Homework 9 Solutions

0 106 t 20 a 4 107 t ma e we can conclude by

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Unformatted text preview: i /2πd and B1′ = μ0i / 2π d ′. Since tan(30º) = 1/ 3 , this leads to 1137 (a) We now make the assumption that wire #2 must be at −π/2 rad (−90º, the bottom of the cylinder) since it would pose an obstacle for the motion of wire #1 (which is needed to make these graphs) if it were anywhere in the top semicircle. (b) Looking at the θ1 = 90º datum in Fig. 29-57(b)), where there is a maximum in Bnet x (equal to +6 μT), we are led to conclude that B1x = 6.0 μ T − 2.0 μ T = 4.0 μ T in that situation. Using Eq. 29-4, we obtain i1 = 2π RB1x μ0 = 2π (0.200 m)(4.0 ×10−6 T) = 4.0 A . 4π ×10−7 T ⋅ m/A (c) The fact that Fig. 29-57(b) increases as θ1 progresses from 0 to 90º implies that wire 1’s current is out of the page, and this is consistent with the cancellation of Bnet y at θ1 = 90° , noted earlier (with regard to Fig. 29-57(c)). (d) Referring now to Fig. 29-57(b) we note that there is no x-component of magnetic field from wire 1 when θ1 = 0, so that plot tells us that B2x = +2.0 μT. Using Eq. 29-4, we find the magnitudes of the current to be i2 = 2π RB2 x μ0 2π (0.200 m)(2.0 × 10−6 T) = = 2.0 A . 4π × 10−7 T ⋅ m/A (e) We can conclude (by the right-hand rule) that wire 2’s current is into the page. 31. (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. We use the result of Problem 29-21 to evaluate the contributions to the field at P, noting that the nearest wire segments (each of length a) produce magnetism into the page at P and the further wire segments (each of length 2a) produce magnetism pointing out of the page at P. Thus, we find (into the page) 2 ( 4p × 10 -7 T ⋅ m A ) (13 A ) ⎛ 2 μ 0 i ⎞ ⎛ 2 μ 0i ⎞ 2 μ 0i BP = 2 ⎜ ⎜ 8pa ⎟ − 2 ⎜ 8p ( 2a ) ⎟ = 8pa = ⎟⎜ ⎟ 8π ( 0.047 m ) ⎝ ⎠⎝ ⎠ = 1.96 × 10 -5 T...
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