Homework 9 Solutions

Thus using eq 29 4 7 0i 0i 4 10 t ma10 a bb 2

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Unformatted text preview: (c) The last remark in the problem statement implies that treating b as a point midway between two infinite wires is a good approximation. Thus, using Eq. 29-4, −7 ⎛ μ0i ⎞ μ0i (4π × 10 T ⋅ m/A)(10 A) Bb = 2 ⎜ = 8.0 × 10−4 T. ⎟= R = π (0.0050 m) ⎝ 2πR ⎠ π (d) This field, too, points out of the page. → 6. With the “usual” x and y coordinates used in Fig. 29-37, then the vector r pointing from a current element to P is r = − s ˆ + R ˆ . Since ds = ds ˆ , then | ds × r | = Rds. i i j Therefore, with r = s 2 + R 2 , Eq. 29-3 gives dB = μ0 iR ds . 2 4π ( s + R 2 )3/ 2 (a) Clearly, considered as a function of s (but thinking of “ds” as some finite-sized constant value), the above expression is maximum for s = 0. Its value in this case is dBmax = μ0i ds / 4π R 2 . (b) We want to find the s value such that dB = dBmax /10 . This is a nontrivial algebra exercise, but is nonetheless straightforward. The result is s = R = 2.00 cm, then we obtain s = 3.82 cm. 10 2/3 − 1 R. If we set 7. (a) Recalling the straight sections discussion in Sample Problem — “Magnetic field at the center of a circular arc of current,” 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 0iθ 4 pb (out of the page) to the field at P. Also, the current in the large radius arc contributes μ 0iθ 4 pa (into the page) to the field there. Thus, the net field at P is μ iθ ⎛ 1 1 ⎞ (4p ×10 -7 T ⋅ m A)(0.411A)(74°⋅π /180°) ⎛ 1 1⎞ B= 0 ⎜ − ⎟ = − ⎜ ⎟ 4 ⎝b a⎠ 4π ⎝ 0.107m 0.135m ⎠ = 1.02 ×10 -7 T. 1129 (b) The direction is out of the page. 8. (a) Recalling the straight sections discussion in Sample Problem — “Magnetic field at the center of a circular arc of current,” we see that the current in segments AH and JD do not contribute to the field at point C. Usi...
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