The Magnetic Field Part 2, Solution Manual

Force between a wire and a loop Consider a little segment in the right-hand side of the square. The current points into the page, and the magnetic field due to the infinite straight wire has magnitude B1 = µ0 I1 /2πR and points down to the left, as shown in Fig. 118. From the righthand rule, the force qv × B on the charges in the current points up to the left, as shown (toward the infinite wire; parallel currents attract). In the left-hand side of the square, the current points out of the page, and the magnetic field due to the infinite wire points up to the left, as shown. The force qv×B on the charges in the current now points down to the left (away from the infinite wire; antiparallel currents repel). The vertical components of the preceding two forces cancel, but the leftward components add. So the net force is leftward, as desired. You can quickly show that the net force on each of the other two sides of the square is zero. wire (current I1 into page) β z Bz F current into page current out of page B l/2 loop Bx R Bz B F Bx Figure 118 In short, it is the vertical component of B that matters, because this component changes sign from the right half to the left half of the square. And the direction of the square’s current into and out of the page also changes sign. So these two negative signs cancel in qv × B, yielding a net leftward force. In contrast, the horizontal component of B does not change sign, so the negation of the current causes a negation of the vertical force. The net vertical force is therefore zero. Quantitatively, the general form of the force on a wire is F = IBℓ. The “B” we are concerned with here is the vertical component, which is B sin β. The force comes from two sides, so the total horizontal force is ) ( µ0 I 1 I 2 ℓ 2 µ0 I1 ℓ/2 · , (458) ℓ= F = 2I2 (B1 sin β)ℓ = 2I2 2πR2 2πR R where we have used sin β = (ℓ/2)/R. The above reasoning shows where the two factors of R in the denominator come from. One comes from the distance to the wire, and the other comes from the fact that the B field becomes more horizontal (which means that the vertical component decreases) as R gets large. We weren’t concerned with torques in this exercise, but from looking at the vertical forces on the left and right sides (which come from the horizontal component of B), it is clear that there is a torque on the square. It will rotate counterclockwise when viewed from the side. This is consistent with conservation of angular momentum, because the straight wire will gain angular momentum (relative to, say, an origin chosen to be the center of the square) as it moves to the right. This angular momentum will have a clockwise sense, consistent with the fact that the total angular momentum of the system remains constant. Helmholtz coils Let the symmetry axis of the setup be the z axis, and let the centers of the rings be located at z = ±b/2. If the currents in the rings are equal and point in the same direction, then from Eq. (6.53) the field along the axis at position z is given by Bz (z) ∝ 1 1 + . [a2 + (z + b/2)2 ]3/2 [a2 + (z − b/2)2 ]3/2 (459) If we expand this function in a Taylor series around z = 0, the first derivative and all other odd derivatives are zero at z = 0, because Bz (z) is an even function of z, due to the symmetry of the setup. So the function will be most uniform near z = 0 if the second derivative is zero there. The deviations will then be of order z 4 . That is, the Taylor series will look like Bz (z) = Bz (0) + Cz4 + · · · . Differentiating the first term in Eq. (459) twice and evaluating the result at z = 0 yields  3(b2 − a2 ) 5(z + b/2)2 − [a2 + (z + b/2)2 ]  . (460) = 2 3  2 2 7/2 [a + (z + b/2) ] [a + b2 /4]7/2 z=0 The second derivative of the second term in Eq. (459) simply involves replacing b/2 with −b/2, so we end up with the same result, because there are no odd powers of b in Eq. (460). We therefore see that the second derivative is zero at z = 0 if a = b. You can show that if a = b, the field halfway from z = 0 to the plane of each ring (that is, at z = ±b/4) is only 0.4% smaller than the field at z = 0. And at z = ±b/8 the field is only 0.03% smaller. Two coils arranged with a = b are called Helmholtz coils. A continuity argument shows why there must exist a point where the second derivative of Bz (z) equals zero. If the rings are far apart (for example, if b = 4a), then the plot of Bz consists of two bumps, as shown in Fig. 119. But if the rings are close together (for example, if b = a/4), then they act effectively like one ring with twice the current, so there is just one bump. The second derivative at z = 0 is positive in the former case, and negative in the latter, so somewhere in between it must be zero. B (in units of µ0I/2a) B 2.0 1.5 ring with radius a 1.0 0.5 -4a -2a B 2.0 2.0 1.5 1.5 1.0 1.0 0.5 0 2a (b = 4a) Figure 119 4a z -4a -2a 0 (b = a) 0.5 2a 4a z -4a -2a 2a (b = a/4) 4a z Field at the tip of a cone Consider a circular strip with width dx, a slant-distance x from the tip. The velocity of any point in this strip is v = ω(x sin θ). The amount of charge in the strip that passes a given point during time dt is dq = σ(dx)(v dt) = σ(dx)(ωx sin θ) dt. The current in the strip is therefore I = dq/dt = σωx sin θ dx. (Equivalently, you can use the general result I = λv, where λ = σ dx is the effective linear charge density of the ring.) dB θ θ x L dx length ds ω Figure 120 From the Biot-Savart law, a small piece of the strip with length ds at the location shown in Fig. 120 produces a dB field at the tip that points up to the right, with magnitude (µ0 /4π)I ds/x2 . When we integrate over the whole strip, the horizontal components of the dB’s cancel, and we are left with only a vertical component. This brings in a factor of sin θ. For a given strip, the ds in the Biot-Savart law integrates up to the length of the strip, which is s = 2π(x sin θ). The contribution to the (vertical) field from a given strip at slant-distance x, with width dx, is therefore ẑ 1 µ0 Is µ0 (σωx sin θ dx)(2πx sin θ) sin θ = ẑ µ0 σω sin3 θ dx. sin θ = ẑ x2 4π x2 4π 2 (461) Integrating from x = 0 to x = L simply gives a factor of L, so the field at the tip is 1 B = ẑ µ0 σωL sin3 θ. 2 (462) If θ = 0 we correctly obtain zero field. If θ = π/2 we obtain B = ẑµ0 σωL/2. In this case we just have a flat disk with radius L, and this is indeed the field at the center; see Exercise 6.49. To check that the units of B are correct, we can compare it with the B due to a wire, which is µ0 I/2πr. And indeed, σωL correctly has the same units as I/r. Note that the result in Eq. (461) is independent of x, so all rings with the same thickness dx give the same contribution to the field. The reason for this is that the larger a ring is, the larger the current I and length s are, and these effects cancel the effect of the x2 in the denominator of the Biot-Savart law. Note also where the three factors of sin θ come from. For given values of the other parameters, a larger θ means a larger velocity (and hence current), a larger circumference s, and a larger vertical component of each of the dB’s. A rotating cylinder Eq. (6.57) gives the magnetic field inside an infinite solenoid as B = µ0 nI, where n is the number of turns per unit length. The surface current density (per unit length) is J = nI, so we can write the field as B = µ0 J . What is the current density in our rotating cylinder? The amount of charge that passes a given segment of length ℓ on the cylinder in a time dt is dq = σℓ(v dt). The current per unit length (that is, the surface current density) is therefore J = (1/ℓ)(dq/dt) = σv . In terms of the angular frequency, J equals σωR. To find the field inside the rotating cylinder, we simply need to replace the current density J = nI in the original solenoid formula with the present current density J = σωR. This yields a field of B = µ0 σωR. Rotating cylinders We must first find the charge per unit length (in the direction of the axis), λ, on the inner cylinder. The electric field between the cylinders ∫ is E(r) = λ/2πϵ0 r. The magnitude of the potential difference is therefore ϕ = E dr = (λ/2πϵ0 ) ln(r2 /r1 ), where r1 and r2 are the radii of the inner and outer cylinders, respectively. So ) ( s2 C2 (1.5 · 104 V) 2π 8.85 · 10−12 kg 2πϵ0 ϕ m3 λ= = 2.9 · 10−6 C/m. (463) = ln(4/3) ln(r2 /r1 ) If the inner cylinder rotates at 30 rev/sec, then it is a solenoidal surface with a current density equal to J = (30 s−1 )(2.9 · 10−6 C/m) = 8.7 · 10−5 C/(s m), because in 1 meter of the cylinder, each revolution carries a charge of 2.9 · 10−6 C. Within the inner cylinder the field is therefore B = µ0 J = 1.09 · 10−10 T. (The continuum limit of the B = µ0 nI expression is B = µ0 J .) If the cylinder rotates counterclockwise as we look along the axis, and if the inner cylinder is the positive one, the field points out of the page (toward us). Outside the inner cylinder (that is, for r > r1 ), the field is zero. If both cylinders rotate at 30 rev/sec in the counterclockwise direction, the outer cylinder produces a field of equal strength but opposite direction in its interior (because it has the opposite charge per unit length, assuming the cylinders started neutral before the potential difference was created). The fields therefore cancel inside the inner cylinder. So B = 0 for r < r1 . And B = 0 for r > r2 , of course. But between the cylinders (that is, for r1 < r < r2 ), the field is B = 1.09 · 10−10 T, pointing into the page. Scaled-down solenoid (a) The resistance of the winding in the small solenoid is 10 times that of the large solenoid. This is true because the resistance is given by R = ρL/A, and the small wire is 1/10 as long, with 1/100 the cross-sectional area. So if we apply the same voltage of 120 V to the small solenoid, we get 1/10 the current. This is just what is needed to produce a magnetic field equal to that in the large solenoid, because the field is proportional to nI, and the small coil has 10 times as many turns per unit length, each with 1/10 the current. Equivalently, the surface current density J is the same in the small coil (it has 10 times as many wires per unit length, with 1/10 the current in each), and J is all that matters for a solenoid, because the field inside is B = µ0 J . Symbolically, we have B ∝ nI = n(V/R) = nV /(ρL/A) = (V /ρ)(nA/L). But the quantity nA/L is dimensionless (the units are m−1 m2 /m), so it can’t depend on the length scale of the solenoid. Therefore, if V and ρ are the same in both setups, then B is also the same. (b) The power is P = IV , so it is smaller by a factor of 10 in the smaller solenoid, because V is the same and I is smaller by a factor of 10. But the smaller solenoid has only 1/100 the surface area (because area is proportional to length squared), so it will be harder to keep it cool; any cooling mechanism operates by interacting with the surface of the wire. Zero field outside a solenoid Let the solenoid extend in the z direction. Consider one of the small patches. The current in this patch flows in some direction in the xy plane. That is, the dl vector in the Biot-Savart law lies in the xy plane. (We can slice the patch into many thin strips represented by dl ’s.) The dl vector gets crossed with the r̂ vector directed to the point P . Now, r̂ has a z component, but the r̂ vector associated with the corresponding patch defined by the thin cone on the other side of P has the opposite z component. These components therefore yield canceling contributions to the total magnetic field. So we need only worry about the component of r̂ that lies in the xy plane. Let’s call this vector r̂xy . We need to compute the cross product of dl and r̂xy , both of which lie in the xy plane. (The resulting cross product will therefore point in the z direction, so we have just proved that the B field from the solenoid must be longitudinal.) In general, dl has a component parallel to r̂xy and a component perpendicular to r̂xy . The parallel component yields zero in the cross product dl × r̂xy , so we need only worry about the component perpendicular to r̂xy . In other words, if we project the area of the patch onto the (vertical) plane orthogonal to r̂xy , then the cross product dl × r̂xy remains the same. We can do the same with the other patch in the same cone. We therefore need to compare the Biot-Savart contributions from the two “projected” patches of area defined by a given cone. If the projected patches are distances r1 and r2 from the point P , then their areas are proportional to r21 and r22, because areas are proportional to length squared, and because the patches cut the line from P at the same angle (perpendicular, by construction). Now, if we imagine a small rectangular patch (any patch can be built up from rectangles), the I dl product in the Biot-Savart law is proportional to the area, because I is proportional to the height dh of the rectangle (since I = J dh), and because dh dl is the area of the rectangle. The numerators in the Biot-Savart law for the corresponding patches are therefore proportional to r12 and r22 . These factors exactly cancel the r2 in the denominator of the Biot-Savart law. So the magnitudes of the contributions from the two patches equal. And since the currents flow in opposite directions in the projected patches, the contributions therefore cancel. The entire solenoid can be considered to be built up from small patches subtended by cones, so the external field is zero, as desired. If the solenoid isn’t convex, then a given cone may define 4, 6, 8, etc., patches. But there will still be equal numbers of patches having currents in each direction (which can be traced to the fact that P has the property of being outside the solenoid), so the sum of the contributions will still be zero. plane (top view) Rectangular torus Loop 1 Loop 2 B1 I B2 Figure 121 I Consider two loops of current that are located symmetrically with respect to a plane through the axis of the torus; a top view is shown in Fig. 121. At any point on this plane (inside or outside the torus) the vector sum of the field due to Loop 1 and the field due to Loop 2 is perpendicular to the plane (or is zero). You can check this by looking at the Biot-Savart contributions from corresponding little pieces of the two loops; the components parallel to the plane cancel. In general, the B1 and B2 vectors shown also have components perpendicular to the page, but you can show that these components are equal and opposite. This result is actually true for two similar loops of any (planar) shape carrying equal currents in the same orientation; the cross section of the torus doesn’t have to be rectangular. The same Biot-Savart reasoning involving corresponding little pieces holds. The entire coil can be decomposed into pairs of loops located symmetrically with respect to a given plane. Hence the total magnetic field at any point must be perpendicular to the plane containing that point and the axis (or be zero). In other words, the field points in the circumferential direction. To find the magnitude of the field, we can use Ampere’s law. By symmetry, the magnetic field must have the same magnitude B everywhere on a circle of radius r around the axis. The line integral of B around this circle equals µ0 times the current enclosed. Since B points in the tangential direction, the line integral equals 2πrB . If the circle doesn’t lie inside the torus, the current enclosed is zero. This is true because either the disk defined by the circle doesn’t intersect the torus, in which case the current enclosed is clearly zero; or the disk does intersect the torus, in which case a current of NI passes through the disk in one direction, but another N I also passes through in the other direction. Therefore, B = 0 everywhere outside the torus. On the other hand, if the circle lies inside the torus, the current enclosed is N I , because the disk defined by the circle intersects only the inner boundary of the torus. Therefore, 2πrB = µ0 N I =⇒ B = µ0 N I/2πr inside the torus. This expression for B holds for a torus of any (uniform) cross section. Note that B depends only on r, and not on the “height” inside the torus. In the limit where b − a ≪ a, the curvature of the torus is negligible, so we essentially have an infinite straight solenoid with rectangular cross section. The field should therefore equal µ0 nI (see the solution to Problem 6.19), where n is the number of turns per unit length. And indeed, in the above result, n = N/2πr is the number of turns per unit length (where r is essentially equal to both a and b), so we do obtain B = µ0 nI. F θ2 E θ0 θ1 Creating a uniform field Since 10 milligauss is about 2% of the earth’s field, we need a compensating field of approximately 0.55 gauss that is uniform to about 2% over the region of interest. Let’s try a solenoid 1 meter long and 50 cm in diameter; see Fig. 122. To see whether it meets the requirement, compare the field at the center C with the field on the axis 15 cm from the center, at D. From Eq. (6.56) we have 2 cos θ0 field at C = . field at D cos θ1 + cos θ2 (464) The various angles are given by θ0 = tan−1 (25/50) =⇒ cos θ0 = .8944, θ1 = tan−1 (25/35) =⇒ cos θ1 = .8137, θ2 = tan−1 (25/65) =⇒ cos θ2 = .9333. (465) The ratio of the fields is therefore 1.024. So the deviation is about 2.4%. This is a little too large for comfort, especially as we have no easy way to estimate the deviation at off-axis points such as E. Let’s lengthen the solenoid to 1.2 meters. The denominators in the above arctans are now 60, 45, and 75, respectively, and you can quickly show that the ratio of the fields is now 1.013. We expect the departure from uniformity in the radial direction to be of the same magnitude, roughly, as the variation in the axial direction. But it has the opposite sign; the field strength at E is larger than that at C. This makes sense, because in the limit where the solenoid is very squat, so that it basically looks like a ring, we know that the field increases (without bound, in fact) as we move away from the center toward the circumference. An exact calculation of the field strength very close to the solenoid at F , which involves an elliptic integral, shows it to be 1.4% greater than the field strength at C, in the case of the 1.2 meter solenoid. C D Figure 122 P2 P1 Figure 123 The number of ampere turns, N I, required to make the field of the solenoid at C equal to the earth’s field, 5.5 · 10−5 T, is found from Eq. (6.56). The number of turns per unit length is n = N/(1.2 m), so we have ( ) µ0 N/(1.2 m) I (2 cos θ0 ) = 5.5 · 10−5 T 2 (1.2 m)(5.5 · 10−5 T) =⇒ N I = ( ) m 4π · 10−7 kg C2 (0.923) = 57 ampere-turns. (466) Solenoids and superposition α2 P2 α1 P1 Figure 124 (a) Imagine adding a similar solenoid on the left, as shown in Fig. 123. This exactly doubles the field strength at P2 . But now the field strengths at P2 and P1 are approximately equal, because both points lie well inside a fairly long solenoid, the field at P2 being slightly stronger. Therefore, the original field at P2 must have been slightly more than half the field at P1 . This “more than half” result is consistent with the extreme case where the solenoid is very short, basically just a ring. In this case the fields at P2 and P1 are essentially equal, both taking on the value of the field at the center of a ring, namely µ0 I/2r. So the field at P2 is certainly more than half of the field at P1 . A less elegant way of solving this exercise is to use Eq. (6.56). The field at the center is proportional to 2 cos α1 , and the field at the end is proportional to cos α2 (plus cos 90◦ , which is zero), where these angles are defined in Fig. 124. For small angles, both of these cosines are essentially equal to 1, hence the ratio of 1/2 in the fields. But cos α2 > cos α1 , hence the “more than half.” (b) Let’s assume (in search of a contradiction) that there exists a field line that crosses the line GH with a vertical component, as shown in Fig. 125(a). Imagine flipping the solenoid upside down to obtain the situation in Fig. 125(b), and then reversing the direction of the current (so that it now has the same direction as in Fig. 125(a)) to obtain the situation in Fig. 125(c). Note that the field at the given point on the line GH has a downward component in both figures (a) and (c) (or upward in both, if we had initially drawn it upward). (a) G H (b) (c) I I G I H G H Figure 125 Now join the two semi-infinite solenoids in figures (a) and (c) end to end, thereby creating an infinite solenoid. By superposition, the fields simply add, so we end up with a downward component at the given point along GH. But this is a contradiction, because we know that the field of an infinite solenoid is zero outside the solenoid. We conclude that the field due to the semi-infinite solenoid at the given point must have had zero vertical component. In other words, it was horizontal, as we wanted to show. (c) The argument used in part (a), applied to the semi-infinite solenoid, shows that the axial component of the field, at any point on the end face is exactly B0 /2, where B0 is the (uniform) field throughout the inside the corresponding infinite solenoid. This is true because adding another semi-infinite solenoid simply doubles the axial field at any point on the end face (see the reasoning in part (b)), and cancels the radial field, resulting in a purely axial field. As far as the flux goes, when calculating the flux through the end face, only the axial field component is involved. Therefore, the flux must be exactly half the interior flux. (d) From the reasoning in part (c), a given flux tube that starts with area A far back in the solenoid must flare out as it approaches the end face, so that its cross section there (where the axial field is half as large as the field far back in the solenoid) has area 2A and thus the same amount of flux. (There can be no net flux into or out of the tube, since div B = 0.) In the special case of an axial tube√with circular cross section everywhere, this√tells us that πr21 = 2πr20 =⇒ r1 = 2 r0 . Of course, this holds only if r0 < R/ 2, where R is the radius of the solenoid. Otherwise, the field line exits the solenoid before it reaches the end. Remark: The arguments used in parts (b) and (c) lead to more general statements about the field of a semi-infinite solenoid. Consider two points P and P ′ symmetrically located with respect to the end plane, as shown in Fig. 126. The fields B and B′ are related as follows: The radial components of B and B′ are equal. The sum of the axial components of B and B′ is equal to B0 if P lies inside the solenoid, or to zero if P lies outside the solenoid (that is, above the top “edge” of the solenoid in the figure). The conclusions of parts (b) and (c) follow in the special case where P and P ′ coincide. Equal magnitudes This setup could be created by taking the setup in Fig. 6.22(a) and superposing a magnetic field pointing out of the page, with magnitude equal to the magnitude of the fields in Fig. 6.22(a) (which is µ0 J /2). The field on the left then points out of the page and down at a 45◦ angle, and the field on the right points out of the page and up at a 45◦ angle. So they are perpendicular, as desired. From Eq. (6.63), the force per unit area on the sheet is (B12 − B 22 )/2µ0 . But the magnitudes B1 and B2 are equal, so the force is zero. This is no surprise, after all, because the B field we superposed pointed out of the page, which was exactly the direction of the current in the sheet in Fig. 6.22(a). So the force on the moving charges is zero. In general, if we superpose a B field that points in the same direction as the current, the force will be zero. And consistent with this, B1 and B2 will be equal, although in only one special case will the two fields be perpendicular. In general, they will simply make equal angles with the direction of the current. Similarly, if we superpose a B field perpendicular to the sheet, the magnitudes of B1 and B2 will be equal. So the force on the sheet will be zero, consistent with the fact that the qv × B force on the charges in the current sheet lies in the plane of the sheet. The only way for the magnitudes of B1 and B2 to be unequal is for the superposed B field to have a component along the direction of the original fields. There will then be a nonzero force on the sheet. Consistent with this, the qv × B force on the charges in the current sheet now has a component perpendicular to the sheet. B' B0 P B Figure 126 P' Proton beam (a) The total energy of each proton is 3 GeV, so√ the γ factor is√ 3 (because the total energy of a particle is γmc2 ). Hence β = 1 − 1/γ 2 = 8/9 = 0.943. The current is 10−3 C/s, so I = λv gives the linear charge density as λ= I 10−3 C/s = 3.53 · 10−12 C/m. = 0.943 · 3 · 108 m/s v (467) The electric field 1 cm from the axis of the beam is E= 3.53 · 10−12 C/m λ = ( ) = 6.35 V/m. s2 C2 2πϵ0 r 2π 8.85 · 10−12 kg m3 (0.01 m) (468) (b) The magnetic field is B= ( m) −3 4π · 10−7 kg C/s) µ0 I C2 (10 = = 2 · 10−8 T. 2πr 2π (0.01 m) (469) (c) In F ′ there is no current because the protons are at rest, so B = 0. The spacing between the protons is “uncontracted,” so the density is smaller; it is λ′ = λ/γ. The electric field is therefore E ′ = E/γ = 2.12 V/m. Fields in a new frame In frame F , the electric field components are Ex = 100 cos 30◦ V/m = 86.6 V/m, Ey = 100 sin 30◦ V/m = 50 V/m, and Ez = 0. And also B = 0. The transformations to frame F ′ are given by Eq. (6.76). The “∥” direction is along the y axis, and the “⊥” direction is in the x-z plane. v is the velocity of F ′ with respect to F , so v = (0.6c)ŷ. Since B = 0 the transformations reduce to: E′∥ = E∥ , E′⊥ = γE⊥ , B∥′ = 0, B′⊥ = −(γ/c2 )v × E⊥ . (470) These yield (with γ = 5/4) E∥′ = ŷEy E′⊥ = γx̂Ex =⇒ E y′ = Ey = 50 V/m. =⇒ Ex′ = γEx = 108.3 V/m, and B∥′ ′ B⊥ 2 =0 = −(γ/c )v × E⊥ Ez′ = 0. =⇒ By′ = 0. =⇒ Bz′ = −(5/4)(1/c2 )(ŷ3c/5) × (x̂Ex ) = (3/4)(Ex /c)ẑ = 2.17 · 10−7 T, ′ and Bx = 0. (471) √ The magnitude of E′ is 108.32 + 502 = 119.3 V/m. The angle that E′ makes with the x̂′ axis is tan−1 (50/108.3) = 24.8◦ . And B′ points directly along the ẑ′ axis with magnitude 2.17 · 10−7 T. Fields from two ions (a) First solution: The electric field can be found via Eq. (5.15) in Chapter 5. If θ = 90◦ , we get an extra factor of γ compared with the static case, so the electric field at (3ℓ, 0, 0) is (with γ = 5/4 and ℓ = 1 m) E = x̂ 1 γe 1 γe 1 25 e = 2 · 10−9 V/m. + x̂ = x̂ 4πϵ0 ℓ2 4πϵ0 (3ℓ)2 4πϵ0 18 ℓ2 (472) Second solution: Let F ′ be the lab frame, and consider the frame F traveling upward with the left ion. Consider the field at (3, 0, 0) due to just the left ion. In frame F there is no B field from the left ion, so the transformations in Eq. (6.76) ′ yield E⊥ = γE⊥ . That is, the electric field due to the left ion is larger in F ′ (the lab frame) by a factor γ. The same reasoning holds for the field due to the right ion (because the direction of the relative velocity of the frames doesn’t matter in γ), so the solution proceeds as above. ′ (b) Consider the frames F and F ′ defined above. Eq. (6.76) gives B⊥ = −(γ/c2 )v × E⊥ , where v is the velocity of F ′ (the lab frame) with respect to F (the ion frame). For the left ion traveling upward, we have v = −vŷ, where v = 3c/5. So the magnetic field in F ′ (the lab frame) at (3ℓ, 0, 0) due to the left ion is (using µ0 = 1/ϵ0 c2 ) ) ( γµ0 ev ˆxe γ ′ . (473) B⊥,left = −ẑ = − 2 (−vŷ) × 2 4π (3ℓ)2 4πϵ0 (3ℓ) c Similarly, the magnetic field due to the right ion (for which F ′ moves with velocity v = vŷ with respect to F ) is ẑγµ0 ev/4πℓ2 . The sum is ) ( e e µ0 2c e µ0 ′ B⊥,total = ẑ (γv) 2 − = 3.2 · 10−18 T. (474) = ˆz 2 4π 3 ℓ2 ℓ 4π (3ℓ) Force on electrons moving together (a) Since there is no transverse length contraction, the distance between the electrons in the frame in which they are at rest is still r . The force on each electron is repulsive, and the magnitude is simply e2 /4πϵ0 r2 . To transform this force to the lab frame, we can use the fact that the transverse force on a particle is always greatest in the rest frame of the particle. It is smaller in any other frame by the factor γ. The repulsive force in the lab frame is therefore (1/γ )(e2 /4πϵ0 r2 ). (b) In the lab frame F ′ , consider the E ′ and B ′ fields at the location of the bottom electron arising from the top electron in Fig. 127. The electric field points upward toward the top electron, with magnitude γe/4πϵ0 r2 . This follows from Eq. (5.15) in Chapter 5, with θ = 90◦ . Alternatively, it follows from the transformations in Eq. (6.76); if F is the rest frame of the electrons, then B = 0, so the E′ field in ′ = γE = γ(e/4πϵ r 2 )ŷ. the lab frame is given by E⊥ ⊥ 0 The transformations in Eq. (6.76) also give the B ′ field. Since B = 0 in the electrons’ frame F , the B′ field in the lab frame F ′ is given by B′⊥ = −γ(v/c2 ) × E⊥ . The v here is the velocity of the lab frame F ′ with respect to the electrons’ ′ frame F . So v points leftward, and the righthand rule gives the direction of B⊥ as pointing out of the page. The magnitude is B ′ = (v/c2 )(γe/4πϵ0 r2 ). All of the fields and forces are shown in Fig. 127 (when calculating the forces, remember that electrons are negatively charged). The resulting force on the electron is the sum of the eE ′ repulsive electric force and the evB ′ attractive (as you can verify from the righthand rule) magnetic force. The net repulsive force is therefore ( ) v γe γe2 e2 γe v2 − ev 2 , (475) eE ′ − evB ′ = e = 1− 2 = 2 2 γ4πϵ 4πϵ0 r2 c 4πϵ0 r2 4πϵ0 r c 0r in agreement with the result in part (a). If you forgot to consider the magnetic field in the lab frame, then you would have a “paradox” where the γ appears in the denominator of the correct force in part (a), but in the numerator in the incomplete (that is, just electric) force in part (b). (lab frame F') y -e x r E' -e FB B' FE Figure 127 (c) In the limit v → c, we have γ → ∞, so the force in the lab frame goes to zero. Relating the forces γλ v r r q v λ Figure 128 Lab frame: In the lab frame (the frame of the top stick and the charge), there is no magnetic force on the charge q, because it is at rest. Label the top stick as “1” and the bottom stick as “2.” The electric field due to the top stick is λ/2πϵ0 r, so lab = λq/2πϵ r downward (assuming both λ and q are positive). the electric force is FE,1 0 Due to length contraction, the bottom stick has charge density γλ in the lab frame, lab = γλq/2πϵ r upward. The total force in the lab frame is so the electric force is FE,2 0 therefore λq γλq λq lab lab + = (γ − 1) F lab . (476) tot = FE,1 + FE,2 = − 2πϵ0 r 2πϵ0 r 2πϵ0 r This is positive, so the total force is upward. Bottom-stick frame: In this frame, the situation is shown in Fig. 128 . The charge q and the top stick are now moving. The density of the top stick is γλ due to length contraction. The top stick produces a current (γλ)v, so the magnetic field is µ0 (γλv)/2πr, and it points into the page at the location of the charge q. The magnetic force qvB therefore points upward with magnitude qv · µ0 (γλv)/2πr. Using bsf = µ0 = 1/ϵ0 c2 , this force in the bottom-stick frame (bsf) can be written as F B,1 (v 2 /c2 )γλq/2πϵ0 r. bsf bsf = The electric forces are quickly found to be F E,1 = γλq/2πϵ0 r downward and FE,2 λq/2πϵ0 r upward. The total force in the bottom-stick frame is therefore bsf Ftot = bsf bsf F bsf B,1 + F E,1 + FE,2 ( ) v 2 γλq γλq λq = + − + 2 2πϵ0 r c 2πϵ0 r 2πϵ0 r ( ( 2 ) ) v λq = γ 2 −1 +1 2πϵ0 r c ( ) 1 λq = − +1 . 2πϵ0 r γ (477) This is positive, so the total force is upward. It is 1/γ times the total force in the lab frame. This is correct, because the force on a particle is largest in the rest frame of the particle (the lab frame here); it is smaller in any other frame by the factor 1/γ . Drifting motion If there is a frame in which the electric field is zero, then we know from Exercise 6.29 that the ion moves in a circle in that frame. Let F be the lab frame, and consider the frame F ′ that moves in the positive y direction with speed v = (0.1 m)/(1 µs) = 105 m/s = c/3000. Since F ′ moves with the average velocity of the ion, it is the only frame in which the ion could possibly be moving in a circle (because in any other frame the ion would drift away). F sees F ′ moving with velocity vŷ, so if we demand that the electric field be zero in F ′ , then Eq. (6.76) tells us how E⊥ and B⊥ in the lab frame F must relate to each other: ′ E⊥ = γ(E⊥ + v × B⊥ ) ( ) =⇒ 0 = γ E⊥ + (vŷ) × (0.6 T)ẑ =⇒ E⊥ = = −(105 m/s)(0.6 T)(ŷ × ẑ) −(6 · 104 V/m)x̂. (478) Note that E⊥ points in the (negative) x direction, whereas the drift of the motion is in the y direction. See Problem 6.26 for more details. Rowland’s experiment With the disk at 10 kV, the electric field strength in the space above and below the disk in Fig. 129 is V 104 V = 1.67 · 106 V/m. (479) E= = 0.006 m d The charge density on each surface (top and bottom) of the disk is therefore ( s2 C2 ) σ = ϵ0 E = 8.85 · 10−12 (1.67 · 106 V/m) = 1.5 · 10−5 C/m2 . kg m3 axis charge on rotating disk charge on electrode (480) stationary electrode stationary glass plate --------------------+++++++++++++++++++++ +++++++++++++++++++++ rotating disk (0.5 cm thick) --------------------- 0.6 cm space 3.9 cm 4.45 cm 10.55 cm 12 cm 19.45 cm Figure 129 The left end of the charged region on the rotating disk is determined by the left end of the electrode on the glass plate, which is at radius 4.45 cm. The right end of the charged region is determined by the right end of the disk, which is at radius 10.55 cm. The mean radius of the charged part of the disk is therefore r = (4.45 cm + 10.55 cm)/2 = 7.5 cm. So the average velocity of the charges is v = 2πrν = 2π(0.075 m)(61 s−1 ) = 28.7 m/s. (481) (Actually, this is the velocity at the average radius, and not the average velocity of all points in the disk. But we’re just doing things roughly, so this distinction isn’t important. For that matter, we could just pick a round number like r ≈ 0.1 m.) The effective surface current density is then J = σv = (1.5 · 10−5 C/m2 )(28.7 m/s) = 4.3 · 10−4 C/(s m). (482) The combined effect of the two current sheets, each with surface current density J (in the same direction), is to produce a horizontal field B = µ0 J both above and below the disk. So ( )( ) (483) B = 4π · 10−7 kg m/C2 4.3 · 10−4 C/(s m) = 5.4 · 10−10 T, or 5.4 · 10−6 gauss. In terms of the various quantities, a few equivalent symbolic expressions for the magnetic field are B= 2πrνµ0 ϵ0 V 2πrνV vV = = 2 . c d d c2 d (484) We have used the speed at the mean radius to estimate the field strength B immediately above the disk in that region. For a more accurate calculation of the field strength to be expected at the location of the magnetometer needle, one could divide the disk into circular rings and integrate over the whole distribution. That is what Rowland did. Transverse Hall field In Gaussian units the relation between current density J and charge carrier velocity v is still J = nqv, so v = J/nq. Here J is measured in esu/(cm2 s), n in cm−3 , q in esu, and v in cm/s. The force on a charge carrier in Gaussian units is q(Et + (v/c) × B) which is zero if J×B v×B . (485) =− Et = − nqc c Hall voltage Our strategy will be to find the current density, then the drift velocity, then the transverse field, then the transverse (Hall) voltage. The current density is I V /R V /(ρL/A) V 1V = 1.25 · 104 A/m2 . = = = = (0.016 ohm-m)(0.005 m) A A A ρL (486) The drift velocity is then J= v= 1.25 · 104 C/s m2 J = 39 m/s. = (2 · 1021 m−3 )(1.6 · 10−19 C) ne (487) The induced electric field is Et = vB = (39 m/s)(0.1 T) = 3.9 V/m. The Hall voltage across the ribbon of width 0.002 m is therefore (3.9 V/m)(0.002 m) = 7.8 · 10−3 V, or 7.8 millivolts. Symbolically, the Hall voltage equals V Bw/ρLne, where w is the width.