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Unformatted text preview: 1139 Chapter 29 1. (a) The field due to the wire, at a point 8.0 cm from the wire, must be 39 μ T and must be directed due south. Since B i r = μ 2 b , i rB = = × × ⋅ = 2 2 39 10 4 16 6 b bfm fMf b Ff ¡k μ m T T m A A. b gc h (b) The current must be from west to east to produce a field which is directed southward at points below it. 2. The straight segment of the wire produces no magnetic field at C (see the straight sections discussion in Sample Problem 291). Also, the fields from the two semicircular loops cancel at C (by symmetry). Therefore, B C = 0. 3. (a) The magnitude of the magnetic field due to the current in the wire, at a point a distance r from the wire, is given by B i r = μ 2 b . With r = 20 ft = 6.10 m, we have B = × ⋅ = × = 4 100 2 33 10 33 6 b Ff bhmFf ¡k T m A A m T T. c hb g b g . . μ (b) This is about onesixth the magnitude of the Earth’s field. It will affect the compass reading. 4. Eq. 291 is maximized (with respect to angle) by setting θ = 90º ( = π /2 rad). Its value in this case is max 2 4 i ds dB R μ π = . From Fig. 2936(b), we have 12 max 60 10 T. B = × We can relate this B max to our dB max by setting “ ds ” equal to 1 × 10 6 m and R = 0.025 m. This allows us to solve for the current: i = 0.375 A. Plugging this into Eq. 294 (for the infinite wire) gives B ∞ = 3.0 μ T. 5. (a) Recalling the straight sections discussion in Sample Problem 291, we see that the current in the straight segments collinear with P do not contribute to the field at that point. Using Eq. 299 (with φ = θ ) and the righthand rule, we find that the current in the semicircular arc of radius b contributes μ θ 4 i b b (out of the page) to the field at P . Also, CHAPTER 29 1140 the current in the large radius arc contributes μ θ 4 i a b (into the page) to the field there. Thus, the net field at P is 1 1 (4 T m A)(0.411A)(74 /180 ) 1 1 4 4 0.107m 0.135m 1.02 T. i B b a μ θ π π × ⋅ °⋅ ° = = = × 7 7 p 10 10 (b) The direction is out of the page. 6. (a) Recalling the straight sections discussion in Sample Problem 291, we see that the current in segments AH and JD do not contribute to the field at point C . Using Eq. 299 (with φ = π ) and the righthand rule, we find that the current in the semicircular arc H J contributes μ 1 4 i R (into the page) to the field at C . Also, arc D A contributes μ 2 4 i R (out of the page) to the field there. Thus, the net field at C is 1 2 1 1 (4 T m A)(0.281A) 1 1 1.67 T. 4 4 0.0315m 0.0780m i B R R μ × ⋅ = = = ×  7 6 p 10 10 (b) The direction of the field is into the page....
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 Summer '10
 Prof.Yang
 Physics, Current

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