Homework 9 Solutions

When wire 3 is at x 004 m the curve passes through

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Unformatted text preview: 69 ×10−4 N)j . 37. We use Eq. 29-13 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 unit-vector 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. 29-64(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. 29-64(b). This means wire 2 is attracted to wire 1, which implies (by the discussion in Section 29-2) 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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