the center of an orbit The time for one revolution is t = 2πr/v, so the average current is I = e/t = ev/2πr. From Eq. (6.54) the field at the center of the orbit is therefore ) ( m −19 C)(0.01 · 3 · 108 m/s) µ0 I 4π · 10−7 kg µ0 ev C2 (1.6 · 10 B= = = = 4.8 T. (451) 4πr2 4π(10−10 m)2 2r Fields from two rings The Biot-Savart law is dB = (µ0 /4π)I dl × r̂/r2 . Consider corresponding pieces of the two rings that subtend the same angle dθ. The dl for the larger piece is twice the dl for the smaller piece. And the I for the larger ring is also twice the I for the smaller ring, because I is proportional to the speed of the ring, which in turn is proportional the radius, because the ω’s are the same. These two powers of 2 in the numerator cancel the two powers of 2 in the r2 in the denominator, so the fields at the centers of the two rings are the same. This reasoning works for any ratio of ring sizes, of course. In terms of the various parameters, you can show that the field at the center is B = µ0 λω/2, which is independent of r, as we just showed. Field at the center of a disk Consider a ring with radius r and thickness dr. The effective linear charge density along the ring is dλ = σ dr. The speed of all points on the ring is v = ωr, so the current in the ring is dI = (dλ)v = (σ dr)(ωr). From the Biot-Savart law, a small piece of the ring with length dl produces a dB field at the center that points perpendicular to the ring and has magnitude (µ0 /4π)I dl/r2 . Integrating over the whole ring turns the dl into 2πr, so the field at the center due to the ring is (µ0 /4π)(σωr dr)(2πr)/r2 = µ0 σω dr/2. Integrating over r (that is, integrating over all the rings in the disk) turns the dr into an R, so the field at the center equals µ0 σωR/2. It points perpendicular to the disk, with the direction determined by the righthand rule. Hairpin field Each of the two straight segments contributes half the field of an infinite wire. (The contributions do indeed add and not cancel.) The semicircle contributes half the field of an entire ring at the center, which is given by Eq. (6.54). The desired field therefore points out of the page and has magnitude ) ( µ0 I 1 µ0 I 1 µ0 I 1 1 µ0 I = (0.409) + + . (452) B =2· = 2 2πr 2 2r r r 2π 4 Current in the earth If the current is essentially all located on the equator of the core, then we can use the expression for the B field due to a ring. From Eq. (6.53) the field along the axis of the ring is Bz = µ0 Ib2 /2(b2 + z 2 )3/2 . In the situation at hand, b = R/2 and z = R, where R is the radius of the earth. So we have µ0 I(R/2)2 µ0 I . Bz = ( )3/2 = √ 5 5R 2 (R/2)2 + R2 (453) If this field equals 0.5 gauss = 5 · 10−5 T, then I must be (taking the earth’s radius to be R = 6 · 106 m) I= √ √ 5 5 RBz 5 5(6 · 106 m)(5 · 10−5 T) = = 2.7 · 109 A, m µ0 4π · 10−7 kg C2 (454) which is huge. For comparison, the peak current in a bolt of lightning is on the order of 105 A. y ( x ,y ) Right-angled wire Consider the contribution from the positive x-axis part of the wire. In the notation of Fig. 115, the Biot-Savart law gives ∫ ∫ ∫ µ0 Iẑ ∞ sin θ dx′ µ0 Iẑ ∞ y dx′ µ0 I dl × r̂ = = . (455) B= 2 2 r r 4π r3 4π 0 4π 0 If y is positive (or negative), then this B points in the positive (or negative) ẑ direction, that is, out of (or into) the page. Writing r in terms of the Cartesian coordinates gives (with x′′ ≡ x′ − x, and using the integral table in Appendix K) Bz = = ∫ µ0 Iy ∞ dx′′ dx′ = 2 + x′′2 ]3/2 2 + (x′ − x)2 ]3/2 [ y 4π [y −x 0 ) ( ∞  µ0 Iy µ0 I 1 x′′ x  = + √ .  2 2 ′′2 1/2 4π y [y + x ] 4π y y x2 + y 2 −x µ0 Iy 4π ∫ I r θ dl x I Figure 115 ∞ (456) If you write out the contribution from the positive y axis, you will find that it takes the same form, except with x and y reversed. We therefore end up with the desired result. We can check a limit: If x ≫ y, we should end up with the field at a distance y from an infinite wire lying along the x axis. And indeed, if x ≫ y, two of the terms in the result in Eq. (6.98) are negligible, and two of the terms are essentially equal to (µ0 I/4π)(1/y). So we end up with µ0 I/2πy, as expected. x (x,y) y Superposing right angles An infinite straight wire carrying current in the positive direction along the x axis is the superposition of the two right-angled wires shown in Fig. 116. So we need to find the fields at (x, y) due to these two wires. The field from the right wire is just the field derived in Exercise 6.52. The field from the left wire is the same as the field at the point shown in the setup in Fig. 117 (we have simply rotated the setup in Fig. 116 by 90◦ ). But this field can be obtained by taking the result from Exercise 6.52 and setting the x value equal to y, and the y value equal to −x, as shown. So the field due to the left right-angled wire is ) ( 1 y −x µ0 I 1 √ + + + √ . (457) Bz = y −x −x y 2 + x2 4π y y 2 + x2 When we add this to the field from the right wire given in Eq. (6.98), the last three terms in Eq. (457) cancel. We therefore end up with Bz = (µ0 I /4π)(2/y) = µ0 I/2πy, as desired, because y is the distance from the wire (the x axis). Figure 116 -x y Figure 117