Homework 9 Solutions

00cm 320 103 n j j and f points toward the wire or

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Unformatted text preview: g glass, we see that all but i2 are directed into the page. Wire 3 is therefore attracted to all but wire 2. Letting d = 0.500 m, we find the net force (per meter length) using Eq. 29-13, with positive indicated a rightward force: |F| = μ0i3 ⎛ i1 i2 i4 i5 ⎞ +++ ⎜− ⎟ 2π ⎝ 2d d d 2d ⎠ which yields | F | / = 8.00 ×10−7 N/m . 40. Using Eq. 29-13, the force on, say, wire 1 (the wire at the upper left of the figure) is along the diagonal (pointing toward wire 3, which is at the lower right). Only the forces (or their components) along the diagonal direction contribute. With θ = 45°, we find the force per unit meter on wire 1 to be ⎛ μ 0i 2 ⎞ μ0i 2 3 ⎛ μ 0i 2 ⎞ = F1 = | F12 + F13 + F14 | = 2 F12 cos θ + F13 = 2 ⎜ ⎟ cos 45° + ⎜ ⎟ 2 2πa 2 2π ⎝ a ⎠ ⎝ 2πa ⎠ −7 3 ( 4π ×10 T ⋅ m A ) (15.0A ) = = 1.12 ×10−3 N/m. −2 2 2π (8.50 ×10 m ) 2 ˆ ˆ j) The direction of F1 is along r = (i − ˆ / 2 . In unit-vector notation, we have (1.12 ×10 -3 N/m) ˆ ˆ ˆ ˆ F1 = (i − j) = (7.94 ×10 -4 N/m)i + (−7.94 ×10 -4 N/m)j 2 41. The magnitudes of the forces on the sides of the rectangle that are parallel to the long straight wire (with i1 = 30.0 A) are computed using Eq. 29-13, but the force on each of the sides lying perpendicular to it (along our y axis, with the origin at the top wire and +y downward) would be figured by integrating as follows: 1142 CHAPTER 29 F⊥ sides = z a +b a i2 μ 0i1 dy. 2 πy Fortunately, these forces on the two perpendicular sides of length b cancel out. For the remaining two (parallel) sides of length L, we obtain F= μ0i1i2 L ⎛ 1 μ0i1i2b 1⎞ ⎜− ⎟= 2π ⎝ a a + d ⎠ 2π a ( a + b ) ( 4π ×10 = −7 T ⋅ m/A ) ( 30.0A )( 20.0A )( 8.00cm ) ( 300 ×10−2 m ) 2π (1.00cm + 8.00cm ) = 3.20 ×10−3 N, j j and F points toward the wire, or + ˆ . That is, F = (3.20 ×10−3 N)ˆ in unit-vector notation. 1 42. The area enclosed by the loop L is A = 2 (4d )(3d ) = 6d 2 . Thus ∫ B ⋅ ds = μ0i = μ0...
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This document was uploaded on 02/26/2014 for the course PHYS 2b at UCSD.

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