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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. 294)
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 unitvector 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 unitvector 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 righthand 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. 2917,
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.
 Fall '08
 schuller
 Magnetism, Work

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