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Unformatted text preview: 69 ×10−4 N)j . 37. We use Eq. 2913 and the superposition of forces: F4 = F14 + F24 + F34 . With θ = 45°,
the situation is as shown on the right.
The components of F4 are given by
F4 x = − F43 − F42 cos θ = − μ 0i 2
2pa − μ 0i 2 cos 45° − μ 0i 2 sin 45° μ 0i 2 2 2 pa =− 3μ 0 i 2
4 pa and
F4 y = F41 − F42 sin θ = μ 0i 2
2pa 2 2 pa = 4 pa . Thus,
⎡ ⎛ 3μ i 2 ⎞ 2 ⎛ μ i 2 ⎞ 2 ⎤
F4 = ( F + F ) = ⎢⎜ − 0 ⎟ + ⎜ 0 ⎟ ⎥
⎢⎝ 4πa ⎠ ⎝ 4πa ⎠ ⎥
⎣
⎦
−4
= 1.32 ×10 N/m
2
4x 2 12
4y 12 10 ( 4π ×10−7 T ⋅ m A ) ( 7.50A )
10μ0i 2
=
=
4πa
4π ( 0.135m ) 2 and F4 makes an angle φ with the positive x axis, where φ = tan −1 FG F IJ = tan FG − 1IJ = 162° .
H 3K
HF K
4y −1 4x In unitvector notation, we have
ˆ
ˆ
F1 = (1.32 ×10 4 N/m)[cos162°ˆ + sin162°ˆ = (−1.25 ×10 4 N/m)i + (4.17 ×10 5 N/m)j
i
j] 38. (a) The fact that the curve in Fig. 2964(b) passes through zero implies that the
currents in wires 1 and 3 exert forces in opposite directions on wire 2. Thus, current i1
points out of the page. When wire 3 is a great distance from wire 2, the only field that
affects wire 2 is that caused by the current in wire 1; in this case the force is negative
according to Fig. 2964(b). This means wire 2 is attracted to wire 1, which implies (by
the discussion in Section 292) that wire 2’s current is in the same direction as wire 1’s 1141
current: out of the page. With wire 3 infinitely far away, the force per unit length is given
(in magnitude) as 6.27 × 10−7 N/m. We set this equal to F12 = μ0i1i2 / 2π d . When wire 3
is at x = 0.04 m the curve passes through the zero point previously mentioned, so the
force between 2 and 3 must equal F12 there. This allows us to solve for the distance
between wire 1 and wire 2:
d = (0.04 m)(0.750 A)/(0.250 A) = 0.12 m.
Then we solve 6.27 × 10−7 N/m= μo i1 i2 /2πd and obtain i2 = 0.50 A.
(b) The direction of i2 is out of the page.
39. Using a magnifyin...
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 Fall '08
 schuller
 Magnetism, Work

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