Unformatted text preview: orizontal current points perpendicularly out of the plane of the paper. Similarly,
the magnetic field due to the vertical current points perpendicularly into the plane of the
paper. Point A is closer to the horizontal wire, so that the effect of the horizontal current
dominates and the net field is directed out of the plane of the paper. Chapter 21 Problems 1173 Using RightHand Rule No. 2, we can see that at point B the magnetic field due to the
horizontal current points perpendicularly into the plane of the paper. Similarly, the magnetic
field due to the vertical current points perpendicularly out of the plane of the paper. Point B
is closer to the horizontal wire, so that the effect of the horizontal current dominates and the
net field is directed into the plane of the paper.
Points A and B are the same distance of 0.20 m from the horizontal wire. They are also the
same distance of 0.40 m from the vertical wire. Therefore, the magnitude of the field
contribution from the horizontal current is the same at both points, although the directions of
the field contributions are opposite. Likewise, the magnitude of the field contribution from
the vertical current is the same at both points, although the directions of the field
contributions are opposite. At either point the magnitude of the net field is the magnitude of
the difference between the two contributions, and this is the same at points A and B.
SOLUTION According to Equation 21.5, the magnitude of the magnetic field from the
current in a long straight wire is B = µ0I/(2π r), where µ0 is the permeability of free space, I
is the current, and r is the distance from the wire. Applying this equation to each wire at
each point, we see that the magnitude of the net field Bnet is Point A Bnet = µ0 I
2π rA, H µ0 I
µ I 1
1
= 0
− 2π rA, V
2π rA, H rA, V 1 24
43 − 1 24
43
Horizontal wire Vertical wire ( 4π ×10−7 T ⋅ m/A ) (5.6 A ) = 1
1
−6
− = 2.8 × 10 T
0.20 m 0.40 m 2π Point B Bnet = µ0 I
2π rB, H µ0 I
µ I 1
1
= 0
− 2π rB, V
2π rB, H rB, V 1 24
43 − 1 24
43
Horizontal wire Vertical wire ( 4π ×10−7 T ⋅ m/A ) (5.6 A ) =
2π 1
1
−6
− = 2.8 × 10 T 0.20 m 0.40 m 63. REASONING The drawing shows an endon view
B1
B2
of the two wires, with the currents directed out of the
B1
Wire 2
Wire 1
plane of the paper toward you. B1 and B2 are the
d
B
individual fields produced by each wire. Applying
B1 2
RHR2 (thumb points in the direction of the current,
B2
fingers curl into the shape of a halfcircle and the
finger tips point in the direction of the field), we obtain the field directions shown in the
drawing for each of the three regions mentioned in the problem statement. Note that it is 1174 MAGNETIC FORCES AND MAGNETIC FIELDS only in the region between the wires that B1 and B2 have opposite directions. Hence, the
spot where the net magnetic field (the vector sum of the individual fields) is zero must lie
between the wires. At this spot the magnitudes of the fields B1 and B2 from the wires are
µI
equal and each is given by B = 0 (Equation 21.5), where µ0 is the permeability of free
2π r
space, I is the current in the wire, and r is the distance from the wire.
SOLUTION The spot we seek is located at a distance d from wire 1 (see the drawing).
Note that the wires are separated by a distance of one meter. Applying Equation 21.5 to
each wire, we have that
µ0 I1
µ0 I 2
B1 = B2 or
=
2π d 2π (1.00 m − d ) Since I1 = 4 I 2 , this result becomes µ0 ( 4 I 2 )
µ0 I 2
=
2π d
2π (1.00 m − d ) or 4
1
=
d 1.00 m − d or 4 (1.00 m − d ) = d Solving for d, we find that d = 0.800 m 64. REASONING
BI
a. The compass needle lines up with the net
horizontal magnetic field B that is the vector sum
BE
BE
B
of the magnetic fields BI (the field at the location
23.0°
of the compass due to the current I in the wire)
compass
BI
and BE (the horizontal component of the earth’s
magnetic field): B = BI + BE. The field lines of
N
the magnetic field created by the current I are
0.28 m
circles centered on the wire. Because the wire is
E
W
perpendicular to the earth’s surface, and the
compass is directly north of the wire, the magnetic
field BI due to the wire must point either due east
S
wire
or due west at the location of the compass,
(perpendicular to the page)
depending on the direction of the current. The
magnetic field BI causes the compass needle to
deflect east of north, so we conclude that BI points due east (see the drawing, which shows
the situation as viewed from above). We will use RightHand Rule No. 2 to determine the
direction of the current I from the direction of BI. Chapter 21 Problems 1175 b. The horizontal component BE of the earth’s magnetic field points north, so it is
perpendicular to the magnetic field BI created by the current in the wire, which points east.
Therefore, the vectors B, BE, and BI form a right triangle with B serving as the hypotenuse
B
(see the drawing). The angle θ between the vectors BE and B, then, is given by tan θ = I
BE
µI
(Equation 1.3). We will use B = 0 (Equation 21...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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