37. 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
36. 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
35. 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
Fx 0i1i2 d 2 (4107 T m/A)(4.00 103 A)(6.80 103 A)(0.050 m) =
34. 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. 29-62(c). This means wire #2 is either at +/2
33. (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 29-21 to evaluate the contributions to
32. By the right-hand rule (which is built-into Eq. 29-3) the field caused by wire 1s 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 2s current will generally have both an
31. 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 = Thus, BP = dBP = 2w d
0di
2 x
=
0idx
2 xw
.
0i
d +w
dx 0i w (4107 T m A)(4.61106 A) 0
30. Initially we have Bi =
0i 0i + 4 R 4 r
using Eq. 29-9. In the final situation we use Pythagorean theorem and write i i 2 B 2 = Bz2 + By = 0 + 0 . f 4 R 4 r
2 2
If we square Bi and divide by Bf , we obtain
Bi B f [(1/ R) + (1/ r )]2 . = (1/ R) 2 + (1
29. 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, i | Bnet | = 2 0 sin 2 r where i = 4.00 A, r = r = d
28. (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,
7 3 1 ( 4 10 T m A ) ( 5.78 10 A ) 1 7 BC = BC1 + BC 2 = 1 + = 1 + = 2.5310 T. 2R 2 ( 0.018
27. The contribution to Bnet from the first wire is (using Eq. 29-4) B1 =
0i1 (4 107 T m/A)(30 A) k= k = (3.0 106 T)k. 2 r1 2 (2.0 m)
The distance from the second wire to the point where we are evaluating Bnet is r2 = 4 m 2 m = 2 m. Thus, B2 =
0i2 (4 107
26. Letting out of the page in Fig. 29-55(a) be the positive direction, the net field is B=
0i1 0i2 4 R 2 ( R / 2)
from Eqs. 29-9 and 29-4. Referring to Fig. 29-55, we see that B = 0 when i2 = 0.5 A, so (solving the above expression with B set equal to ze
25. 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
= tan1(B2 /B1) = tan1(i2 /i1) = 53.13.
The accomplish the net field rotation de
24. 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!). Dividi
23. (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 ) 0 ( 0.80 A
22. Using the Pythagorean theorem, we have i i B = B + B = 0 1 + 0 2 4 R 2 R
2 2 2 1 2 2 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 1010) and the part of the se
21. 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
20. Using the law of cosines and the requirement that B = 100 nT, we have
2 B12 + B2 B 2 = cos = 144 , 2 B1 B2
1
where Eq. 29-10 has been used to determine B1 (168 nT) and B2 (151 nT).
19. 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 2 a . The fields
18. We consider Eq. 29-6 but with a finite upper limit (L/2 instead of ). This leads to
B=
0i
L/2
2R ( L / 2) 2 + R 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
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
16. Initially, we have 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) i B1 B3 = 2 0 tan(30) 2 d
where B3 = o i /2d and B1 = 0i /
15. 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
14. The fact that By = 0 at x = 10 cm implies the currents are in opposite directions. Thus
By =
0i2 0i2 4 1 = 2 ( L + x) 2 x 2 L + x x
0i1
using Eq. 29-4 and the fact that i1 = 4i2 . To get the maximum, we take the derivative with respect to x and set e
13. 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 proton with charge q is B = ( 0i / 2 d ) k. Thus, F = qv B =
0iq
2 d
ev kj.
12. With the usual x and y coordinates used in Fig. 29-43, then the vector r pointing i from a current element to P is r = s + R . Since ds = ds , then | ds r | = Rds. i j Therefore, with r = s 2 + R 2 , Eq. 29-3 gives dB =
0 iR ds . 2 4 ( s + R 2 )3/ 2
(
11. (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 i
10. (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
9. (a) BP1 = 0i1 / 2 r1 where i1 = 6.5 A and r1 = d1 + d2 = 0.75 cm + 1.5 cm = 2.25 cm, and BP2 = 0i2 / 2 r2 where r2 = d2 = 1.5 cm. From BP1 = BP2 we get
r 1.5 cm i2 = i1 2 = ( 6.5 A ) = 4.3A. 2.25 cm r1
(b) Using the right-hand rule, we see that the cu
8. (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