28.11: a At the point exactly midway between the wires, the two magnetic fields are in opposite directions and cancel... 28.12: The total magnetic field is the vector sum of the constan
Trang 128.1: For a charge with velocity v(8.00106 m s ˆj, the magnetic field produced at
a position r away from the particle is ˆ
r
.ˆˆ
)T101.92(
ˆ)
m0.50(
)sm108.0)(
C10(6.04
ˆ
5 2
6 6
0 2
ˆˆˆˆ)m0.500
ˆˆˆˆ)m0.500(ˆm0.500
ˆ2
ˆ22
ˆ4
0 0
qv B
B B
page
theintoT,1038.4
)m120.0(
)sm100.9)(
C100.3()
m120.0(
)sm105.4)(
C100.8(4
4
2
6 6
2
6 6
Trang 228.3: 0 3
q π
is
T101.31)
m0.500(
)sm106.80)(
C104.80)(
CsN1014
6
6 2
5 6
2 2 7 2
4.62
)m0.7071(
)sm106.80)(
m0.500(C)104.80)(
CsN1014
7 7
3
5 6
2 2 7 3
m)0.500(
)sm106.80)(
C104.80)(
CsN1014
6 6
2
5 6
2 2 7 2
theonforcetheanduppointschargeupper
theonforce(theN10
1.69
)m0.240(
m109.00)(
sm104.50)(
C103.00)(
C108.00()AmT10(4
3
2
6 6
6 6
7 2
CmN108.99
2 12
m) (0.240
C 10 0) (8.00)(3.0 2
charge points up and the force on the lower charge points down)
The ratio of the Coulomb force to the magnetic force is
2
2
v c
10 1.69 N 3.75 3 2.22 10 b) The magnetic forces are reversed when the direction of only one velocity is
reversed but the magnitude of the force is unchanged
Trang 328.5: The magnetic field is into the page at the origin, and the magnitude is
page
theintoT,101.64
m)0.400(
)sm108.0)(
C101.5()
m0.300(
)sm102.0)(
C104.0(44
6
2
5 6
2
5 6
0
2 2 0
r
v q r
qv π
μ B B
q q into the page; 0 2
4πd
v q μ
qv μ B
v v
4π d
v v q μ
0 2
0
2 2
2
)2(4
,)2(
F
F d
πε
q F
d π
v q μ F
C
B C
B
.101.00 6
)m0.500)(
m0.010)(
A125(4ˆm)0.500(4
ˆ
μ r
dl I π
μ r
d I π
μ
.ˆT104.3
d
Trang 428.8: The magnetic field at the given points is:
2 0
2
0 2
0
6 2
0 2
0
6 2
0 2
0
6 2
0 2
0
sin4
.0)0(sin4
sin4
T
102.00)
m0.100(
)m0.000100(
A)200(4
sin4
T.100.705)
m0.100(2
45sin)m0.000100(
A)(2004
sin4
T
102.00)
m0.100(
)m0.000100(
A)200(4
sin4
r
θ dl I π
μ dB
r
dl I π
μ r
θ dl I π
μ dB
π
μ r
θ dl I π
μ dB
π
μ r
θ dl I π
μ dB
π
μ r
θ dl I π
μ dB
3
2)
m100.0(3
)m00100.0(A)200(4
6
2 0
Trang 528.9: The wire carries current in the z-direction The magnetic field of a small piece of
d I π
90sinm)105()A00.4(4ˆ
sin4
11 2
4 0
θ dl I π
ˆ)
m00.2(
)90(sinm)105()A00.4(4ˆ
sin4
11
2
4 0
2 0
i
i i
θ dl I π
μ d
2
1ˆˆˆm00.2(ˆm00
1077
1
)
ˆ2
1m)2.00(m)00.2(
m)10(5.0A)00.4(4)
ˆ2
1sin4
11
2 2
4 0
2 0
i j
i j i
j B
θ dl I π
μ d
d) r(2.00m)kˆlˆrˆ0
3
43
8223
12
12
:2
0 0
0
πd
I μ d
π
I μ d
d π
I μ B
3
4 0
, in the jˆ direction
28.11: a) At the point exactly midway between the wires, the two magnetic fields are in
opposite directions and cancel
b) At a distance a above the top wire, the magnetic fields are in the same
3
2ˆ)3(2
ˆ2
ˆ2
ˆ2
0 0
0 2
0 1
0
a
I a
I a
I r
I r
Trang 628.12: The total magnetic field is the vector sum of the constant magnetic field and the
wire’s magnetic field So:
a) At (0, 0, 1 m):
.ˆ)T100.1(
ˆ)m00.1(2
)A00.8(ˆT1050.1(
ˆ2
7 0
6 0
T,102.19ˆT)10(1.6ˆT)1050.1(
ˆ)m00.1(2
)A00.8(ˆT1050.1(
ˆ2
6 6
6
0 6
0 0
θ π
μ πr
I μ
k i
B
k i
k B
B
from x to z.
)m25.0(2
)A00.8(ˆT1050.1(
ˆ2
0 6
0 0
π
μ πr
24
)(
4)
(
0 2
1 2 2 2
0 2 3 2 2
0
a x x
a π
I μ y
x x
y π
Ix μ y
x
xdy π
I μ
μ
π μ
πrB I
πr
I μ B
2m)0.080(
so,2
4 0
πr
I μ
B
4m)160.0
B
m)(5.502
A)800(2
5 0
π
μ πr
I μ
b) Since the magnitude of the earth’s magnetic filed is 5.00105 T, to the north, the total magnetic field is now 30o east of north with a magnitude of 5.78105 T This could be a problem!
Trang 728.16: a) B = 0 since the fields are in opposite directions.
a b
I πr
I μ πr
I μ B B
22
2
0 0
T6.67T1067.6
m0.2
1m0.3
12
)A(4.0)ATm04(
θ B θ B B
a
b a
cos2
coscos
0.0(2
)A(4.0)ATm104(2
cos2
2
2 2
7 0
I B
a
,T53.7T1053
Trang 8
28.17: The only place where the magnetic fields of the two wires are in opposite
directions is between the wires, in the plane of the wires
Consider a point a distance x from the wire carrying I = 75.0 A 2 Btot will be zero where B1 B2
A0.75A,
0.25
;)m400.0(
2)m400.0(2
2 1
1 2
2 0 1
x I x I
πx
I μ x π
I μ
x = 0.300 m; Btot 0 along a line 0.300 m from the wire carrying 75.0 A amd 0.100 m from the wire carrying current 25.0 A
b) Let the wire with I1 25.0 A be 0.400 m above the wire with I = 75.0 A 2
The magnetic fields of the two wires are in opposite directions in the plane of the wires and at points above both wires or below both wires But to have B1 B2 must be closer
to wire #1 since I1 < I , so can have 2 Btot 0 only at points above both wires
Consider a point a distance x from the wire carrying I1 25.0 A B will be tot
zero where B1 B2
m200.0);
m400.0(
)m400.0(22
1 2
2 0 1
x I
x π
I μ πx
I μ
0
tot
B along a line 0.200 m from the wire carrying 25.0 A and 0.600 m from the wire carrying current I2 75.0 A
Trang 928.18: (a) and (b) B = 0 since the magnetic fields due to currents at opposite corners of
the square cancel
(c)
left
the toT,100.4
45cosm)
210.0(2
A)(100)ATm104(4
m20.10cm210cm)(10cm)(10
45cos2
445cos4
45cos45
cos45
cos45
cos
4
7
2 2
r
πr
I μ B
B B
B B
B
a
d c
b a
Trang 10r
I B
T1080.0T,
5 1
0
T1000.2,
T1080.0T,
1000
1
6 3
2 1
4
4 3
2
z
5 z
5 z
5 z
z
z z z
B B B
B
B B
B
B
B B
B
To give B 4 in the direction the current in wire 4 must be toward the bottom of the page
A0.2)
AmT10(2
T)100.2(m)200.0()2(
rB I
112
2 0
2 0
d
I d
d
I L
On the middle wire, the magnetic fields cancel so the force is zero
42
112
2 0
2 0
d
I d
d
I L
L I Lg
λ22
λ
2 0
2 0
Trang 1128.22: a) 6.00 10 N,
)m400.0(2
)m20.1()A00.2()A00.5(2
6 0
2 1
repulsive since the currents are in opposite directions
b) Doubling the currents makes the force increase by a factor of four to
.N10
)m0250.0(2)mN100.4(
2
5 1
0 2
2 1
F I r
I I L
F
b) The two wires repel so the currents are in opposite directions
28.24: There is no magnetic field at the center of the loop from the straight sections
The magnetic field from the semicircle is just half that of a complete loop:
,42
2
12
loop
R
I R
I B
28.25: As in Exercise 28.24, there is no contribution from the straight wires, and now we
have two oppositely oriented contributions from the two semicircles:
,2
2
1)
2
R B
28.26: a) The field still points along the positive x-axis, and thus points into the loop
from this location
b) If the current is reversed, the magnetic field is reversed At point P the field would then point into the loop
c) Point the thumb of your right hand in the direction of the magnetic moment, under the given circumstances, the current would appear to flow in the direction that your
fingers curl (i.e., clockwise).
Trang 1228.27: a) 2.77A
)800()AmT104(
)T0580.0()m024.0(22
so,
aB I
a NI μ
x
b) At the center, B c 0 NI 2a. At a distance x from the center,
m0184.0so,m024.0with ,4
)
(
2
1)(
means
)(
)(
2)
(
2
6 3
2
2
2 2 2
3 2
1
2 2 2
3 2
2 2
3 0
2 2 2
2 0
a a
x
a x
a B
B
a x
a B
a x
a a
NI a
)m020.0(2
)A500.0()600(2
3 0
0
a
NI B
b) From Eq (29-16),
.T1034.1))m020.0()m080.0((
2
)m020.0()A500.0()600()m08.0()
2 0
2 2 2
2 2
2
)(
2)
(
Ia μ
a x x B N a
x
NIa μ x
)m06.0()m06.0()T1039.6(2
2 0
2 2 2
b) 4.0A (4.0A)5.0310 6Tm
0 1
2 1
I I I
Using Ampere’s Law in each case, the sign of the line integral was determined by using the right-hand rule This determines the sign of the integral for a counterclockwise path
Trang 1328.32: Consider a coaxial cable where the currents run in OPPOSITE directions.
22
0 0
πr
I μ B I μ πr B I μ d I
I b r
1 0 1
0 1
I μ B I μ r B I μ d I
I b r
πr
I I μ
28.34: Using the formula for the magnetic field of a solenoid:
.T0402.0)
m150.0(
)A00.8()600(0 0
L
NI μ nI μ
B
)A0.12(
)m400.0()T0270.0(0 0
I μ
BL N L
NI μ B
1790turns m
m400.0
turns716
b) The length of wire required is 2rN 2(0.0140 m )(116) 63 m
28.36:
L
N I
B0
N
BL I
)4000)(
ATm104(
)m40.1()T150.0(7
Trang 1428.37: a) 3.72 10 A
)2(so,2
6 0
π
Br I
πr
I μ B
2
5 0
N μ
aB I
, a
NI μ
B
a) r = 0.12 m, which is outside the toroid, so B = 0.
)m160.0(2
)A50.8()250(2
3 0
c) r = 0.20 m, which is outside the toroid, so B = 0
)m070.0(2
)A650.0()600(2
3 0
)m060.0(2
)A25.0()400()80(2
1400(
)T350.0()m0290.0(22
0
μ
π N μ K
B πr I
πr
NI μ K
m m
If Km I Ipart(a)
)A400.2()500(
)T940.1()m2500.0(22
πrB K
πr
NI μ K
b) Xm K m 12020
Trang 1528.43: a) The magnetic field from the solenoid alone is:
(i) (6000m )(0.15A) 1.13 10 3T
0
1 0
mNT
1)
slope
(
0 0
i r
v i
B
v
4)(
sin4
ˆ(
ˆ4
ˆ
2
0 2
μ r
θ v q π
qv r
q π
μ qv
4.0)
m500.0(
s)m1050.6)(
sm1000.9)(
C1000.5)(
C1000.8
(
4 4
6 6
Trang 1628.47:
m)045.0(
A)s)(2.50m
1000.6)(
C1060.1(22
4 19
I μ qv qvB
F
1.0710 19 N
Let the current run left to right, the electron moves in the opposite direction, below the wire, then the magnetic field at the electron is into the page, and the electron feels a force upward, toward the wire, by the right-hand rule (remember the electron is negative)
ev m
qvB m
F a
2
)m020.0)(
2)(
kg1011.9(
)A25)(
ATm104)(
sm000,250)(
C106.1(
31
7 17
b) The electric force must balance the magnetic force
eE = eVB
r
i v vB E
2
A)Tm/A)(2510
m/s)(4000
,250
π
π
62.5N/C, away from the wire
(c) mg(9.1110 31 kg)(9.8m/s2)10 29 N
gravity
neglectcan
weso,10
N10N/C)C)(62.510
6.1(grav
12 el
17 19
el
F F
eE F
Trang 1728.49: Let the wire connected to the 25.0 resistor be #2 and the wire connected to the 10.0 resistor be #1 Both I1 and I2 are directed toward the right in the figure, so at the location of the proton I2 isand I1 ⊙
T1080.4and
T1020.3T,10
with,2
and
2
5 2
1
5 2
5 1
2 0 2 1
B
πr
I B πr
I
μ
and in the direction ⊙
Force is to the right
N105.00T)
10m/s)(4.8010
C)(65010
602.1
)(
222
x R
IR B
B B
T1075.5]
m)125.0(m)20.0[(
m)A)(0.2050
.1(Tm/A)10
4
2 / 3 2 2
2 7
perpendicular to the line aband to the velocity
28.51: a)
Along the dashed line, B1 and B2 are in opposite directions
If the line has slope 1.00then r1 r2 and
.0so
, tot2
1 B B
B
Trang 1828.52: a)
001
ˆˆˆ4
ˆ
40 02 0 2 v0x v0y v0z
r
q π
μ r
q π
C1020.7(
m)T)(0.2510
00.6(4
T1000.6
|
|4and00
4
ˆT)1000.6(ˆˆ4
3 0
2 6
0
6 0
2
0 0
0 2 0
6 0
0 2 0
v r
q π v
v r
q π μ
v v r
q π μ
z
z y
y
y
m/s.607m/s)
521()m/s800(
0
2 0
2 0
v
k j i r
v
4010
ˆˆˆ4
ˆ4
)0,m250.0,0
0 0 0 2
0 2
0 0
z x z
y
r
q π
μ v
v v r
q π r
q π
250.0(
)C1020.7(4
|
|40)m,250.0,0
2
3 0
0 2
q π
μ B
28.53: Choose a cube of edge length L , with one face on the y-z plane Then:
,0ˆ
3 0 0
L B d a
x B d
d
L x L
x L
Trang 193 2
2
r
I θ r
I r
I π μ
B
j i
B
ˆ)16(ˆ)3.3312(2
ˆ8.0(m)050.0(
ˆ)6.0(m)050.0(m)030.0(2
3 2
3 0
3 3
2 0
I I
I π μ
I I
I π
j i
ˆT1028.1ˆT1072.3
ˆA)(16)(4.00ˆ
A))0(33.3)(2.0A)
00.4)(
12(2
5 6
c) FI1lBI1lB x jˆI1lB y iˆ
ckwisecounterclo2
.16,N1033.1
;ˆT1028.1ˆT1072.3
]T)1028.1(ˆT)1072.3[(
m010.0A00.1
7 7
8
5 6
i j
from +x-axis.
Trang 2028.55: a) If the magnetic field at point P is zero, then from Figure (28.46) the current I2
must be out of the page, in order to cancel the field from I Also:1
m)50.1(
m)500.0(A00.62
2 1 2 2
2 0 1
1 0 2
r
r I I πr
I μ πr
I μ B B
b) Given the currents, the field at Q points to the right and has magnitude
m50.1
A00.2m500.0
A00.622
6 0
2
2 1
I r
I π
μ
B Q
c) The magnitude of the field at S is given by the sum of the squares of the two
fields because they are at right angles So:
.T101.2m80.0
A00.2m
60.0
A00.622
6
2 0
2
2 2 2
1
1 0 2 2
I r
I π
μ B
Ia μ B
a x
a a
x π
I μ πr
I μ B
0
2 2 2 2 0 0
22
Trang 212 2
0 net
2 2 2 2 0 0
net
a x π
Ix μ B
a x
x a
x π
I μ θ
πr
I μ B
x
C dx
2 2
0
x I
B a
x which is just like a wire carrying current 2I
Trang 2228.58: a) Wire carrying current into the page, so it feels a force downward from the
other wires, as shown at right
0
400000
2 2
2 0
m A μ L
F
a x π
Ia μ I IB L
F
b) If the wire carries current out of the page then the forces felt will be the opposite of part (a) Thus the force will be 1.1110 5 N m, upward
28.59: The current in the wires is I R45.0V 0.50090.0A The currents
in the wires are in opposite directions, so the wires repel The force each wire exerts on the other is
m50.3A0.90AN1022
2 2
To hold the wires at rest, each spring exerts a force of 0.189 N on each wire
F kx so k F x0.189N 0.0050m37.8N m
28.60: a) Note that the Earth’s magnetic field exerts no force on wire B, since the
current in wire B is parallel to the Earth’s magnetic field Thus, for equilibrium, the remaining two forces that act on wire B must cancel Assuming that the length of wire B
is L and that wire A carries a current I we obtain
0m)(0.1002
)AA)(3.0(1.0
05002
m) π(
A)L I(
μ
So
A1.5m0.100
m0.050A)
0.3
Trang 23The wires are in equilibrium, so:
mg θ T y θ T F
x: sin and : cos
lB
mg mg
T IlB
sin(6.00m)
tan2
2
But
3 0 0
I lμ
θ mg πr I r
sm(9.80m)kg(0.0125m)
1036.8(2
0
2 3
28.62: The forces on the top and bottom segments cancel, leaving the left and right sides:
.ˆ)N10(7.97
ˆm0.026
1m
0.100
12
A)m)(14.0A)(0.200
(5.00
ˆ112
ˆ22
ˆ(ˆ(
5 0
wire 0 wire
0 wire 0
i i
F
i i
i i
r r π
IlI μ πr
I μ πr
I μ Il IlB IlB
l r r
l r
l r
l
x
Ia Nμ a
x
Ia Nμ B
a
2)
(
2 0 2
/ 3 2 2
2 0
2
sinsin
2)(
x
a a I πI N N θ x
Ia Nμ A I N
2)(
cos)
2 2 0 3
2 0 2
x
a a I μ N N x
Ia Nμ a
I N θ μB
c) Having x allows us to simplify the form of the magnetic field, whereas a
assuming x a means we can assume that the magnetic field from the first loop is constant over the second loop
Trang 2428.64: 1 ,outof thepage.
4
1122
I b
a
I μ B
B
Trang 2528.65: a) Recall for a single loop: 2 2 3 / 2.
0
) (
2x a a I
B Here we have two loops, each of
N turns, and measuring the field along the x-axis from between them means that the
"
" x in the formula is different for each case:
))2((
2
2 0
a a
x
NIa μ B
a x
2
2 0
a a
x
NIa μ B
a x
1)
)2((
1
2 0
x a
a x
NIa μ B
b) Below left: Total magnetic field Below right: Magnetic field from right coil
1)
)2((
12
0,
a a
a a
NIa μ B x
P
.5
4)
45(
0 2 2
2
2 0
a
NI μ a
NIa μ
A)(300)(6.005
45
)2(
3)
)2((
)2(
3
a x a
a x
a x NIa
2 2
2 2
2 0 2
2
2 2 2 2
2 2
2 0 0
))2((
)25()2(
6)
)2((
32
.0)
)2((
)2(3)
)2((
)2(32
a a
x
a x a
a x
NIa μ dx
B
d
a a
a a
a
a NIa
μ dx
2((
)25()2(
6)
)2((
3
a a
x
a x a
a x
Trang 2628.66: A wire of length l produces a field 0 2 2.
) 2 / (
2()2()2(4and
2
1)
2()2()2(4and
2
2 2
0 2 2
0 top
2 2
0 2 2
0 left
b a b
a π
I μ a
b b
a π
I μ B a l b
x
b a a
b π
I μ b
a a
b π
I μ B b l a
1
2 2
πab
I μ b a b
a a
b π
I μ
and is out of the page
28.68: The horizontal wire yields zero magnetic field since l r 0
d The vertical current provides the magnetic field of HALF of an infinite wire (The contributions from all infinitesimal pieces of the wire point in the same direction, so there is no vector addition or components to worry about.)
,22
and is out of the page
2
33
22
3 2
παR dr
r π α αrrdrdθ JdA
I
r I dr r π πR
I drdθ r πR
I I
R r
i
3 2
3
2 3
2
32
3)
3 0 encl 0
πR
Ir μ B R
r I μ I μ πr B
)
0 encl 0
I B I I
r B d I
I R r
Trang 2728.70: 2
2 0 encl 0 2
r I A
A I I a r
foundwaswhatjustiswhich2
,
πa
I μ B a
12
1)
2 2
2 2 0 2
2
2 2 0 2 2
2 2 0
2 2
2 2 encl
b c
r c πr
I μ B b
c
r c I μ b c
b r I μ πr B d
B
b c
b r I A
A I I I c
r
b
b
c b
r b
l
(b)
part 28.32,
Exercise
Ex
in asjust 0,,
at and(a),part 28.32,ExerciseEx
in asjust ,2,
πb
I μ B
b
r
28.71: If there is a magnetic field component in the z-direction, it must be constant
because of the symmetry of the wire Therefore the contribution to a surface integral over
a closed cylinder, encompassing a long straight wire will be zero: no flux through the barrel of the cylinder, and equal but opposite flux through the ends The radial field will have no contribution through the ends, but through the barrel:
.00
)(
2 2
2 2 2
2
2 2 encl
a b
a r I ) a π(b
) a π(r I A
A I I b r a
b a
r a
)(
2)
(
)(
2 2
2 2 0
a b
a r πr
I μ B a
b
a r I μ πr B d
πr
I μ B I μ πr B d B I I b