Chapter-29 - 1139 Chapter 29 1. (a) The field due to the...

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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 29-1). Also, the fields from the two semi-circular 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 one-sixth the magnitude of the Earth’s field. It will affect the compass reading. 4. Eq. 29-1 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. 29-36(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. 29-4 (for the infinite wire) gives B ∞ = 3.0 μ T. 5. (a) Recalling the straight sections discussion in Sample Problem 29-1, we see that the current in the straight segments collinear with P do not contribute to the field at that point. Using Eq. 29-9 (with φ = θ ) and the right-hand 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 29-1, we see that the current in segments AH and JD do not contribute to the field at point C . Using Eq. 29-9 (with φ = π ) and the right-hand 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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Chapter-29 - 1139 Chapter 29 1. (a) The field due to the...

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