1127 Chapter 291. (a) The magnitude of the magnetic field due to the current in the wire, at a point a distance rfrom the wire, is given by Bir=μ02p. With r= 20 ft = 6.10 m, we have B=×⋅=×=−410023310336p10p 6.10-7T m AAmTT.chbgbg..μ(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/2rad). Its value in this case is 0max24i dsdBRμπ=. From Fig. 29-34(b), we have 12max6010T.B−=×We can relate this Bmaxto our dBmaxby 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 μT. 3. (a) The field due to the wire, at a point 8.0 cm from the wire, must be 39 μT and must be directed due south. Since Bir=μ02p, irB==××⋅=−22391041606pp 0.080p10-7μmTT m AA.bgch(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 sectionsdiscussion 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, BC= 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= πrad). The direction of GBis out of the page, as can be checked by referring to Fig. 29-6(c). The magnitude of GBat point ais therefore
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