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C Use the same coordinated system as in 8E. Then F = ( e0 iv=2d)( k v), as obtained there. (a) Now v = v( i) so iev j = (1:26 10 6 T m/A)(50 A)(1:6 10 19 C)(1:0 107 m/s)j F = 0 2d 2(5:0 10 2 m) = (3:2 10 16 N) j : (c) Now v = v k so F / k v = 0:
e e e 10E (a) The straight segment of the wire produces no magnetic eld at C . (b) For the semicircular loop 11P 0 Z jds rj = 0 i Z r ds B = 4 4 loop r3 loop r3 0 i)( 0 = (4Rr) = Ri : 2 4
C 804 CHAPTER 30 AMPERE'S LAW B points into the page. (c) The same as (b), since the straight segments do not contribute to B. Since sections AH and JD do not contribute to B , we only need to consider the two arcs. For the smaller arc
C 12P BC 0 1 = 4 Z i ds1 r = 0 i Z R1 ds1 e = 0 ie ; 3 r3 4 R1 4R1
BC 2 = where e is a unit vector pointing into the page. Similarly, for the larger arc 0 ie : 4R2 Thus BC R2 = B 1 + B 2 = 0 i(4R R R1 )e : 1 2
C C First nd the magnetic eld of a circular arc at its center. Let ds be an innitesimal segment of the arc and r be the vector from the segment to the arc center. ds and r are perpendicular to each other, so the contribution of the segment to the eld at the center has magnitude 13P ds r dB = 0 ids : 4r2
P Here s is the arc length and is the angle (in radians) subtended by the arc at its center. The second expression was obtained by replacing s with r. must be in radians for this expression to be valid. The eld is into the page if the current is from left to right in the diagram and out of the page if the current is from right to left. All segments contribute magnetic elds in the same direction. Furthermore r is the same for all of them. Thus the magnitude of the total eld at the center is given by 0 is B = 4r2 = 0 i : 4r CHAPTER 30 AMPERE'S LAW 805 Now consider the circuit of Fig. 30{41a. The magnetic eld produced by the inner arc has magnitude 0 i=4b and is out of the page. The eld produced by the outer arc has magnitude 0 i=4a and is into the page. The two straight segments of the circuit do not produce elds at the center of the arcs because the vector r from any point on them to the center is parallel or antiparallel to the current at that point. If the positive direction is out of the page then the total magnetic eld at the center is 0 i 1 1 : B = 4 b a Since b < a the total eld is out of the page.
14P Label the various sections of the wires as shown. Firstly, the sections a2 and c2 do not coutribute to B at point O, as point O lies on the straight line coinciding with a2 and c2 . Secondly, the Beld due to the curved sections b1 and b2 cancel each other at point O. This leaves us just a1 and c1 . Finally, if we relocate c1 to c01 , its contribution to the Beld at O will not change. Note that a1 and c01 together do form an innite straight wire carrying a current i to the left.
15P c1 ' b1
R O a1 a2 b2 i i c1 c2 (a) The contribution to B due to the straight sections of the wire is
a 2R where e is a unit vector which points out of the page. For the bent section (see 12P) ie Ba1 = 0 ; ie Ba2 = 0 : Thus
Ba 4R 0 1 1 = B 1 + B 2 = Ri + 2 e 2 7...
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This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.
 Fall '07
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