Homework 9 Solutions

1133 the contribution from the circular loop is bc 2

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Unformatted text preview: e small radius is rsmall = 2.00 cm. 19. The contribution to Bnet from the first wire is (using Eq. 29-4) B1 = μ0i1 ˆ (4π× 10−7 T ⋅ m/A)(30 A) ˆ ˆ k= k = (3.0 × 10−6 T)k. 2π r1 2π (2.0 m) The distance from the second wire to the point where we are evaluating Bnet is r2 = 4 m − 2 m = 2 m. Thus, B2 = μ0i2 ˆ (4π× 10−7 T ⋅ m/A)(40 A) ˆ ˆ i= i = (4.0 × 10−6 T)i. 2π r2 2π (2.0 m) and consequently is perpendicular to B1 . The magnitude of Bnet is therefore | Bnet | = (3.0 ×10−6 T) 2 + (4.0 × 10−6 T) 2 = 5.0 ×10−6 T . 20. (a) The contribution to BC from the (infinite) straight segment of the wire is BC1 = μ 0i 2 πR . 1133 The contribution from the circular loop is BC 2 = μ 0i 2R . Thus, −7 −3 1 ⎞ ( 4π ×10 T ⋅ m A ) ( 5.78 ×10 A ) ⎛ 1 ⎞ −7 BC = BC1 + BC 2 = ⎜1 + ⎟ = ⎜ 1 + ⎟ = 2.53×10 T. π⎠ 2R ⎝ π ⎠ 2 ( 0.0189 m ) ⎝ μ0i ⎛ BC points out of the page, or in the +z direction. In unit-vector notation, ˆ B = (2.53×10−7 T)k C (b) Now, BC1 ⊥ BC 2 so 2 2 BC = BC1 + BC 2 = μ 0i 2R 1+ -7 −3 1 ( 4p ×10 T ⋅ m A ) ( 5.78 ×10 A ) 1 = 1 + 2 = 2.02 ×10−7 T. 2 π π 2 ( 0.0189 m ) and BC points at an angle (relative to the plane of the paper) equal to ⎛B ⎞ ⎛1⎞ tan −1 ⎜ C1 ⎟ = tan −1 ⎜ ⎟ = 17.66° . ⎝π ⎠ ⎝ BC 2 ⎠ In unit-vector notation, ˆ ˆ ˆ i BC = 2.02 ×10−7 T(cos17.66°ˆ + sin17.66°k) = (1.92 ×10−7 T)i + (6.12 ×10−8 T)k . → 21. Using the right-hand rule (and symmetry), we see that B net points along what we will refer to as the y axis (passing through P), consisting of two equal magnetic field ycomponents. Using Eq. 29-17, μi | Bnet | = 2 0 sin θ 2π r where i = 4.00 A, r = r = d 22 + d12 / 4 = 5.00 m, and ⎛ d2 ⎞ −1 ⎛ 4.00 m ⎞ −1 ⎛ 4 ⎞ ⎟ = tan ⎜ ⎟ = tan ⎜ ⎟ = 53.1° . ⎝ 6.00 m / 2 ⎠ ⎝3⎠ ⎝ d1 / 2 ⎠ θ = tan −1 ⎜ Therefore, | Bnet | = μ0i (4π ×10−7 T ⋅ m A)(4.00 A) sin θ = sin 53.1° = 2.56 ×10−7 T . πr π (5.00 m) 22. The fact t...
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This document was uploaded on 02/26/2014 for the course PHYS 2b at UCSD.

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