Homework 9 Solutions

Homework 9 Solutions - Chapter 29 1(a The magnitude of the...

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1127 Chapter 29 1. (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 0 2 p . With r = 20 ft = 6.10 m, we have B = ×⋅ = 4 100 2 33 10 33 6 p1 0 p6 .10 -7 TmA A m TT . ch b g b g .. (b) This is about one-sixth the magnitude of the Earth’s field. It will affect the compass reading. 2. Equation 29-1 is maximized (with respect to angle) by setting θ = 90º ( = π /2 rad). Its value in this case is 0 max 2 4 i ds dB R = . From Fig. 29-34(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 T. 3. (a) The field due to the wire, at a point 8.0 cm from the wire, must be 39 T and must be directed due south. Since Bi r = 0 2 p , i rB == × = 2 23 9 1 0 4 16 0 6 p p 0.080 0 -7 mT A. bg (b) The current must be from west to east to produce a field that is directed southward at points below it. 4. The straight segment of the wire produces no magnetic field at C (see the straight sections discussion in Sample Problem — “Magnetic field at the center of a circular arc of current”). Also, the fields from the two semicircular loops cancel at C (by symmetry). Therefore, B C = 0. 5. (a) We find the field by superposing the results of two semi-infinite wires (Eq. 29-7) and a semicircular arc (Eq. 29-9 with φ = rad). The direction of G B is out of the page, as can be checked by referring to Fig. 29-6(c). The magnitude of G B at point a is therefore
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CHAPTER 29 1128 7 3 000 1 1 (4 10 T m/A)(10 A) 1 1 2 1.0 10 T 4 2 2 2(0.0050 m) 2 a iii B RR R μμ π μ ππ ×⋅ ⎛⎞ =+ = + = + = × ⎜⎟ 4 ⎝⎠ upon substituting i = 10 A and R = 0.0050 m. (b) The direction of this field is out of the page, as Fig. 29-6(c) makes clear. (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 4 00 (4 10 T m/A)(10 A) 28 . 0 1 0 T . 2 (0.0050 m) b ii B == = = × 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 ˆˆ ij . rs R =− + G Since ˆ i ds ds = G , then ||. ds r Rds ×= GG Therefore, with 22 R , Eq. 29-3 gives 0 3 / 2 4( ) iR ds dB sR = + . (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 2 max 0 /4 dB i ds R = . (b) We want to find the s value such that max /10 dB dB = . This is a nontrivial algebra exercise, but is nonetheless straightforward. The result is s = 10 2/3 1 R . If we set 2.00 cm, R = then we obtain s = 3.82 cm. 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 μθ 0 4 ib p (out of the page) to the field at
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Homework 9 Solutions - Chapter 29 1(a The magnitude of the...

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