Solution manual fundamentals of physics extended, 8th editionch29

1. (a) 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=

µ 0i
2πr

.

With r = 20 ft = 6.10 m, we have

c4π × 10
B=

hb
2 π b6.10 mg
−7

T ⋅ m A 100 A

g = 3.3 × 10

−6

T = 3.3 µT.

(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 µT and must
be directed due south. Since B = µ 0i 2 πr ,
i=

2 πrB

µ0

=

b

gc

2 π 0.080 m 39 × 10−6 T
4 π × 10 −7 T ⋅ m A

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 φ = π) and the right-hand rule, we find that the current in the semicircular arc H J

contributes µ 0i 4 R1 (into the page) to the field at C. Also, arc D A contributes µ 0i 4 R2
(out of the page) to the field there. Thus, the net field at C is

B=

µ 0i § 1

¨
4 © R1

−

1 · (4π ×10 −7 T ⋅ m A)(0.281A) §
1
1
·
−6
−
¸ =
¨
¸ = 1.67 ×10 T.
R2 ¹
4
© 0.0315m 0.0780m ¹

(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 φ = θ) and the right-hand rule, we find that the current in the
semicircular arc of radius b contributes µ 0iθ 4 πb (out of the page) to the field at P. Also,
the current in the large radius arc contributes µ 0iθ 4 πa (into the page) to the field there.
Thus, the net field at P is
B=

µ 0iθ § 1 1 ·

¨ − ¸ =
4 ©b a¹
= 1.02 ×10 −7 T.

(4π ×10 −7 T ⋅ m A)(0.411A)(74°⋅π /180°) § 1
1 ·
−
¨
¸
4π
© 0.107m 0.135m ¹

(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 φ = π) indicates that the current in the semicircular arc contributes µ 0i 4 R
to the field at C. Thus, the magnitude of the magnetic field is

B=

µ 0i
4R

=

(4π ×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, µ0i/2πr. Thus the total
field at the midpoint has magnitude B = µ0i/πr and
i=

πrB

µ0

=

π ( 0.040 m ) ( 300 ×10−6 T )
4π × 10 −7 T ⋅ m A

= 30 A.

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

µ 0i
µ 0 (3i )
d 16.0 cm
=
Ÿ r= =
= 4.0 cm.
2π r 2π (d − r )
4
4
(b) Doubling the currents does not change the location where the magnetic field is zero.


9. (a) BP1 = µ0i1/2πr1 where i1 = 6.5 A and r1 = d1 + d2 = 0.75 cm + 1.5 cm = 2.25 cm, and
BP2 = µ0i2/2πr2 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 ¹
© r1 ¹

(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 =

µo
i R ds
2
2 3/2 .
4π (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 = µo i ds /4πR2.
1
dBmax. This is a non-trivial algebra
(b) We want to find the s value such that dB = 10

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
G µi
 Thus,
proton with charge q is B = 0 k.
2πd
G
G G µ iq G
F = qv × B = 0 v × k .
2πd

e

j

G
In this situation, v = v − j (where v is the speed and is a positive value), and q > 0. Thus,

e j

G µ iqv
F= 0
2πd

(( ) )

−ˆj × kˆ = −

= (−7.75 ×10

−23

ˆ
N)i.

µ 0iqv ˆ
2πd

i =−

(4π ×10 −7 T ⋅ m A)(0.350A)(1.60 ×10−19 C)(200m/s) ˆ
i
2π (0.0289 m)


12. The fact that By = 0 at x = 10 cm implies the currents are in opposite directions. Thus
By =

µo i1
µo i2 µo i2 § 4
1·
–
=
¨L + x – x¸ .
2π (L + x) 2 π x 2π ©
¹

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: µo i2 /2πL. 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 µo i2 /2πL = 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 µ 0i 4 πR (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

µ 0iφ

§ µi ·
= 2.00 ¨ 0 ¸
4πR
© 4πR ¹

which yields φ = 2.00 rad.

14. Initially, Bnet y = 0, and Bnet x = B2 + B4 = 2(µo i /2πd) using Eq. 29-4, where d = 0.15
m. To obtain the 30º condition described in the problem, we must have
Bnet y = Bnet x tan(30º)
B1′ – B3 = 2(µo i /2πd) tan(30º)
where B3 = µo i /2πd and B1′ = µo i /2πd′. Since tan(30º) = 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 = µ0i/2πr, 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 = µ 0i 2πa .
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

−7
2 µ 0i 2 ( 4π × 10 T ⋅ m A ) ( 20 A )
cos 45° =
=

= 8.0 ×10−5 T.
πa
π ( 0.20 m )
2πa

µ 0i

In the calculation cos 45° was replaced with 1
G
ˆ
the +y direction. Thus, Btotal = (8.0 ×10−5 T)j.

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=

µo i
2πR

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
L
≈ 14.1
≈ 14.1R Ÿ
R
201


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 =

µ 0i sin θ
4π r 2

dx

where θ (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

x 2 + R 2 and sin θ with

x 2 + R 2 , we integrate from x = –L/2 to x = L/2. The total field is

B=

µ 0iR
4π

³

dx

L2

−L 2

( 4π ×10
=

−7

(x

=

µ 0iR 1

x

4π R ( x
)
T ⋅ m A ) ( 0.0582 A )

2

+R

2 32

2π ( 0.131 m )

2

2

L2

+R

)

2 1 2 −L 2

=

µ 0i

L

2πR

L + 4R2
2

0.180m

(0.180m) + 4(0.131m)
2

2

= 5.03 × 10−8 T.


18. Using the law of cosines and the requirement that B = 100 nT, we have
B12 + B22 – B2·
= 144º .
–2B1B2
¹

θ = cos−1 §
©

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 =

µ 0i sin θ

4π r 2

dx

where θ (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 θ with

x 2 + R 2 , we integrate from x = –L to x = 0. The total field is

B=

µ 0iR
4π

³

dx

0

−L

( 4π ×10
=

(x
−7

2

+R

)

2 32

=

µ 0iR 1
4π R

T ⋅ m A ) ( 0.693 A )

4π ( 0.251 m )

2

x

(x

2

0

+R

)

2 1 2 −L

=

µ 0i

L

4πR

L + R2
2

0.136m
(0.136m) + (0.251m)
2

2

= 1.32 × 10−7 T.


20. In the one case we have Bsmall + Bbig = 47.25 µT, and the other case gives Bsmall – Bbig
= 15.75 µT (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.
−
rsmall rbig
The solution of this is straightforward: rsmall = 12 rbig.

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 =

µ 0i
2 πR

The contribution from the circular loop is BC 2 =

.

µ 0i
2R

. Thus,

−7
−3
1 · ( 4π ×10 T ⋅ m A )( 5.78 ×10 A ) § 1 ·

−7
BC = BC1 + BC 2 =
¨1 + ¸ =
¨1 + ¸ = 2.53 ×10 T.
2R © π ¹
2 ( 0.0189 m )
© π¹

µ 0i §

G
BC points out of the page, or in the +z direction. In unit-vector notation,
G
BC = (2.53×10−7 T)kˆ
G
G
(b) Now BC1 ⊥ BC 2 so
BC = B + B
2
C1

2
C2

−7
−3
1 ( 4π ×10 T ⋅ m A )( 5.78 ×10 A )
1
=
1+ 2 =

1 + 2 = 2.02 ×10−7 T.
2R
π
2 ( 0.0189 m )
π

µ 0i

G
and BC points at an angle (relative to the plane of the paper) equal to
§B ·
§1·
tan −1 ¨ C1 ¸ = tan −1 ¨ ¸ = 17.66° .
© π¹
© BC 2 ¹

In unit-vector notation,
G
ˆ = (1.92 ×10−7 T)iˆ + (6.12 ×10−8 T)kˆ
BC = 2.02 ×10−7 T(cos17.66°ˆi + sin17.66°k)


22. Letting “out of the page” in Fig. 29-50(a) be the positive direction, the net field is
B=

µo i1 φ
µo i2
–
4πR 2π(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

φ = 4(i2 /i1) = 4(0.5/2) = 1.00 rad (or 57.3º).


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 =

µ 0di
2 πx

=

µ 0idx
2 πxw

.

Thus,
BP = ³ dBP =

µ 0i

d +w

2πw ³d

−7

−6
dx µ 0i § w · ( 4π ×10 T ⋅ m A )( 4.61×10 A ) § 0.0491 ·
ln ¨1 + ¸ =
ln ¨1 +
=
¸
2π ( 0.0491 m )
x 2πw © d ¹
© 0.0216 ¹

= 2.23×10−11 T.

G
G
and BP points upward. In unit-vector notation, BP = (2.23×10−11 T)ˆj


24. Initially we have
Bi =

µo i φ
µo i φ
+
4πR
4πr

using Eq. 29-9. In the final situation we use Pythagorean theorem and write
Bf

2

2

2

= Bz + By

2

2

§µo i φ· §µo i φ·
=¨
¸ +¨
¸ .
© 4πR ¹ © 4πr ¹

2

If we square Bi and divide by Bf , we obtain
2

§ 1 1·
¨R + r ¸
©
¹
§ Bi ·
¨B ¸ = 1 1 .
© f¹
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)
2 ( 4π × 10 −7 T ⋅ m A ) (13 A )
§ 2 µ 0i · § 2 µ 0i ·
2 µ 0i
=
BP = 2 ¨¨
¸¸ − 2 ¨¨
¸¸ =
8π ( 0.047 m )
© 8πa ¹ © 8π ( 2 a ) ¹ 8πa
= 1.96 × 10 −5 T ≈ 2.0 × 10 −5 T.

(b) The direction of the field is into the page.