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96 Pages

### ch29

Course: PHYS 214, Spring 2008
School: Kansas State
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Word Count: 9861

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(a) 1. The magnitude of the magnetic field due to the current in the wire, at a point a distance r from the wire, is given by B= 0i 2r . With r = 20 ft = 6.10 m, we have c4 10 B= hb 2 b6.10 mg -7 T m A 100 A g = 3.3 10 -6 T = 3.3 T. (b) This is about one-sixth the magnitude of the Earth's field. It will affect the compass reading. 2. The straight segment of the wire produces no magnetic field at C...

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(a) 1. The magnitude of the magnetic field due to the current in the wire, at a point a distance r from the wire, is given by B= 0i 2r . With r = 20 ft = 6.10 m, we have c4 10 B= hb 2 b6.10 mg -7 T m A 100 A g = 3.3 10 -6 T = 3.3 T. (b) This is about one-sixth the magnitude of the Earth's field. It will affect the compass reading. 2. The straight segment of the wire produces no magnetic field at C (see the straight sections discussion in Sample Problem 29-1). Also, the fields from the two semi-circular loops cancel at C (by symmetry). Therefore, BC = 0. 3. (a) The field due to the wire, at a point 8.0 cm from the wire, must be 39 T and must be directed due south. Since B = 0i 2 r , i= 2 rB = 2 0.080 m 39 10-6 T 4 10 -7 T m A b gc 0 h = 16 A. (b) The current must be from west to east to produce a field which is directed southward at points below it. 4. (a) Recalling the straight sections discussion in Sample Problem 29-1, we see that the current in segments AH and JD do not contribute to the field at point C. Using Eq. 29-9 (with = ) and the right-hand rule, we find that the current in the semicircular arc H J contributes 0i 4 R1 (into the page) to the field at C. Also, arc D A contributes 0i 4 R2 (out of the page) to the field there. Thus, the net field at C is B= 0i 4 1 1 - R1 R2 = (4 10 -7 T m A)(0.281A) 1 1 - = 1.67 10 -6 T. 4 0.0315m 0.0780m (b) The direction of the field is into the page. 5. (a) Recalling the straight sections discussion in Sample Problem 29-1, we see that the current in the straight segments collinear with P do not contribute to the field at that point. Using Eq. 29-9 (with = ) and the right-hand rule, we find that the current in the semicircular arc of radius b contributes 0i 4 b (out of the page) to the field at P. Also, the current in the large radius arc contributes 0i 4 a (into the page) to the field there. Thus, the net field at P is B= 0i 4 1 1 - b a = (4 10 -7 T m A)(0.411A)(74 /180) 1 1 - 4 0.107m 0.135m = 1.02 10 -7 T. (b) The direction is out of the page. 6. (a) Recalling the straight sections discussion in Sample Problem 29-1, we see that the current in the straight segments collinear with C do not contribute to the field at that point. Eq. 29-9 (with = ) indicates that the current in the semicircular arc contributes 0i 4 R to the field at C. Thus, the magnitude of the magnetic field is B= 0i 4R = (4 10 -7 T m A)(0.0348A) = 1.18 10 -7 T. 4(0.0926m) (b) The right-hand rule shows that this field is into the page. 7. (a) The currents must be opposite or antiparallel, so that the resulting fields are in the same direction in the region between the wires. If the currents are parallel, then the two fields are in opposite directions in the region between the wires. Since the currents are the same, the total field is zero along the line that runs halfway between the wires. (b) At a point halfway between they have the same magnitude, 0i/2r. Thus the total field at the midpoint has magnitude B = 0i/r and i= rB = ( 0.040 m ) ( 300 10-6 T ) 4 10 -7 T m A = 30 A. 0 8. (a) Since they carry current in the same direction, then (by the right-hand rule) the only region in which their fields might cancel is between them. Thus, if the point at which we are evaluating their field is r away from the wire carrying current i and is d r away from the wire carrying current 3.00i, then the canceling of their fields leads to 0i 0 (3i ) = 2 r 2 (d - r ) r= d 16.0 cm = = 4.0 cm. 4 4 (b) Doubling the currents does not change the location where the magnetic field is zero. 9. (a) BP1 = 0i1/2r1 where i1 = 6.5 A and r1 = d1 + d2 = 0.75 cm + 1.5 cm = 2.25 cm, and BP2 = 0i2/2r2 where r2 = d2 = 1.5 cm. From BP1 = BP2 we get i2 = i1 r2 1.5 cm = ( 6.5 A ) = 4.3A. r1 2.25 cm (b) Using the right-hand rule, we see that the current i2 carried by wire 2 must be out of the page. 10. With the "usual" x and y coordinates used in Fig. 29-40, then the vector r pointing ^ ^ ^ from a current element to P is r = -s i + R j . Since ds = ds i , then | ds r | = R ds. Therefore, with r = s2 + R2 , Eq. 29-3 becomes dB = o i R ds 2 2 3/2 . 4 (s + R ) (a) Clearly, considered as a function of s (but thinking of "ds" as some finite-sized constant value), the above expression is maximum for s = 0. Its value in this case is dBmax = o i ds /4R2. 1 (b) We want to find the s value such that dB = 10 dBmax. This is a non-trivial algebra exercise, but is nonetheless straightforward. The result is s = 102/3 - 1 R. If we set R = 2.00 cm, then we obtain s = 3.82 cm. 11. We assume the current flows in the +x direction and the particle is at some distance d in the +y direction (away from the wire). Then, the magnetic field at the location of a i proton with charge q is B = 0 k. Thus, 2d F = qv B = 0iq 2d ev kj. In this situation, v = v - j (where v is the speed and is a positive value), and q > 0. Thus, e j F= 0iqv 2d (( ) ) -23 -^ k = - j ^ ^ N)i. 0iqv ^ 2d i =- (4 10 -7 T m A)(0.350A)(1.60 10-19 C)(200m/s) ^ i 2 (0.0289 m) = (-7.75 10 12. The fact that By = 0 at x = 10 cm implies the currents are in opposite directions. Thus By = o i1 o i2 o i2 4 1 = x . 2 (L + x) 2 x 2 L + x using Eq. 29-4 and the fact that i1 = 4 i2. To get the maximum, we take the derivative 2 with respect to x and set equal to zero. This leads to 3x2 2Lx L = 0 which factors and becomes (3x + L)(x - L) = 0, which has the physically acceptable solution: x = L . This produces the maximum By: o i2 /2L. To proceed further, we must determine L. Examination of the datum at x = 10 cm in Fig. 29-42(b) leads (using our expression above for By and setting that to zero) to L = 30 cm. (a) The maximum value of By occurs at x = L = 30 cm. (b) With i2 = 0.003 A we find o i2 /2L = 2.0 nT. (c) and (d) Fig. 29-42(b) shows that as we get very close to wire 2 (where its field strongly dominates over that of the more distant wire 1) By points along the y direction. The right-hand rule leads us to conclude that wire 2's current is consequently is into the page. We previously observed that the currents were in opposite directions, so wire 1's current is out of the page. 13. Each of the semi-infinite straight wires contributes 0i 4 R (Eq. 29-7) to the field at the center of the circle (both contributions pointing "out of the page"). The current in the arc contributes a term given by Eq. 29-9 pointing into the page, and this is able to produce zero total field at that location if Barc = 2.00 Bsemiinfinite , or 0i 4R = 2.00 0i 4R which yields = 2.00 rad. 14. Initially, Bnet y = 0, and Bnet x = B2 + B4 = 2(o i /2d) using Eq. 29-4, where d = 0.15 m. To obtain the 30 condition described in the problem, we must have Bnet y = Bnet x tan(30) B1 B3 = 2(o i /2d) tan(30) where B3 = o i /2d and B1 = o i /2d. Since tan(30) = 1/ 3 , this leads to d = 3d . 3+2 (a) With d = 15.0 cm, this gives d = 7.0 cm. Being very careful about the geometry of the situation, then we conclude that we must move wire 1 to x = -7.0 cm. (b) To restore the initial symmetry, we would have to move wire 3 to x = +7.0 cm. 15. Each wire produces a field with magnitude given by B = 0i/2r, where r is the distance from the corner of the square to the center. According to the Pythagorean theorem, the diagonal of the square has length 2a , so r = a 2 and B = 0i 2a . The fields due to the wires at the upper left and lower right corners both point toward the upper right corner of the square. The fields due to the wires at the upper right and lower left corners both point toward the upper left corner. The horizontal components cancel and the vertical components sum to Btotal = 4 0i -7 2 0i 2 ( 4 10 T m A ) ( 20 A ) cos 45 = = = 8.0 10-5 T. a ( 0.20 m ) 2a In the calculation cos 45 was replaced with 1 ^ the +y direction. Thus, B = (8.0 10-5 T)j. total 2 . The total field points upward, or in 16. We consider Eq. 29-6 but with a finite upper limit (L/2 instead of ). This leads to B= o i 2R L/2 . R + (L/2)2 2 In terms of this expression, the problem asks us to see how large L must be (compared with R) such that the infinite wire expression B (Eq. 29-4) can be used with no more than a 1% error. Thus we must solve B B B = 0.01 . This is a non-trivial algebra exercise, but is nonetheless straightforward. The result is L= 200 R 14.1R 201 L 14.1 R 17. Our x axis is along the wire with the origin at the midpoint. The current flows in the positive x direction. All segments of the wire produce magnetic fields at P1 that are out of the page. According to the Biot-Savart law, the magnitude of the field any (infinitesimal) segment produces at P1 is given by dB = 0i sin 4 r 2 dx where (the angle between the segment and a line drawn from the segment to P1) and r (the length of that line) are functions of x. Replacing r with R r=R B= x 2 + R 2 and sin with x 2 + R 2 , we integrate from x = L/2 to x = L/2. The total field is 0iR 4 L2 -L 2 -7 dx ( 4 10 = (x 2 ( 0.131 m ) 4 R ( x ) T m A ) ( 0.0582 A ) 2 +R 2 32 = 0iR 1 x 2 L2 2 1 2 -L 2 2 +R ) = 0i 2R 2 L L + 4R2 = 5.03 10-8 T. 0.180m 2 2 (0.180m) + 4(0.131m) 18. Using the law of cosines and the requirement that B = 100 nT, we have = cos-1 B12 + B22 B2 = 144 . 2B1B2 where Eq. 29-10 has been used to determine B1 (168 nT) and B2 (151 nT). 19. Our x axis is along the wire with the origin at the right endpoint, and the current is in the positive x direction. All segments of the wire produce magnetic fields at P2 that are out of the page. According to the Biot-Savart law, the magnitude of the field any (infinitesimal) segment produces at P2 is given by dB = 0i sin 4 r 2 dx where (the angle between the segment and a line drawn from the segment to P2) and r (the length of that line) are functions of x. Replacing r with R r=R x 2 + R 2 and sin with x 2 + R 2 , we integrate from x = L to x = 0. The total field is B= 0iR 4 0 -L dx ( 4 10 = (x -7 2 +R 2 32 4 ( 0.251 m ) T m A ) ( 0.693 A ) ) = 0iR 1 4 R 2 x 0 2 1 2 -L (x 2 +R ) = 0i 4R 2 L L + R2 = 1.32 10-7 T. 0.136m (0.136m) + (0.251m) 2 2 20. In the one case we have Bsmall + Bbig = 47.25 T, and the other case gives Bsmall Bbig = 15.75 T (cautionary note about our notation: Bsmall refers to the field at the center of the small-radius arc, which is actually a bigger field than Bbig!). Dividing one of these equations by the other and canceling out common factors (see Eq. 29-9) we obtain 1 1 rsmall + rbig 1 1 = 3. -r rsmall big The solution of this is straightforward: rsmall = 1 rbig. 2 Using the given fact that the big radius 4.00 cm, then we conclude that the small radius is 2.00 cm. 21. (a) The contribution to BC from the (infinite) straight segment of the wire is BC1 = 0i 2 R . The contribution from the circular loop is BC 2 = 0i 2R . Thus, ( 4 10 -7 T m A )( 5.78 10-3 A ) 1+ 1 = 2.5310-7 T. 1 1+ = BC = BC1 + BC 2 = 2R 2 ( 0.0189 m ) 0i BC points out of the page, or in the +z direction. In unit-vector notation, ^ B = (2.5310-7 T)k C (b) Now BC1 BC 2 so BC = B + B 2 C1 2 C2 -7 -3 1 ( 4 10 T m A )( 5.78 10 A ) 1 = 1+ 2 = 1 + 2 = 2.02 10-7 T. 2R 2 ( 0.0189 m ) 0i and BC points at an angle (relative to the plane of the paper) equal to tan -1 In unit-vector notation, ^ ^ ^ BC = 2.02 10-7 T(cos17.66^ + sin17.66k) = (1.92 10-7 T)i + (6.12 10-8 T)k i BC1 1 = tan -1 = 17.66 . BC 2 22. Letting "out of the page" in Fig. 29-50(a) be the positive direction, the net field is B= o i1 o i2 4R 2(R/2) from Eqs. 29-9 and 29-4. Referring to Fig. 29-50, we see that B = 0 when i2 = 0.5 A, so (solving the above expression with B set equal to zero) we must have = 4(i2 /i1) = 4(0.5/2) = 1.00 rad (or 57.3). 23. Consider a section of the ribbon of thickness dx located a distance x away from point P. The current it carries is di = i dx/w, and its contribution to BP is dBP = 0di 2 x = 0idx 2 xw . Thus, BP = dBP = 0i 2w d +w d -7 -6 0.0491 dx 0i w ( 4 10 T m A )( 4.6110 A ) ln 1 + ln 1 + = = 2 ( 0.0491 m ) 0.0216 x 2w d = 2.2310-11 T. and BP points upward. In unit-vector notation, BP = (2.2310-11 T)^ j 24. Initially we have Bi = o i o i + 4R 4r using Eq. 29-9. In the final situation we use Pythagorean theorem and write Bf 2 = Bz + By 2 2 2 o i = 4R 2 o i + 4r 2 . If we square Bi and divide by Bf , we obtain 2 Bi Bf 1 1 R+r = 1 1 . R2 + r2 2 From the graph (see Fig. 29-52(c) note the maximum and minimum values) we estimate Bi /Bf = 12/10 = 1.2, and this allows us to solve for r in terms of R: r=R 1 1.2 2 1.22 1.22 1 = 2.3 cm or 43.1 cm. Since we require r < R, then the acceptable answer is r = 2.3 cm. 25. (a) Recalling the straight sections discussion in Sample Problem 29-1, we see that the current in the straight segments collinear with P do not contribute to the field at that point. We use the result of problem 16 to evaluate the contributions to the field at P, noting that the nearest wire-segments (each of length a) produce magnetism into the page at P and the further wire-segments (each of length 2a) produce magnetism pointing out of the page at P. Thus, we find (into the page) BP = 2 2 ( 4 10 -7 T m A ) (13 A ) 2 0i 2 0i 2 0i -2 = = 8a 8 ( 2 a ) 8a 8 ( 0.047 m ) = 1.96 10 -5 T 2.0 10 -5 T. (b) The direction of the field is into the page. 26. By the right-hand rule (which is "built-into" Eq. 29-3) the field caused by wire 1's current, evaluated at the coordinate origin, is along the +y axis. Its magnitude B1 is given by Eq. 29-4. The field caused by wire 2's current will generally have both an x and a y component which are related to its magnitude B2 (given by Eq. 29-4) and sines and cosines of some angle. A little trig (and the use of the right-hand rule) leads us to conclude that when wire 2 is at angle 2 (shown in Fig. 29-54) then its components are B2x = B2 sin2, B2y = B2 cos2. The magnitude-squared of their net field is then (by Pythagoras' theorem) the sum of the square of their net x-component and the square of their net y-component: B2 = (B2 sin2)2 + (B1 B2 cos2)2 = B12 + B22 2 B1 B2 cos2. (since sin2 + cos2 =1), which we could also have gotten directly by using the law of cosines. We have B1 = o i1 /2R = 60 nT and B2 = o i2 /2R = 40 nT, so with the requirement that the net field have magnitude B = 80 nT, we find 2 = cos-1 B12 + B22 B2 = cos -1 (-1/ 4) = 104 , 2B1B2 where the positive value has been chosen. 27. Eq. 29-13 gives the magnitude of the force between the wires, and finding the xcomponent of it amounts to multiplying that magnitude by cos = the x-component of the force per unit length is o i1 i2 d2 Fx = = 8.84 10-11 N/m . L 2 (d12 + d22 ) d12 d2 + d22 . Therefore, 28. Using a magnifying glass, we see that all but i2 are directed into the page. Wire 3 is therefore attracted to all but wire 2. Letting d = 0.500 m, we find the net force (per meter length) using Eq. 29-13, with positive indicated a rightward force: | F| = 0i3 FG - i + i + i + i IJ 2 H 2d d d 2d K 1 2 4 5 which yields | F | / = 8.00 10-7 N/m . 29. We label these wires 1 through 5, left to right, and use Eq. 29-13. Then, (a) The magnetic force on wire 1 is F1 = 0i 2l 2 -7 1 1 1 1 ^ 25 0i 2l ^ 25 ( 4 10 T m A ) ( 3.00A ) (10.0m) ^ + + + j= j= j d 2d 3d 4d 24d 24 ( 8.00 10-2 m ) 2 ^ = (4.69 10-4 N)j (b) Similarly, for wire 2, we have F2 = 0i 2l 2 1 1 ^ 5 0i 2l ^ ^ + j= j=(1.88 10-4 N)j. 2d 3d 12d (c) F3 = 0 (because of symmetry). ^ (d) F4 = - F2 =( - 1.88 10 -4 N)j , and ^ (e) F5 = - F1 = -(4.69 10-4 N)j . 30. Using Eq. 29-13, the force on, say, wire 1 (the wire at the upper left of the figure) is along the diagonal (pointing towards wire 3 which is at the lower right). Only the forces (or their components) along the diagonal direction contribute. With = 45, we find F1 =| F12 + F13 + F14 |= 2 F12 cos + F13 = 2 2 0i 2 2a cos 45 + 0i 2 2 2a = 3 2 2 0i 2 a -7 3 ( 4 10 T m A ) (15.0A ) = = 1.12 10 -3 N/m. -2 2 2 (8.50 10 m ) ^ ^ j) The direction of F1 is along r = (i - ^ / 2 . In unit-vector notation, we have F1 = (1.12 10 -3 N/m) ^ ^ ^ ^ (i - j) = (7.94 10 -4 N/m)i + (-7.94 10 -4 N/m)j 2 31. We use Eq. 29-13 and the superposition of forces: F4 = F14 + F24 + F34 . With = 45, the situation is as shown next: The components of F4 are given by F4 x = - F43 - F42 cos = - and F4 y = F41 - F42 sin = Thus, F4 = ( F + F 2 4x 2 12 4y 0i 2 2a - 0i 2 cos 45 2 2a =- 3 0 i 2 4a 0i 2 2a - 0i 2 sin 45 0i 2 2 2a = 4a . ) = 3 0 i 2 - 4a 2 + 0i 2 4a 2 12 10 ( 4 10 -7 T m A ) ( 7.50A ) 10 0i 2 = = 4a 4 ( 0.135m ) 2 = 1.32 10-4 N/m. and F4 makes an angle with the positive x axis, where = tan -1 In unit-vector notation, we have FG F IJ = tan FG - 1IJ = 162 . H 3K HF K 4y 4x -1 ^ ^ F1 = (1.32 10 -4 N/m)[cos162^ + sin162^ = (-1.25 10 -4 N/m)i + (4.17 10 -5 N/m)j i j] 32. (a) The fact that the curve in Fig. 29-57(b) passes through zero implies that the currents in wires 1 and 3 exert forces in opposite directions on wire 2. Thus, current i1 points out of the page. When wire 3 is a great distance from wire 2, the only field that affects wire 2 is that caused by the current in wire 1; in this case the force is negative according to Fig. 29-57(b). This means wire 2 is attracted to wire 1, which implies (by the discussion in section 29-2) that wire 2's current is in the same direction as wire 1's current: out of the page. With wire 3 infinitely far away, the force per unit length is given (in magnitude) as 6.27 10-7 N/m. We set this equal to F12 = o i1 i2 /2d. When wire 3 is at x = 0.04 m the curve passes through the zero point previously mentioned, so the force between 2 and 3 must equal F12 there. This allows us to solve for the distance between wire 1 and wire 2: d = (0.04 m)(0.750 A)/(0.250 A) = 0.12 m. Then we solve 6.27 10-7 N/m= o i1 i2 /2d and obtain i2 = 0.50 A. (b) The direction of i2 is out of the page. 33. The magnitudes of the forces on the sides of the rectangle which are parallel to the long straight wire (with i1 = 30.0 A) are computed using Eq. 29-13, but the force on each of the sides lying perpendicular to it (along our y axis, with the origin at the top wire and +y downward) would be figured by integrating as follows: F sides = z a +b a i2 0i1 dy. 2 y Fortunately, these forces on the two perpendicular sides of length b cancel out. For the remaining two (parallel) sides of length L, we obtain F= 0i1i2 L 1 2 ( 410 = and F 0i1i2b 1 = a a+d 2a ( a + b ) - -7 T m/A ) ( 30.0A )( 20.0A )( 8.00cm ) ( 300 10-2 m ) 2 (1.00cm + 8.00cm ) = 3.20 10-3 N, points toward the wire, or + ^ . In unit-vector notation, we have j F = (3.20 10-3 N)^ j 34. A close look at the path reveals that only currents 1, 3, 6 and 7 are enclosed. Thus, noting the different current directions described in the problem, we obtain B ds = 0 ( 7i - 6i + 3i + i ) = 5 0i = 5 ( 410-7 T m/A )( 4.50 10-3 A ) = 2.8310-8 T m. 35. (a) Two of the currents are out of the page and one is into the page, so the net current enclosed by the path is 2.0 A, out of the page. Since the path is traversed in the clockwise sense, a current into the page is positive and a current out of the page is negative, as indicated by the right-hand rule associated with Ampere's law. Thus, z B ds = - 0i = - 2.0 A 4 10-7 T m A = -2.5 10-6 T m. b gc h (b) The net current enclosed by the path is zero (two currents are out of the page and two are into the page), so B ds = 0ienc = 0 . z 36. We use Ampere's law: z B ds = 0i , where the integral is around a closed loop and i is the net current through the loop. (a) For path 1, the result is 1 B ds = 0 ( -5.0A + 3.0A ) = ( -2.0A ) ( 4 10 -7 T m A ) = -2.5 10-6 T m. (b) For path 2, we find 2 B ds = 0 ( -5.0A - 5.0A - 3.0A ) = ( -13.0A ) ( 4 10 -7 T m A ) = -1.6 10-5 T m. 37. We use Eq. 29-20 B = 0ir / 2 a 2 for the B-field inside the wire ( r < a ) and Eq. 29-17 B = 0i / 2 r for that outside the wire (r > a). (a) At r = 0, B = 0 . (b) At r = 0.0100m , B = 0ir (410-7 T m/A)(170A)(0.0100m) = = 8.50 10-4 T. 2 2 2 a 2 (0.0200m) 0ir (410-7 T m/A)(170A)(0.0200m) = = 1.70 10-3 T. 2 a 2 2 (0.0200m) 2 (c) At r = a = 0.0200m , B = (d) At r = 0.0400m , B = 0i (410-7 T m/A)(170A) = = 8.50 10-4 T. 2 r 2 (0.0400m) 38. (a) The field at the center of the pipe (point C) is due to the wire alone, with a magnitude of BC = 2 ( 3R ) 0iwire = 0iwire 6R . For the wire we have BP, wire > BC, wire. Thus, for BP = BC = BC, wire, iwire must be into the page: BP = BP ,wire - BP ,pipe = 0iwire 2 R - 0i 2 2 R b g. Setting BC = BP we obtain iwire = 3i/8 = 3(8.00 10-3 A) / 8 = 3.00 10-3 A . (b) The direction is into the page. 39. For r a , B (r ) = (a) At r = 0, B = 0 . 0ienc 2r = 0 2r r 0 J ( r ) 2rdr = 0 2 r 0 J0 0 J 0 r 2 r 2rdr = . a 3a (b) At r = a / 2 , we have B (r ) = (c) At r = a, 0 J 0 r 2 3a = (410-7 T m/A)(310A/m 2 )(3.110-3 m / 2) 2 = 1.0 10-7 T. -3 3(3.110 m) B (r = a) = 0 J 0 a (410-7 T m/A)(310A/m 2 )(3.110-3 m) 3 = 3 = 4.0 10-7 T. 40. It is possible (though tedious) to use Eq. 29-26 and evaluate the contributions (with the intent to sum them) of all 1200 loops to the field at, say, the center of the solenoid. This would make use of all the information given in the problem statement, but this is not the method that the student is expected to use here. Instead, Eq. 29-23 for the ideal solenoid (which does not make use of the coil radius) is the preferred method: B = 0in = 0i FG N IJ H K where i = 3.60 A, = 0.950 m and N = 1200. This yields B = 0.00571 T. 41. It is possible (though tedious) to use Eq. 29-26 and evaluate the contributions (with the intent to sum them) of all 200 loops to the field at, say, the center of the solenoid. This would make use of all the information given in the problem statement, but this is not the method that the student is expected to use here. Instead, Eq. 29-23 for the ideal solenoid (which does not make use of the coil diameter) is the preferred method: B = 0in = 0i FG N IJ H K where i = 0.30 A, = 0.25 m and N = 200. This yields B = 0.0030 T. 42. We find N, the number of turns of the solenoid, from the magnetic field B = 0in = oiN / : N = B / 0i. Thus, the total length of wire used in making the solenoid is . 2 2.60 10-2 m 23.0 10-3 T 130 m 2 rB 2 rN = = = 108 m. -7 0i 2 4 10 T m / A 18.0 A c hc hb g c hb g 43. (a) We use Eq. 29-24. The inner radius is r = 15.0 cm, so the field there is B= 0iN 2 r c4 10 = c4 10 = -7 T m / A 0.800 A 500 . 2 0150 m hb b g gb g = 5.33 10 -4 T. (b) The outer radius is r = 20.0 cm. The field there is B= 0iN 2 r -7 T m / A 0.800 A 500 2 0.200m hb b g gb g = 4.00 10 -4 T. 44. The orbital radius for the electron is r= mv mv = eB e 0ni which we solve for i: ( 9.1110-31 kg ) ( 0.0460 ) ( 3.00 108 m s ) mv = i= e 0 nr (1.60 10-19 C )( 410-7 T m A ) (100 0.0100m ) ( 2.30 10-2 m ) = 0.272A. 45. (a) We denote the B -fields at point P on the axis due to the solenoid and the wire as Bs and Bw , respectively. Since Bs is along the axis of the solenoid and Bw is perpendicular to it, Bs Bw respectively. For the net field B to be at 45 with the axis we then must have Bs = Bw. Thus, Bs = 0is n = Bw = 0iw 2d , which gives the separation d to point P on the axis: d= iw 6.00 A = = 4.77 cm . 2 is n 2 20.0 10-3 A 10 turns cm c hb g (b) The magnetic field strength is B = 2 Bs = 2 ( 410-7 T m A )( 20.0 10-3 A ) (10turns 0.0100m ) = 3.55 10-5 T. 46. As the problem states near the end, some idealizations are being made here to keep the calculation straightforward (but are slightly unrealistic). For circular motion (with speed v which represents the magnitude of the component of the velocity perpendicular to the magnetic field [the field is shown in Fig. 29-19]), the period is (see Eq. 28-17) T = 2r/v = 2m/eB. Now, the time to travel the length of the solenoid is t = L / v where v|| is the component of the velocity in the direction of the field (along the coil axis) and is equal to v cos where = 30. Using Eq. 29-23 (B = 0in) with n = N/L, we find the number of revolutions made is t /T = 1.6 106. 47. The magnitude of the magnetic dipole moment is given by = NiA, where N is the number of turns, i is the current, and A is the area. We use A = R2, where R is the radius. Thus, = 200 0.30 A 0.050 m = 0.47 A m2 . b gb gb g 2 48. (a) We set z = 0 in Eq. 29-26 (which is equivalent using to Eq. 29-10 multiplied by the number of loops). Thus, B(0) i/R. Since case b has two loops, Bb 2i Rb 2 Ra = = = 4.0 . Ba i Ra Rb (b) The ratio of their magnetic dipole moments is b 2iAb 2 Rb2 1 = = 2 =2 Ra a iAa 2 2 = 1 = 0.50. 2 49. (a) The magnitude of the magnetic dipole moment is given by = NiA, where N is the number of turns, i is the current, and A is the area. We use A = R2, where R is the radius. Thus, = NiR 2 = 300 4.0 A 0.025 m = 2.4 A m2 . (b) The magnetic field on the axis of a magnetic dipole, a distance z away, is given by Eq. 29-27: B= b gb gb g 2 0 2 z 3 . We solve for z: F IJ = FG c4 10 T m Ahc2.36 A m hIJ z=G H 2 B K H 2 c5.0 10 Th K 13 -7 2 0 -6 13 = 46 cm . 50. We use Eq. 29-26 and note that the contributions to BP from the two coils are the same. Thus, BP = 2 0iR 2 N 2 R2 + ( R 2) 2 32 -7 8 0 Ni 8 ( 410 T m/A ) (200) ( 0.0122A ) = = = 8.78 10-6 T. 5 5R 5 5 ( 0.25m ) BP is in the positive x direction. 51. (a) To find the magnitude of the field, we use Eq. 29-9 for each semicircle ( = rad), and use superposition to obtain the result: B= 0i 0i 0i 1 1 4 + 4b -7 = 4 a b + = (410-7 T m/A) ( 0.0562A ) 1 1 + 4 0.0572m 0.0936m = 4.97 10 T. (b) By the right-hand rule, B points into the paper at P (see Fig. 29-6(c)). (c) The enclosed area is A = (a 2 + b 2 ) / 2 which means the magnetic dipole moment has magnitude | |= i 2 (a 2 + b 2 ) = (0.0562A) 2 [(0.0572m) 2 + (0.0936m) 2 ] = 1.06 10-3 A m 2 . (d) The direction of is the same as the B found in part (a): into the paper. 52. By imagining that each of the segments bg and cf (which are shown in the figure as having no current) actually has a pair of currents, where both currents are of the same magnitude (i) but opposite direction (so that the pair effectively cancels in the final sum), one can justify the superposition. (a) The dipole moment of path abcdefgha is = bc f gb + abgha + cde f c = ( ia 2 ) ^ - ^ + ^ = ia 2 ^ j i i j 2 = ( 6.0 A )( 0.10 m ) ^ = (6.0 10-2 A m 2 ) ^ . j j ( ) (b) Since both points are far from the cube we can use the dipole approximation. For (x, y, z) = (0, 5.0 m, 0) B(0, 5.0 m, 0) 0 2 y 3 = (1.26 10-6 T m/A)(6.0 10-2 m 2 A) ^ j = (9.6 10-11 T ) ^ . j 3 2(5.0 m) 53. (a) We denote the large loop and small coil with subscripts 1 and 2, respectively. B1 = 0i1 2 R1 c4 10 = -7 T m A 15 A 2 012 m . b g hb g = 7.9 10 -5 T. (b) The torque has magnitude equal to =| 2 B1 |= 2 B1 sin 90 = N 2i2 A2 B1 = N 2i2 r22 B1 = ( 50 )(1.3A ) ( 0.82 10-2 m ) ( 7.9 10-5 T ) = 1.110-6 N m. 2 54. Using Eq. 29-26, we find that the net y-component field is o i1 R o i2 R By = 2 2 3/2 2 2 3/2 , 2(R + z1 ) 2(R + z2 ) where z12 = L2 (see Fig. 29-68(a)) and z22 = y2 (because the central axis here is denoted y instead of z). The that fact there is a minus sign between the two terms, above, is due to the observation that the datum in Fig. 29-68(b) corresponding to By = 0 would be impossible without it (physically, this means that one of the currents is clockwise and the other is counterclockwise). (a) As y , only the first term contributes and (with By = 7.2 10-6 T given in this case) we can solve for i1. We obtain i1 = (45/16) 0.90 A. (b) With loop 2 at y = 0.06 m (see Fig. 29-68(b)) we are able to determine i2 from o i1 R o i2 R 2 2 3/2 = 2 2 3/2 . 2(R + L ) 2(R + y ) We obtain i2 = (117 13 /50) 2.7 A. 2 2 2 2 55. (a) We find the field by superposing the results of two semi-infinite wires (Eq. 29-7) and a semicircular arc (Eq. 29-9 with = rad). The direction of B is out of the page, as can be checked by referring to Fig. 29-6(c). The magnitude of B at point a is therefore Ba = 2 FG i IJ + i = i FG 1 + 1 IJ . H 4R K 4R 2 R H 2 K 0 0 0 With i = 10 A and R = 0.0050 m, we obtain Ba = 10 10-3 T . . (b) The direction of this field is out of the page, as Fig. 29-6(c) makes clear. (c) The last remark in the problem statement implies that treating b as a point midway between two infinite wires is a good approximation. Thus, using Eq. 29-4, Bb = 2 FG i IJ = 8.0 10 H 2R K 0 -4 T. (d) This field, too, points out of the page. 56. Using the Pythagorean theorem, we have B = B1 + B2 2 2 2 o i1 = 4R 2 o i2 + 2R 2 which, when thought of as the equation for a line in a B2 versus i22 graph, allows us to identify the first term as the "y-intercept" (1 10-10) and the part of the second term which multiplies i22 as the "slope" (5 10-10). The latter observation leads to the conclusion that R = 8.9 mm, and then our observation about the "y-intercept" determines the angle subtended by the arc: = 1.8 rad. 57. We refer to the center of the circle (where we are evaluating B ) as C. Recalling the straight sections discussion in Sample Problem 29-1, we see that the current in the straight segments which are collinear with C do not contribute to the field there. Eq. 29-9 (with = /2 rad) and the right-hand rule indicates that the currents in the two arcs contribute 0i 2 4R b g - i b 2g = 0 0 4R to the field at C. Thus, the non-zero contributions come from those straight-segments which are not collinear with C. There are two of these "semi-infinite" segments, one a vertical distance R above C and the other a horizontal distance R to the left of C. Both contribute fields pointing out of the page (see Fig. 29-6(c)). Since the magnitudes of the two contributions (governed by Eq. 29-7) add, then the result is B=2 FG i IJ = i H 4 R K 2 R 0 0 exactly what one would expect from a single infinite straight wire (see Eq. 29-4). For such a wire to produce such a field (out of the page) with a leftward current requires that the point of evaluating the field be below the wire (again, see Fig. 29-6(c)). 58. We use Eq. 29-4 to relate the magnitudes of the magnetic fields B1 and B2 to the currents (i1 and i2, respectively) in the two long wires. The angle of their net field is = tan-1(B2 /B1) = tan-1(i2 /i1) = 53.13. The accomplish the net field rotation described in the problem, we must achieve a final angle = 53.13 20 = 33.13. Thus, the final value for the current i1 must be i2 /tan = 61.3 mA. 59. Using the right-hand rule (and symmetry), we see that B net points along what we will refer to as the y axis (passing through P), consisting of two equal magnetic field ycomponents. Using Eq. 29-17, | B net | = 2 0i sin 2r where i = 4.00 A, r = r = d 22 + d12 / 4 = 5.00 m, and = tan-1 Therefore, | B net | = 2.56 10-7 T. d2 1 2d1 = 53.1 . 60. The radial segments do not contribute to B (at the center) and the arc-segments ^ contribute according to Eq. 29-9 (with angle in radians). If k designates the direction "out of the page" then 0 i 2 rad ^ 0 i 2 rad ^ 0i( rad) ^ B = k + k - k 4(4.00 m) 4(2.00 m) 4(2.00 m) where i = 2.00 A. This yields B = (1.57 10-7 T) k , or | B |= 1.57 10-7 T . ^ 61. (a) The magnetic field at a point within the hole is the sum of the fields due to two current distributions. The first is that of the solid cylinder obtained by filling the hole and has a current density that is the same as that in the original cylinder (with the hole). The second is the solid cylinder that fills the hole. It has a current density with the same magnitude as that of the original cylinder but is in the opposite direction. If these two situations are superposed the total current in the region of the hole is zero. Now, a solid cylinder carrying current i which is uniformly distributed over a cross section, produces a magnetic field with magnitude B= 0ir 2R 2 at a distance r from its axis, inside the cylinder. Here R is the radius of the cylinder. For the cylinder of this problem the current density is J= i i = , 2 A a - b2 c h where A = (a2 b2) is the cross-sectional area of the cylinder with the hole. The current in the cylinder without the hole is I1 = JA = Ja 2 = ia 2 a 2 - b2 and the magnetic field it produces at a point inside, a distance r1 from its axis, has magnitude B1 = 0 I1r1 2 a 2 = 0ir1a 2 2 a a - b 2 2 c 2 h = 0ir2 2 a 2 - b2 c h. The current in the cylinder that fills the hole is ib 2 I 2 = Jb = 2 a - b2 2 and the field it produces at a point inside, a distance r2 from the its axis, has magnitude B2 = 0 I 2 r2 2 b 2 = 0ir2b 2 2 b a - b 2 2 c 2 h = 0ir2 2 a 2 - b2 c h. At the center of the hole, this field is zero and the field there is exactly the same as it would be if the hole were filled. Place r1 = d in the expression for B1 and obtain (410-7 T m/A) ( 5.25A ) (0.0200m) B= = = 1.53 10-5 T 2 2 2 2 2 [(0.0400m) - (0.0150m) ] 2 ( a - b ) 0id for the field at the center of the hole. The field points upward in the diagram if the current is out of the page. (b) If b = 0 the formula for the field becomes B= 0id 2a 2 . This correctly gives the field of a solid cylinder carrying a uniform current i, at a point inside the cylinder a distance d from the axis. If d = 0 the formula gives B = 0. This is correct for the field on the axis of a cylindrical shell carrying a uniform current. (c) Consider a rectangular path with two long sides (side 1 and 2, each with length L) and two short sides (each of length less than b). If side 1 is directly along the axis of the hole, then side 2 would be also parallel to it and also in the hole. To ensure that the short sides do not contribute significantly to the integral in Ampere's law, we might wish to make L very long (perhaps longer than the length of the cylinder), or we might appeal to an argument regarding the angle between B and the short sides (which is 90 at the axis of the hole). In any case, the integral in Ampere's law reduces to z z side1 rectangle B ds = 0ienclosed B ds = 0iin hole B ds + z side 2 dB side1 - Bside2 L = 0 i where Bside 1 is the field along the axis found in part (a). This shows that the field at offaxis points (where Bside 2 is evaluated) is the same as the field at the center of the hole; therefore, the field in the hole is uniform. 62. We note that when there is no y-component of magnetic field from wire 1 (which, by the right-hand rule, relates to when wire 1 is at 90 = /2 rad), the total y-component of magnetic field is zero (see Fig. 29-76(c)). This means wire #2 is either at +/2 rad or -/2 rad. (a) We now make the assumption that wire #2 must be at -/2 rad (-90, the bottom of the cylinder) since it would pose an obstacle for the motion of wire #1 (which is needed to make these graphs) if it were anywhere in the top semicircle. (b) Looking at the 1 = 90 datum in Fig. 29-76(b)) where there is a maximum in Bnet x (equal to +6 T) we are led to conclude that B1x = 6.0 T 2.0 T = +4.0 T in that situation. Using Eq. 29-4, we obtain i1 = B1x 2R /o = 4.0 A . (c) The fact that Fig. 29-76(b) increases as 1 progresses from 0 to 90 implies that wire 1's current is out of the page, and this is consistent with the cancellation of Bnet y at 1 = 90, noted earlier (with regard to Fig. 29-76(c)). (d) Referring now to Fig. 29-76(b) we note that there is no x-component of magnetic field from wire 1 when 1 = 0, so that plot tells us that B2x = +2.0 T. Using Eq. 29-4, we have i2 = B2x 2R /o = 2.0 A for the magnitudes of the currents. (e) We can conclude (by the right-hand rule) that wire 2's current is into the page. 63. Using Eq. 29-20 and Eq. 29-17, we have | B1 |= 0 i 2R 2 r1 | B2 |= 0i 2r2 where r1 = 0.0040 m, B1 = 2.8 10-4 T, r2 = 0.010 m and | B2 | = 2.0 10-4 T. Point 2 is known to be external to the wire since | B2 | < | B1 | . From the second equation, we find i = 10 A. Plugging this into the first equation yields R = 5.3 103 m. 64. Eq. 29-1 is maximized (with respect to angle) by setting = 90 ( = /2 rad). Its value in this case is dBmax = o i ds /4R2. From Fig. 29-77(b), we have Bmax = 60 10-12 T. We can relate this Bmax to our dBmax by setting "ds" equal to 1 10-6 m and R = 0.025 m. This allows us to solve for the current: i = 0.375 A. Plugging this into Eq. 29-4 (for the infinite wire) gives B = 3.0 T. 65. Eq. 29-4 gives i= 2 RB = 2 0.880 m 7.30 10-6 T 4 10 T m A -7 b gc 0 h = 32.1A . 66. (a) By the right-hand rule, the magnetic field B1 (evaluated at a) produced by wire 1 (the wire at bottom left) is at = 150 (measured counterclockwise from the +x axis, in the xy plane), and the field produced by wire 2 (the wire at bottom right) is at = 210. By symmetry B1 = B2 we observe that only the x-components survive, yielding d i B1 + B2 = 2 0i 2 ^ cos 150 ^ = (-3.46 10-5 T)i i where i = 10 A, = 0.10 m, and Eq. 29-4 has been used. To cancel this, wire b must carry current into the page (that is, the - k direction) of value ib = 3.46 10-5 c h 2r = 15 A 0 where r = 3 2 = 0.087 m and Eq. 29-4 has again been used. (b) As stated above, to cancel this, wire b must carry current into the page (that is, the - z direction) 67. (a) The field in this region is entirely due to the long wire (with, presumably, negligible thickness). Using Eq. 29-17, B= 0 iw 2 r = 4.8 10-3 T where iw = 24 A and r = 0.0010 m. (b) Now the field consists of two contributions (which are anti-parallel) -- from the wire (Eq. 29-17) and from a portion of the conductor (Eq. 29-20 modified for annular area): | B |= 0 iw 2r - 0 ienc 2r = 0 iw 2r - 0 ic 2r r 2 - Ri2 R02 - Ri2 where r = 0.0030 m, Ri = 0.0020 m, Ro = 0.0040 m and ic = 24 A. Thus, we find | B |= 9.3 10-4 T. (c) Now, in the external region, the individual fields from the two conductors cancel completely (since ic = iw): B = 0. 68. (a) We designate the wire along y = rA = 0.100 m wire A and the wire along y = rB = 0.050 m wire B. Using Eq. 29-4, we have Bnet = BA + BB = - 2rA 0 iA ^ 0 iB ^ ^ k- k = (-52.0 10-6 T)k. 2rB (b) This will occur for some value rB < y < rA such that 0 iA 2 rA - y Solving, we find y = 13/160 0.0813 m. b g = 0 iB 2 y - rB b g. (c) We eliminate the y < rB possibility due to wire B carrying the larger current. We expect a solution in the region y > rA where 0 iA 2 y - rA Solving, we find y = 7/40 0.0175 m. b g = 0 iB 2 y - rB b g. 69. (a) As illustrated in Sample Problem 29-1, the radial segments do not contribute to BP and the arc-segments contribute according to Eq. 29-9 (with angle in radians). If k designates the direction "out of the page" then B= 0 0.40 A rad g k - b0.80 Agb radg k 4 b0.050 mg 4 b0.040 mg b gb 0 2 3 ^ which yields B = -1.7 10-6 kT , or | B |= 1.7 10-6 T . ^ (b) The direction is - k , or into the page. (c) If the direction of i1 is reversed, we then have B=- 0 0.40 A rad g k - b0.80 Agb radg k 4 b0.050 mg 4 b0.040 mg b gb 0 2 3 ^ which yields B = (-6.7 10-6 T)k , or | B |= 6.7 10-6 T. ^ (d) The direction is - k , or into the page. 70. We note that the distance from each wire to P is r = d the current is i = 100 A. 2 = 0.071 m. In both parts, (a) With the currents parallel, application of the right-hand rule (to determine each of their contributions to the field at P) reveals that the vertical components cancel and the horizontal components add, yielding the result: B=2 0i 2r cos 45.0 = 4.00 10-4 T ^ and directed in the x direction. In unit-vector notation, we have B = (-4.00 10-4 T)i . (b) Now, with the currents anti-parallel, application of the right-hand rule shows that the horizontal components cancel and the vertical components add. Thus, B=2 0i 2r sin 45.0 = 4.00 10-4 T ^ and directed in the +y direction. In unit-vector notation, we have B = (4.00 10-4 T)j . 71. Since the radius is R = 0.0013 m, then the i = 50 A produces B= 0i 2 R = 0.0077 T at the edge of the wire. The three equations, Eq. 29-4, Eq. 29-17 and Eq. 29-20, agree at this point. 1 72. The area enclosed by the loop L is A = 2 (4d )(3d ) = 6d 2 . Thus c B ds = 0i = 0 jA = ( 4 10 -7 T m A ) (15 A m 2 ) ( 6 ) ( 0.20m ) = 4.5 10-6 T m. 2 73. (a) With cylindrical symmetry, we have, external to the conductors, B= 0 ienc 2 r which produces ienc = 25 mA from the given information. Therefore, the thin wire must carry 5.0 mA. (b) The direction is downward, opposite to the 30 mA carried by the thin conducting surface. 74. (a) All wires carry parallel currents and attract each other; thus, the "top" wire is pulled downward by the other two: F = 0 L 5.0 A 3.2 A b gb g + Lb5.0 Agb5.0 Ag 2 b010 mg . 2 b0.20 mg 0 . where L = 3.0 m. Thus, F = 17 10-4 N. (b) Now, the "top" wire is pushed upward by the center wire and pulled downward by the bottom wire: | F |= 0 L ( 5.0A )( 3.2A ) 2 ( 0.10m ) - 0 L ( 5.0A )( 5.0A ) 2 ( 0.20m ) = 2.110-5 N . 75. We use B x , y , z = 0 4 is r r 3 , where s = sj and r = x i + yj + zk . Thus, B x, y, z = b g b g b - g FGH 4 IJK i sxj +e xi + yj + zk j = 4 xi +seyzi + zxkj c y +z h c h 0 0 2 2 2 32 2 s 2 32 . (a) The field on the z axis (at z = 5.0 m) is B ( 0, 0, 5.0m ) ( 4 10 = -7 T m/A ) ( 2.0A ) ( 3.0 10-2 m ) ( 5.0m ) ^ i 4 0 + 0 + ( 5.0m ) 2 2 ( 2 3/ 2 ) ^ = (2.4 10-10 T)i. (b) B (0, 6.0 m, 0), since x = z = 0. (c) The field in the xy plane, at (x, y) = (7,7), is B ( 7.0m,7.0m,0 ) = ^ (4 10 -7 T m/A)(2.0 A)(3.0 10-2 m)(-7.0 m)k 4 ( 7.0m ) + ( 7.0m ) + 0 2 2 ( 2 ) 3/ 2 ^ = (-4.3 10-11 T)k. (d) The field in the xy plane, at (x, y) = (3, 4), is B ( -3.0 m, - 4.0m, 0 ) = ^ (4 10 -7 T m/A)(2.0 A)(3.0 10-2 m)(3.0 m)k 4 ( -3.0 m ) + ( -4.0 m ) + 0 2 2 ( 2 ) 3/ 2 ^ = (1.4 10-10 T )k. 76. (a) The radial segments do not contribute to BP and the arc-segments contribute ^ according to Eq. 29-9 (with angle in radians). If k designates the direction "out of the page" then BP = 0 i 7 4 rad 4(4.00 m) k ^ 0 i 7 4 rad 4(2.00 m) ^ k ^ where i = 0.200 A. This yields B = -2.75 10-8 k T, or | B | = 2.75 10-8 T. ^ (b) The direction is - k , or into the page. 77. The contribution to B net from the first wire is (using Eq. 29-4) B1 = 0(30 A) ^ ^ k = (3.0 10-6T) k . 2(2.0 m) The distance from the second wire to the point where we are evaluating B net is 4 m - 2 m = 2 m. Thus, B2 = 0(40 A) ^ i = ( 4.0 10-6 T) ^ i 2(2 m) The magnitude of | B net| and consequently is perpendicular to B1 . 3.0 + 4.0 = 5.0 T. 2 2 is therefore 7 8. Using Eq. 29-20, | B | = 0 i r, we find that r = 0.00128 m gives the desired field 2R2 value. 79. The points must be along a line parallel to the wire and a distance r from it, where r i satisfies Bwire = 0 = Bext , or 2r r= 0i 2 Bext c1.26 10 T m Ahb100 Ag = 4.0 10 = 2 c5.0 10 Th -6 -3 -3 m. 80. (a) The magnitude of the magnetic field on the axis of a circular loop, a distance z from the loop center, is given by Eq. 29-26: B= N 0iR 2 , 2( R 2 + z 2 ) 3/ 2 where R is the radius of the loop, N is the number of turns, and i is the current. Both of the loops in the problem have the same radius, the same number of turns, and carry the same current. The currents are in the same sense, and the fields they produce are in the same direction in the region between them. We place the origin at the center of the lefthand loop and let x be the coordinate of a point on the axis between the loops. To calculate the field of the left-hand loop, we set z = x in the equation above. The chosen point on the axis is a distance s x from the center of the right-hand loop. To calculate the field it produces, we put z = s x in the equation above. The total field at the point is therefore N 0iR 2 1 1 B= + 2 . 2 2 3/ 2 2 2 (R + x ) ( R + x - 2 sx + s 2 ) 3/ 2 Its derivative with respect to x is 3x 3( x - s) dB N 0iR 2 . =- + 2 2 2 5/ 2 2 (R + x ) ( R + x 2 - 2 sx + s 2 ) 5/ 2 dx When this is evaluated for x = s/2 (the midpoint between the loops) the result is 3s / 2 3s / 2 dB N 0iR 2 =0 =- - 2 2 2 2 5/ 2 2 dx s / 2 ( R + s / 4) ( R + s / 4 - s 2 + s 2 ) 5/ 2 independent of the value of s. (b) The second derivative is 3 15x 2 d 2 B N 0iR 2 = - 2 + 2 2 ( R + x 2 ) 5/ 2 ( R + x 2 ) 7 / 2 dx 2 - 3 15( x - s) 2 + 2 ( R 2 + x 2 - 2 sx + s 2 ) 5/ 2 ( R + x 2 - 2 sx + s 2 ) 7 / 2 LM N OP Q LM N OP Q LM N OP Q LM N OP Q At x = s/2, N 0iR 2 d 2B 6 30 s 2 / 4 = - 2 2 + 2 2 dx 2 s / 2 2 ( R + s / 4)5 / 2 ( R + s / 4)7 / 2 N 0 R 2 -6( R 2 + s 2 / 4) + 30 s 2 / 4 s2 - R2 2 = = 3 N 0iR . 2 ( R 2 + s 2 / 4)7 / 2 ( R 2 + s 2 / 4)7 / 2 Clearly, this is zero if s = R. 81. The center of a square is a distance R = a/2 from the nearest side (each side being of length L = a). There are four sides contributing to the field at the center. The result is Bcenter = 4 2 ( a 2 ) 0i a a2 + 4 ( a 2) 2 = 2 2 0 i . a 82. We refer to the side of length L as the long side and that of length W as the short side. The center is a distance W/2 from the midpoint of each long side, and is a distance L/2 from the midpoint of each short side. There are two of each type of side, so the result of problem 11 leads to B=2 0i L L +4 W 2 2 2 W 2 b g b g 2 +2 0i W W +4 L 2 2 2 L 2 b g b g 2 . The final form of this expression, shown in the problem statement, derives from finding the common denominator of the above result and adding them, while noting that L2 + W 2 W +L 2 2 = W 2 + L2 . 83. We imagine the square loop in the yz plane (with its center at the origin) and the evaluation point for the field being along the x axis (as suggested by the notation in the problem). The origin is a distance a/2 from each side of the square loop, so the distance from the evaluation point to each side of the square is, by the Pythagorean theorem, R= ba 2g + x 2 2 = 1 2 a + 4x2 . 2 Only the x components of the fields (contributed by each side) will contribute to the final result (other components cancel in pairs), so a trigonometric factor of a2 a = 2 R a + 4x2 multiplies the expression of the field given by the result of problem 11 (for each side of length L = a). Since there are four sides, we find B x =4 i b g FGH 2R IJK FGH 0 a a2 + 4 R2 IJ FG KH a a2 + 4x2 IJ = K 2 b g e 1 2 4 0 i a 2 a2 + 4x2 j 2 a2 + 4 a 2 + 4x2 b g 2 which simplifies to the desired result. It is straightforward to set x = 0 and see that this reduces to the expression found in problem 12 (noting that 42 = 2 2 ). 84. Using the result of problem 12 and Eq. 29-10, we wish to show that 2 2 0i 0i 4 2 1 > , or > , a 2R a R but to do this we must relate the parameters a and R. If both wires have the same length L then the geometrical relationships 4a = L and 2R = L provide the necessary connection: 4a = 2 R a= R . 2 Thus, our proof consists of the observation that 4 2 8 2 1 = 2 > , a R R as one can check numerically (that 8 2 2 > 1 ). 85. The two small wire-segments, each of length a/4, shown in Fig. 29-83 nearest to point P, are labeled 1 and 8 in the figure below. ^ Let - k be a unit vector pointing into the page. We use the results of problem 19 to calculate BP1 through BP8: B P1 = B P 8 = B P 4 = B P5 = BP2 = BP7 = 2 0i 2 0i = , 8 a 4 2 a b g 2 0i 2 0i = , 8 3a 4 6 a b g 0i 4 a 4 b g b3 a 4 g + ba 4 g 2 3a 4 2 12 = 3 0i 10a , and BP3 = BP6 = 0i 4 3a 4 b g ba 4 g + b3 a 4 g 2 a4 2 12 = 0i 3 10a . Finally, BP = = i ^ BPn (-k) = 2 0 a n =1 8 2 2 3 1 ^ + + + (- k) 2 6 10 3 10 2 2 3 1 ^ + + + (-k) 2 6 10 3 10 2 ( 4 10 -7 T m A ) (10A ) ( 8.0 10 -2 m ) ^ = ( 2.0 10-4 T ) (-k). 86. (a) Consider a segment of the projectile between y and y + dy. We use Eq. 29-12 to find the magnetic force on the segment, and Eq. 29-7 for the magnetic field of each semiinfinite wire (the top rail referred to as wire 1 and the bottom as wire 2). The current in rail 1 is in the + i direction, and the current in rail 2 is in the - i direction. The field (in the region between the wires) set up by wire 1 is into the paper (the - k direction) and that set up by wire 2 is also into the paper. The force element (a function of y) acting on the segment of the projectile (in which the current flows in the - j direction) is given below. The coordinate origin is at the bottom of the projectile. dF = dF1 + dF2 = idy -^ B1 + dy -^ B2 = i [ B1 + B2 ] ^dy = i j j i ( ) ( ) 4 ( 2 R + w - y ) 0i + 4y 0i ^ idy. Thus, the force on the projectile is i 2 0 F = dF = 4 R+w R 2 1 1 ^ = 0i ln 1 + w ^ dy i i. + R 2R + w - y y 2 (b) Using the work-energy theorem, we have 1 K = 2 mv 2 = Wext = F ds = FL. f z Thus, the final speed of the projectile is vf F 2W IJ = LM 2 i lnFG1 + w IJ LOP =G H m K N m 2 H R K Q LM 2c4 10 T m / Ahc450 10 Ah lnb1 + 12 cm / 6.7 cmgb4.0 mg OP . = MN PQ 2 c10 10 kgh 1/ 2 2 1/ 2 ext 0 -7 3 2 1/ 2 -3 = 2.3 103 m / s. 87. We take the current (i = 50 A) to flow in the +x direction, and the electron to be at a point P which is r = 0.050 m above the wire (where "up" is the +y direction). Thus, the field produced by the current points in the +z direction at P. Then, combining Eq. 29-4 with Eq. 28-2, we obtain Fe = - e 0i 2r v k . b ge j (a) The electron is moving down: v = - vj (where v = 1.0 107 m/s is the speed) so Fe = -e 0iv ^ -i = (3.2 10-16 N) ^ , i 2r ( ) or | Fe |= 3.2 10-16 N . (b) In this case, the electron is in the same direction as the current: v = v i so Fe = -e 0iv ^ - j = (3.2 10-16 N) ^ , j 2r ( ) or | Fe |= 3.2 10-16 N . (c) Now, v = vk so Fe k k = 0. 88. Eq. 29-17 applies for each wire, with r = R 2 + d / 2 b g 2 (by the Pythagorean theorem). The vertical components of the fields cancel, and the two (identical) horizontal components add to yield the final result B=2 0i 2 r 0id d /2 = = 1.25 10-6 T , 2 2 r 2 R + ( d / 2 ) ( ) where (d/2)/r is a trigonometric factor to select the horizontal component. It is clear that this is equivalent to the expression in the problem statement. Using the right-hand rule, we find both horizontal components point in the +x direction. Thus, in unit-vector ^ notation, we have B = (1.25 10-6 T)i . 89. The "current per unit x-length" may be viewed as current density multiplied by the thickness y of the sheet; thus, = Jy. Ampere's law may be (and often is) expressed in terms of the current density vector as follows z B ds = 0 J dA z where the area integral is over the region enclosed by the path relevant to the line integral (and J is in the +z direction, out of the paper). With J uniform throughout the sheet, then it is clear that the right-hand side of this version of Ampere's law should reduce, in this problem, to 0JA = 0Jyx = 0x. (a) Figure 29-86 certainly has the horizontal components of B drawn correctly at points P and P' (as reference to Fig. 29-4 will confirm [consider the current elements nearest each of those points]), so the question becomes: is it possible for B to have vertical components in the figure? Our focus is on point P. Fig. 29-4 suggests that the current element just to the right of the nearest one (the one directly under point P) will contribute a downward component, but by the same reasoning the current element just to the left of the nearest one should contribute an upward component to the field at P. The current elements are all equivalent, as is reflected in the horizontal-translational symmetry built into this problem; therefore, all vertical components should cancel in pairs. The field at P must be purely horizontal, as drawn. (b) The path used in evaluating z B ds is rectangular, of horizontal length x (the horizontal sides passing through points P and P' respectively) and vertical size y > y. The vertical sides have no contribution to the integral since B is purely horizontal (so the scalar dot product produces zero for those sides), and the horizontal sides contribute two equal terms, as shown next. Ampere's law yields 2 Bx = 0 x B= 1 0 . 2 90. In this case L = 2r is roughly the length of the toroid so B = 0i0 FG N IJ = ni H 2r K 0 0 This result is expected, since from the perspective of a point inside the toroid the portion of the toroid in the vicinity of the point resembles part of a long solenoid. 91. (a) For the circular path L of radius r concentric with the conductor z FG r Thus, B = 2ca - b h H 0i 2 2 L B ds = 2 rB = 0ienc = 0i c c a r 2 - b2 2 h. -b h 2 2 - b2 . r IJ K (b) At r = a, the magnetic field strength is FG a 2 c a - b h H 0i 2 2 2 - b2 i = 0 . a 2 a IJ K At r = b, B r 2 - b 2 = 0 . Finally, for b = 0 B= which agrees with Eq. 29-20. (c) The field is zero for r < b and is equal to Eq. 29-17 for r > a, so this along with the result of part (a) provides a determination of B over the full range of values. The graph (with SI units understood) is shown below. 0i r 2 2 a r 2 = 0ir 2 a 2 92. (a) Eq. 29-20 applies for r < c. Our sign choice is such that i is positive in the smaller cylinder and negative in the larger one. B= 0ir 2c 2 , r c. (b) Eq. 29-17 applies in the region between the conductors. B= 0i 2r , c r b. (c) Within the larger conductor we have a superposition of the field due to the current in the inner conductor (still obeying Eq. 29-17) plus the field due to the (negative) current in that part of the outer conductor at radius less than r. The result is B= 0i 2r - 0i r 2 - b 2 2r a 2 - b 2 , b < r a. If desired, this expression can be simplified to read B= FG 2 r H a 0i a 2 - r 2 2 IJ . -b K 2 (d) Outside the coaxial cable, the net current enclosed is zero. So B = 0 for r a. (e) We test these expressions for one case. If a and b (such that a > b) then we have the situation described on page 696 of the textbook. (f) Using SI units, the graph of the field is shown below: 93. We use Ampere's law. For the dotted loop shown on the diagram i = 0. The integral B ds is zero along the bottom, right, and top sides of the loop. Along the right side the z field is zero, along the top and bottom sides the field is perpendicular to ds . If is the length of the left edge, then direct integration yields B ds = B , where B is the z magnitude of the field at the left side of the loop. Since neither B nor is zero, Ampere's law is contradicted. We conclude that the geometry shown for the magnetic field lines is in error. The lines actually bulge outward and their density decreases gradually, not discontinuously as suggested by the figure.
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Kansas State - PHYS - 214
1. The amplitude of the induced emf in the loop is m = A 0 ni0 = (6.8 10-6 m 2 )(4 10 -7 T m A)(85400 / m)(1.28 A)(212 rad/s)= 1.98 10-4 V.2. (a) =d B d = 6.0t 2 + 7.0t = 12t + 7.0 = 12 2.0 + 7.0 = 31 mV. dt dtchb g(b) Appealing to
Kansas State - PHYS - 214
1. (a) All the energy in the circuit resides in the capacitor when it has its maximum charge. The current is then zero. If Q is the maximum charge on the capacitor, then the total energy is2.90 10-6 C Q2 U= = = 117 10-6 J. . -6 2C 2 3.60 10 Fc
FSU - HIST - 08758
Prcis 3-28-08The Supreme Court's decision on Brown v. Board of Education of Topeka (1954) served as a turning point and as a milestone in the start of the civil rights movement. It reviewed the &quot;separate but equal&quot; principle decided by the case of
FSU - HIST - 08758
Prcis 4-2-08George C. Wallace displayed his prominence among white conservatives as Governor of Alabama in a speech on&quot; The Civil Rights Movement: Fraud, Sham, and a Hoax&quot; (1964). In June of 1963, Wallace defiantly prevented the first black student
FSU - HIST - 08758
Prcis 4-4-08Martin Luther King's Letter from Birmingham Jail (1963)is a letter Dr. King wrote after being jailed in Birmingham Alabama after organizing a economic boycott of white businesses, using a smuggled pen and piece of paper. King's statemen
FSU - GLY - 01636
Origin of the Universe: The Big Bang: Supernova: Life elements: Leon Sinks: Karst topography: Caves, caverns: Ground water: Porosity: Permeability Water table: Springs: Sink holes: o Wet Sinks: o Dry sinks: o Sink vs. Swamp Natural Bridge: Evolution
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 1 ~ Eras of Roman History Reading: BHR 2-4.Spring 2008Roman history is traditionally divided into three phases: monarchy, republic and empire. The dates are as follows (keep in mind that the very earliest date
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 2: Roman Society and Roman Values Readings: BHR 129-131; Shelton 128, 163-4, 171, 194-5, 197Spring 2008Roman Values and Virtues Roman virtues public, not private: virtuous man acts on behalf of the state Virtu
FSU - CLA - 2123
Classics 2123: The Roman Way Notes to Virgil's Aeneid, Books I-IVSpring 2008Virgil's Aeneid, strongly modeled on Homer's Iliad and Odyssey, was considered even in antiquity the great epic of Rome. It tells the story of the Trojan Aeneas who, afte
FSU - CLA - 2123
Classics 2123: The Roman Way Notes to Virgil's Aeneid, Books V-VIIISpring 2008Book V provides a transition between the high emotion of Book IV and the sombre majesty of the descent to the underworld in Book VI. Most of the book is taken up with g
FSU - CLA - 2123
Classics 2123: The Roman Way Notes to Virgil's Aeneid, Books IX-XIISpring 2008Book IX. War finally breaks out, the full-scale battles spoken of in Book VII. The book divides into three sections: (1) Turnus and the Rutulians attack the Trojan ship
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 5. Roman Imperialism and Expansion Readings: BHR 44-67; Shelton 291-293 I. Roman Imperialism - definition of `imperialism' - older notion of `defensive imperialism': how realistic? - was Rome more warlike/aggressi
FSU - CLA - 2123
Classics 2123: The Roman Way Roman Names A Roman had three names: praenomen (first name) nomen (name of the gens or clan) cognomen (family branch) Thus for: Publius praenomen Cornelius nomen Scipio cognomenSpring 2008his given name was &quot;Publius,&quot;
Stevens - PEP - 111
AC CIRCUITS35.1. Model: A phasor is a vector that rotates counterclockwise around the origin at angular frequency w. Solve: (a) Refemng to the phasor in Figure Ex35.1, the phase angle isU? = 180'n rad - 30&quot; = 150 x -= 2.618 rad180&quot;w=2*618ra
Stevens - PEP - 111
15.1. Solve: The density of the liquid is=m 0.120 kg 0.120 kg = = = 1200 kg m 3 V 100 mL 100 10 -3 10 -3 m 3Assess: The liquid's density is more than that of water (1000 kg/m3) and is a reasonable number.15.2. Solve: The volume of the helium
Stevens - PEP - 111
16.1. Solve: The mass of lead mPb = Pb VPb = (11,300 kg m 3 )(2.0 m 3 ) = 22,600 kg . For water to have thesame mass its volume must beVwater =mwater 22,600 kg = = 22.6 m 3 water 1000 kg m 316.2. Solve: The volume of the uranium nucleus isV
Stevens - PEP - 111
17.1. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules. That is, Eth =Kmicro. Solve: The number of atoms isN=M 0.0020 kg = = 3.01 10 23 m 6.64 10 -27 kgBecause helium atoms have an atomic mass number A
Stevens - PEP - 111
18.1. Solve: We can use the ideal-gas law in the form pV = NkBT to determine the Loschmidt number (N/V):1.013 10 5 Pa N p = 2.69 10 25 m -3 = = V kB T (1.38 10 -23 J K )(273 K )()18.2. Solve: Nitrogen is a diatomic molecule, so r 1.0 10-1
Stevens - PEP - 111
19.1. Model: The heat engine follows a closed cycle, starting and ending in the original state. The cycleconsists of three individual processes. Visualize: Please refer to Figure Ex19.1. Solve: (a) The work done by the heat engine per cycle is the a
Stevens - PEP - 111
20.1. Model: This is a wave traveling at constant speed. The pulse moves 1 m to the right every second.Visualize: Please refer to Figure Ex20.1. The snapshot graph shows the wave at all points on the x-axis at t = 0 s. You can see that nothing is h
Stevens - PEP - 111
21.1. Model: The principle of superposition comes into play whenever the waves overlap.Visualize:The graph at t = 1 s differs from the graph at t = 0 s in that the left wave has moved to the right by 1 m and the right wave has moved to the left by
Stevens - PEP - 111
22.1. Visualize: Please refer to Figure Ex22.1.Solve: (a)(b) The initial light pattern is a double-slit interference pattern. It is centered behind the midpoint of the slits. The slight decrease in intensity going outward from the middle indicates
Stevens - PEP - 111
23.1. Model: Light rays travel in straight lines.Solve: (a) The time ist=x 1.0 m = = 3.33 10 -9 s = 3.33 ns c 3 10 8 m / s(b) The refractive indices for water, glass, and zircon are 1.33, 1.50, and 1.96, respectively. In a time of 3.33 ns, l
Stevens - PEP - 111
24.1. Model: Balmer's formula predicts a series of spectral lines in the hydrogen spectrum.Solve: Substituting into the formula for the Balmer series,=91.18 nm 91.18 nm = = 410.3 nm 1 1 1 1 - 2 - 2 2 22 n 2 6where n = 3, 4, 5, 6, . and wher
Stevens - PEP - 111
ELECTROMAGNETIC AND WAVES FIELDSw.1. Model: The net magnetic flux over a closed surface is zero. Visualize: Please refer to Ex34.1. Solve: Because we can't enclose a &quot;net pole&quot; within a surface, Q, = f B . d i = 0 . Since the magnetic field isunif
FSU - CLA - 2123
CLA 2123: The Roman Way First Exam. February 8, 2007Name _Please read all directions carefully; no credit will be given for doing more than is required in each section. Part I. Identifications (35 points). Choose FIVE of the following and identif
Stevens - PEP - 111
14.1. Solve: The frequency generated by a guitar string is 440 Hz. The period is the inverse of the frequency, henceT= 1 1 = = 2.27 10 -3 s = 2.27 ms f 440 Hz14.2. Solve: Your pulse or heart beat is 75 beats per minute. The frequency of your hear
Stevens - PEP - 111
1.1.Solve:1.2.Solve:Solve: (a) The basic idea of the particle model is that we will treat an object as if all its mass is concentrated into a single point. The size and shape of the object will not be considered. This is a reasonable approxim
Stevens - PEP - 111
2.1.Solve:Model: The car is represented by the particle model as a dot. (a) Time t (s) Position x (m) 0 1200 1 975 2 825 3 750 4 700 5 650 6 600 7 500 8 300 9 0(b)2.2. Solve:Diagram (a) (b) (c)Position Negative Negative PositiveVelocity
Stevens - PEP - 111
3.1. Solve: (a) If one component of the vector is zero, then the other component must not be zero (unless the whole vector is zero). Thus the magnitude of the vector will be the value of the other component. For example, if Ax = 0 m and Ay = 5 m, the
Stevens - PEP - 111
4.1. Solve: A force is basically a push or a pull on an object. There are five basic characteristics of forces. (i) A force has an agent that is the direct and immediate source of the push or pull. (ii) Most forces are contact forces that occur at a
Stevens - PEP - 111
5.1.Model: We can assume that the ring is a single massless particle in static equilibrium. Visualize:Solve:Written in component form, Newton's first law is( Fnet ) x = Fx = T1x + T2 x + T3 x = 0 NT1 x = - T1T1y = 0 N Using Newton's first l
Stevens - PEP - 111
6.1. Model: We will assume motion under constant-acceleration kinematics in a plane.Visualize:Instead of working with the components of position, velocity, and acceleration in the x and y directions, we will use the kinematic equations in vector f
Stevens - PEP - 111
7.1. Solve: (a) From t = 0 s to t = 1 s the particle rotates clockwise from the angular position +4 rad to -2 rad. Therefore, = -2 - ( +4 ) = -6 rad in one sec, or = -6 rad s . From t = 1 s to t = 2 s, = 0 rad/s. From t = 2 s to t = 4 s the partic
Stevens - PEP - 111
8.1. Visualize:Solve: Figure (i) shows a weightlifter (WL) holding a heavy barbell (BB) across his shoulders. He is standing on a rough surface (S) that is a part of the earth (E). We distinguish between the surface (S), which exerts a contact forc
Stevens - PEP - 111
Solve: (a) The momentum p = mv = (1500 kg)(10 m /s) = 1.5 10 4 kg m /s . (b) The momentum p = mv = (0.2 kg)( 40 m /s) = 8.0 kg m /s .9.1. Model: Model the car and the baseball as particles.9.2. Model: Model the bicycle and its rider as a particl
Stevens - PEP - 111
10.1. Model: We will use the particle model for the bullet (B) and the bowling ball (BB).Visualize:Solve:For the bullet,KB =For the bowling ball,1 1 2 mB vB = (0.01 kg)(500 m /s) 2 = 1250 J 2 2 1 1 2 mBB vBB = (10 kg)(10 m / s) 2 = 500 J 2
Stevens - PEP - 111
11.1. Visualize:r Please refer to Figure Ex11.1. rSolve: (b) (c)(a) A B = AB cos = ( 4)(5)cos 40 = 15.3. r r C D = CD cos = (2)( 4)cos120 = -4.0. r r E F = EF cos = (3)( 4)cos 90 = 0.11.2. Visualize:r Please refer to Figure Ex11.2. rSolve
Stevens - PEP - 111
12.1.Solve: (b)Model: Model the sun (s), the earth (e), and the moon (m) as spherical. (a)Fs on e =Gms me (6.67 10 -11 N m 2 / kg 2 )(1.99 10 30 kg)(5.98 10 24 kg) = 3.53 10 22 N = (1.50 1011 m ) 2 rs2 e -Fm on e =GMm Me (6.67 10 -1
Stevens - PEP - 111
13.1. Model: The crankshaft is a rotating rigid body.Solve: The crankshaft at t = 0 s has an angular velocity of 250 rad/s. It gradually slows down to 50 rad/s in 2 s, maintains a constant angular velocity for 2 s until t = 4 s, and then speeds up
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 3. The Early History of Rome and Roman Republican Government Readings: BHR 15-41; Shelton 2-4, 7-8, 251-3, 255-259, 262, 264-5Spring 2008Early History of Rome Regal period: dimly known, mostly through legends;
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 4. Roman Family LifeSpring 2008Readings: BHR 129-31; Shelton nos. 15, 17-23, 25-27, 30-37, 44-45, 50, 54-56, 59-61, 63-67, 72, 75, 119-120, 124-5I. Definition of the family - familia - power of the father (p
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 4. Roman Family LifeSpring 2008Readings: BHR 129-31; Shelton nos. 15, 17-23, 25-27, 30-37, 44-45, 50, 54-56, 59-61, 63-67, 72, 75, 119-120, 124-5I. Definition of the family - familia - power of the father (p
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 6. Internal Disorders; Roman Slavery Readings: BHR 82-92; Shelton 207-209, 219, 227-229, 317-318Spring 2008I. Establishment of Roman provincial government Provincia: sphere of action of a magistrate with imper
UMiami - ACC - 212
Chapter 10Standard CostingAccounting 21210 - 1Learning Objective 1 Describe standard costing and indicate why standard costing is important.Accounting 212 10 - 2Why is Standard Costing Used?A standard is a preestablished benchmark for des
FSU - CLA - 2123
Classics 2123: The Roman Way Outline for Lecture 7. Roman Religion Readings: BHR 41-44; Shelton nos. 402-419, 423-428Spring 2008Problems Studying Ancient Religious Systems - time and culture (modern politics separated from religion) - vocabulary
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 8. The Strains of Empire (I) Readings: BHR 72-77, 92-110; Shelton 187-189, 266, 317-318Spring 2008I. Continuation of Roman Imperialism Macedonian Wars - Rome fights four wars in Greece, first against the Maced
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 9. The Strains of Empire (II) Readings: BHR 111-128, 132-140Spring 2008I. New Developments in the Roman State in the Late Republic Breakdown of concordia in the late Republic the result of many things, includi
UMiami - ACC - 212
Chapter 9The Operating Budget2004 Prentice Hall Business Publishing Introduction to Management Accounting , 2/e Werner/Jones9-1Learning Objective 1Describe some of the benefits of the operating budget.2004 Prentice Hall Business Publishing
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 10. The Fall of the Roman Republic Readings: BHR 124-179.Fall 2008I. Career of Pompey (Gn. Pompeius Magnus) (106-48 BCE) - begins as supporter of Sulla: raises legions from his father's troops (client army); a
UMiami - ACC - 212
Chapter 11Evaluating PerformanceAccounting 21211 - 1Learning Objective 1Describe centralized and decentralized management styles.Accounting 212 11 - 2Centralized ManagementTop management makes most of the decisions.The most experience
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 11. Augustus' `Restored' Republic Readings: BHR 167-199; Shelton 38-40, 77-78, 267, 271, 274-276, 294-305.Spring 2008I. From Octavian to `Augustus' - 31 BCE victory over Antony and Cleopatra at Actium; after h
FIU - GEG - 211
Earth 1. When was Big bang and when did our Earth form? 13.5 billion years ago big bang. 4.53 billion years. 2. Know earth's radius/diameter, thicknesses of crusts and lithosphere 6,370km radius, crust is 40 km, lithosphere is 100 km 3. Know the phys
UMiami - ACC - 212
HAPTER 1MANAGEMENT ACCOUNTING: ITS ENVIRONMENT AND FUTURESOLUTIONS TO CHAPTER 1 QUICK QUIZ 1. 2. 3. 4. 5. C B D C D 6. 7. 8. 9. 10. C B D B D 2004 Prentice Hall, Inc.M1 - 2Chapter 1 Management Accounting: Its Environment and FutureQUICK
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 12: Virgil's Aeneid, Books I-IVSpring 2008I. Publius Vergilius Maro (Virgil or Vergil) - born ca. 70, died ca. 19 BCE - from northern Italy; ancient tradition that his family lost land in the proscriptions of
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 12. Virgil's Aeneid, Books II-VIIISpring 2008I. Dido and Aeneas - Dido's story: the wrong done her husband; her exile; her foundation of Carthage - her welcoming of the Trojans (the concern of Jupiter) - the p
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 14: Virgil's Aeneid, Books VII-XII I. War in the Aeneid war always considered glorious in epic, but Virgil here exploring a civil war war also in the world of the Aeneid not an end, but a means to an endSpring 2
FSU - CLA - 2123
Classics 2123: The Roman Way Ovid's AmoresSpring 2008Readings: Amores (given by Book and poem number): I. 3, 5, 7, 9, 14, 15; II. 1, 4, 7, 13, 14; III. 7, 8, 15 I. Ovid (P. Ovidius Naso) - other great poet of the Augustan age - poet of love par e
FSU - CLA - 2123
Classics 2123: The Roman Way Ovid's Ars Amatoria Readings: Ars Amatoria Books IIIISpring 2008I. Precedents and Heritage - AA meant to be seen as a didactic poem: long history of didactic poetry in Greek and Roman literature - first practitioner H
FSU - CLA - 2123
Classics 2123: The Roman WaySpring 2008Lecture 17. The Julio-Claudians; Imperial Women and the Imperial HouseholdReadings: BHR 201-221; Shelton 78, 429-431 JulioClaudian Emperors: Name given to the first five emperors: Augustus (ruled 27 BCE14 C
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 18: Social Life in the Early EmpireSpring 2008Readings: BHR 241-5; Shelton 9-13, 41-42, 134-139, 141-157, 160, 220-226, 232-250, 377-81, 383-88, 391-401 I. Education and Schools (Shelton 134-139, 141-157, 160)