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Unformatted text preview: add them as vectors to get the net field. Since they
are perpendicular, we can use the Pythagorean theorem to calculate the magnitude of the net
field.
SOLUTION Using Equation 21.6 and the Pythagorean theorem, we find that the magnitude
of the net magnetic field at the common center of the two loops is 1170 MAGNETIC FORCES AND MAGNETIC FIELDS N
N
N
Fµ I I +Fµ I I = 2Fµ I I
G2 R J G2 R J G2 R J
H KH K H K
2 bg π × 10 T ⋅ m / A h .7 A g
1c
4
1
b = 3.8 × 10
2b
0.040 m g
2 B net = 2 0 0 0 −7 = −5 T 58. REASONING Each of the four wires makes a contribution to the net magnetic field B at
µI
the center of the square. The magnitude of each wire’s magnetic field is given by B = 0
2π r
(Equation 21.5), where I is the current in a wire, r is the radial distance from the wire to the
center of the square, and µ 0 = 4π × 10 −7 T ⋅ m/A is the permeability of free space. For all
four wires, the radial distance r is half of the length s of one of the sides of the square (r = 1 s) .
2 In order to add up the four magnetic fields, we must determine the direction of each one.
RightHand Rule No. 2 predicts that, at the center of the square, the currents I1 = 3.9 A,
I2 = 8.5 A, and I3 = 4.6 A all produce fields pointing into the page, while the magnetic field
B of the unknown current I points out of the page. Therefore, the magnitude Bnet of the net
magnetic field is given by Bnet = B1 + B2 + B3 − B
SOLUTION Applying B = (1) µ0 I
(Equation 21.5) to the magnetic fields on the right side of
2π r Equation (1), we obtain
Bnet = µ0 I1 µ0 I 2 µ 0 I 3 µ0 I
µ
+
+
−
= 0 ( I1 + I 2 + I 3 − I )
2π r 2π r 2π r 2π r 2π r (2) Solving Equation (2) for I yields
2π rBnet µ0 = I1 + I 2 + I 3 − I or I = I1 + I 2 + I 3 − 2π rBnet µ0 (3) The length s of a side of the square is s = 0.050 m, so the radial distance r from each wire to
the center of the square is half as large: r = 1 s = 0.025 m. Therefore, Equation (3) yields
2 I = 3.9 A + 8.5 A + 4.6 A − ( 2 π ( 0.025 m ) 61×10−6 T
4 π ×10 −7 T ⋅ m/A ) = 9.4 A Chapter 21 Problems 1171 59. REASONING AND SOLUTION Let the current in the lefthand wire be labeled I1 and that
in the righthand wire I2.
a. At point A: B1 is up and B2 is down, so we subtract them to get the net field. We have
B1 = µ 0I1/(2π d1) = µ 0(8.0 A)/[2π (0.030 m)]
B2 = µ 0I2/(2π d2) = µ 0(8.0 A)/[2π (0.150 m)]
So the net field at point A is
BA = B1 − B2 = 4.3 × 10 −5 T
b. At point B: B1 and B2 are both down so we add the two. We have
B1 = µ0(8.0 A)/[2π (0.060 m)]
B2 = µ0(8.0 A)/[2π (0.060 m)]
So the net field at point B is
BB = B1 + B2 = 5.3 × 10 −5 T 60. REASONING At the center of the loop, the magnitude B of the magnetic field due to the
current I in the long, straight wire is equal to the magnitude Bloop of the magnetic field due
to the current Iloop in the circular wire loop. We know this because the vector sum of the two
magnetic fields at that location is zero. We will determine the magnitude of the magnetic
µ0 I loop
field due to the current in the circular loop from Bloop =
(Equation 21.6, with
2R
N = 1), where µ0 = 4π ×10−7 T ⋅ m/A is the permeability of free space, and R is the radius of
the loop. The magnitude B of the magnetic field of the long, straight wire is given by
µI
B = 0 (Equation 21.5), where r is the radial distance from the wire to the center of the
2π r
loop. Because the wire is tangent to the loop, the radial distance r is equal to the radius R of
the loop, and we have that B= µ0 I
2π R (1) 1172 MAGNETIC FORCES AND MAGNETIC FIELDS SOLUTION The two magnetic fields have equal magnitudes at the center of the circular
µ0 Iloop
loop, so Bloop =
(Equation 21.6) and Equation (1) together yield
2R Bloop = µ0 I loop = 2R µ0 I
=B
2π R (2) Solving Equation (2) for Iloop, we obtain µ0 I loop
2R = µ0 I
2π R I loop = or I π = 0.12 A π = 0.038 A 61. REASONING The magnitude Bi of the magnetic field at the center of the inner coil is given
by Equation 21.6 as Bi = µ 0 Ii Ni /(2 Ri ) , where Ii, Ni, and Ri are, respectively, the current,
the number of turns, and the radius of the inner coil. The magnitude Bo of the magnetic field
at the center of the outer coil is Bo = µ 0 I o N o /(2 Ro ) . In order that the net magnetic field at
the common center of the two coils be zero, the individual magnetic fields must have the
same magnitude, but opposite directions. Equating the magnitudes of the magnetic fields
produced by the inner and outer coils will allow us to find the current in the outer coil. SOLUTION Setting Bi = Bo gives µ 0 I i Ni
2 Ri = µ0 I o No
2 Ro Solving this expression for the current in the outer coil, we have
I o = Ii No Ni Ro Ri 140 turns 0.023 m = 8.6 A 180 turns 0.015 m = ( 7.2 A ) In order that the two magnetic fields have opposite directions, the current in the outer coil
must have an opposite direction to the current in the inner coil. 62. REASONING Using RightHand Rule No. 2, we can see that at point A the magnetic field
due to the h...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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