Physics I Mechanics (Class notes for PH111)
Massimiliano Bonamente Physics Department University of Alabama in Huntsville
(version of September 7, 2009)
Contents 1 Physical quantities, measurements and units of measure 1.1 1.2 Physical quantities, standar

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

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

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

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

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

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

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

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

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

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

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

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

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) =

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

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

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).

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 cu

6. (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 semicircul

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. Sin

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

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

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

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

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
(

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.

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

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