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ch29 - Chapter 29 1(a The magnitude of the magnetic field...

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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 0 i 2 r . With r = 20 ft = 6.10 m, we have B 4 10 7 T m A 100A 2 6.10m 3.3 10 6 T 3.3 T. (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 = 90º ( = / 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 B i r 0 2 , i 2 rB 0 2 0.080m 39?10 -6 T 4p?10 -7 T m A 16A. (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.
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CHAPTER 29 1128 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 = rad). The direction of B is out of the page, as can be checked by referring to Fig. 29-6(c). The magnitude of B at point a is therefore 7 3 0 0 0 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 i i i B R R R   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 0 0 (4 10 T m/A)(10 A) 2 8.0 10 T. 2 (0.0050 m) b i i B R R (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 ? i j. r s R   Since ˆ i ds ds , then | | . ds r Rds Therefore, with 2 2 r s R , Eq. 29-3 gives 0 2 2 3/2 4 ( ) iR ds dB s R . (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 .
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