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解答五 27.51.(a) IDENTIFY: Use Eq. (27.2) to relate v ,B and F . SET UP: The directions of v1 and F1 are shown in Figure 27.55a. F qv B says that F is perpendicular to v and B. The information given here means that B can have no z-component. Figure 27.55a The directions of v 2 and F2 are shown in Figure 27.55b. F is perpendicular to v and B, so B can have no x-component. Figure 27.55b Both pieces of information taken together say that B is in the y-direction; B By ˆj. EXECUTE: Use the information given about F2 to calculate By : F2 F2iˆ,v2 v2kˆ ,B By ˆj. F2 qv2 B says F2iˆ qv2 By kˆ ˆj qv2 By (iˆ) and F2 qv2 By By F2/(qv2 ) F2/(qv1) B has the magnitude F2/(qv1) and is in the y-direction. (b) F1 qvB sin qv1 B y / 2 F2 / 2 EVALUATE: v1 v2 . v2 is perpendicular to B whereas only the component of v1 perpendicular to B contributes to the force, so it is expected that F2 F1, as we found. 28.43.IDENTIFY: Apply Ampere’s law. SET UP: To calculate the magnetic field at a distance r from the center of the cable, apply Ampere’s law to a circular path of radius r. By symmetry, B dl B(2 r ) EXECUTE: (a) For a r b,Iencl I B dl 0 I B2 r 0 I B for such a path. 0 I . 2 r (b) For r c, the enclosed current is zero, so the magnetic field is also zero. EVALUATE: A useful property of coaxial cables for many applications is that the current carried by the cable doesn’t produce a magnetic field outside the cable. IDENTIFY: Apply Ampere’s law to calculate B. (a) SET UP: For a r b the end view is shown in Figure 28.46a. Apply Ampere’s law to a circle of radius r, where a r b. Take currents I1 and I 2 to be directed into the page. Take this direction to be positive, so go around the integration path in the clockwise direction. Figure 28.46a EXECUTE: B dl 0 Iencl B dl B(2 r ),Iencl I1 Thus B(2 r ) 0 I1 and B 0 I1 2 r (b) SET UP: r c: See Figure 28.46b. Apply Ampere’s law to a circle of radius r, where r c. Both currents are in the positive direction. Figure 28.46b EXECUTE: B dl 0 Iencl B dl B(2 r ),Iencl I1 I 2 Thus B(2 r ) 0 ( I1 I 2 ) and B 0 ( I1 I 2 ) 2 r EVALUATE: For a r b the field is due only to the current in the central conductor. For r c both currents contribute to the total field. 28.62.IDENTIFY: Find the vector sum of the magnetic fields due to each wire. SET UP: For a long straight wire B 0 I . The direction of B is given by the right-hand rule 2 r and is perpendicular to the line from the wire to the point where the field is calculated. EXECUTE: (a) The magnetic field vectors are shown in Figure 28.68a. (b) At a position on the x-axis Bnet 2 0 I 0 I sin 2 r x2 a2 a x a 2 2 0 Ia , in the ( x2 a2 ) positive x-direction. (c) The graph of B versus x/a is given in Figure 28.68b. EVALUATE: (d) The magnetic field is a maximum at the origin, x 0. (e) When x a, B 0 Ia . x2 Figure 28.68 28.73.IDENTIFY: Use what we know about the magnetic field of a long, straight conductor to deduce the symmetry of the magnetic field. Then apply Ampere’s law to calculate the magnetic field at a distance a above and below the current sheet. SET UP: Do parts (a) and (b) together. Consider the individual currents in pairs, where the currents in each pair are equidistant on either side of the point where B is being calculated. Figure 28.83a shows that for each pair the z-components cancel, and that above the sheet the field is in the – x-direction and that below the sheet it is in the x-direction. Figure 28.83a Also, by symmetry the magnitude of B a distance a above the sheet must equal the magnitude of B a distance a below the sheet. Now that we have deduced the symmetry of B, apply Ampere’s law. Use a path that is a rectangle, as shown in Figure 28.83b. B dl 0 Iencl Figure 28.83b I is directed out of the page, so for I to be positive the integral around the path is taken in the counterclockwise direction. EXECUTE: Since B is parallel to the sheet, on the sides of the rectangle that have length 2a, B dl 0. On the long sides of length L, B is parallel to the side, in the direction we are integrating around the path, and has the same magnitude, B, on each side. Thus B dl 2BL. n conductors per unit length and current I out of the page in each conductor gives I encl InL. Ampere’s law then gives 2BL 0 InL and B 12 0 In. EVALUATE: Note that B is independent of the distance a from the sheet. Compare this result to the electric field due to an infinite sheet of charge (Example 22.7).