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# At a point halfway between they have the same

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Unformatted text preview: at which B = 0 form a line parallel to both currents passing through the line joining the two wires, as shown. Note that B 1 / i1 =r1 , and B 2 / i2 =(d r1 ). Let B 1 = B 2 to obtain i1 =r1 = i2 =(d r1 ). Solve for r1 : P P P P P i1 =3i r1 d d-r 1 i1=i 3 1 r1 = i i+di = 3i id i = 34d : + 1 2 The current i2 carried by wire 2 must be out of the page. Since B 1 / i1 =r1 where i1 = 6:5 A and r1 = 0:75 cm + 1:5 cm = 2:25 cm, and B 2 / i2 =r2 where r2 = 1:5 cm, from B 1 = B 2 we get r2 = (6:5 A) 1:5 cm = 4:3 A : i2 = i1 r 2:25 cm P P P P 29E 1 CHAPTER 30 AMPERE'S LAW 811 30E Lable these wires 1 through 5, left to right. Then i2 1 + 1 + 1 + 1 j = 250 i2 j F1 = 0 2 d 2d 3d 4d 24d 7 T m/A)(3 00 A)2 = (13)(4 10 (8:00 10 :2 m) (1:00 m)j 24 = (4:69 10 5 T) j ; 50 i2 j = (1:88 10 5 T) j ; j= 12d F3 = 0 (because of symmetry); F4 = F2 ; and F5 = F1 . i2 1 + 1 F2 = 0 2 2d 3d 31E Consider, for example, x &gt; d. Then in Fig. 30-10 the direction of B is reversed, and b i B (x) = B (x) B (x) = 2(0+ x) 2(0 i d) = (d0 idx2 ) : 2 d x a b 32E (a) Refer to the gure to the right. We have B = jB + B j = 2B cos i 0p cos =2 2d 2 p (4 10 7 T m/A)(100 A)(cos 45 ) 2 = (10 10 2 m) = 4:0 10 4 T : P L R L BL B P d/ 21/2 BR d BP points to the left. (b) Reverse the direction of B in the gure above but keep its magnitude unchanged. Obviously B still has a magnitude of 4:0 10 4 T but now points up the page. R P 33P For 0 &lt; x &lt; d i B (x) = B (x) B (x) = 2(0+ x) 2(0 i x) = (x0 ixd2 ) ; 2 d d a b 812 CHAPTER 30 AMPERE'S LAW and for x &gt; d i B (x) = B (x) + B (x) = 2(0+ d) + 2(0 i d) = (x0 ixd2 ) : 2 x x a b 20 10 B(x) (mT) 0 10 20 2 1.5 1 0.5 0 0.5 1 1.5 2 x (cm) The same expression for B (x) is valid for x &lt; 0 due to symmetry. Each wire produces a eld with magnitude given by B = 0 i=2r, where r is the distance from the corner of the square to the center. According to the Pythagorean theorem the p p p diagonal of the square has length 2a, so r = a= 2 and B = 0 i= 2a. The elds due to the wires at the upper left and lower right corners both point toward the upper right corner of the square. The elds due to the wires at the upper right and lower left corners both point toward the upper left corner. The horizontal components cancel and the vertical components sum to 34P 0 B total = 4 p 0 i cos 45 = 2ai 2a 7 = 2(4 10 (0:T m/A)(20 A) = 8:0 10 5 T : 20 m) In the calculation cos 45 was replaced with 1= 2. The total eld points up the page. p CHAPTER 30 AMPERE'S LAW 813 35P Refer to the gure as shown. For example, the force on wire 1 is 1 F12 =45 o 2 F1 = jF12 + F13 + F14 j = 2F12 cos + F13 0 i2 cos 45 + p0 i2 = 2 2a 2 2a 2 = 0:338 0 i : a F1 F13 F14 points from wire 1 to the center of the square. 4 3 36P Use F4 = F14 + F24 + F34 : The components of F4 are then given by y F4x = F43 2 2 cos = 0 i 0 i p 45 2a 2 2a 2 0 i = 34a F43 F42 cos 1 2 F41 =45 o and p #1 2 2 30 i2 2 + 0 i2 2 = 100 i ; F4 = (F42 + F42 )1 2 = 4a 4a 4a and F4 makes an angle with the positive x axis, where &quot; = = x y Thus F4 = F41 F42 sin 2 2 = 0 i 0 i psin 45 2a 2 2a 2 = 0 i : 4a y 4 3 x F42 F = tan 1 F4 4 y x = tan 1 1 = 162 : 3 814 CHAPTER 30 AMPERE'S LAW 37P (a) From the gure to the right 1 B = jB1 + B2 j = 2B1 cos 0 i d=2 = 2 2r r 0 = 2[R2+idd=2)2 ] ( id = (420+ d2 ) : 2 R P r B1 P d r BP B2 (b) B points to the right, as shown. P 2 38P The forces on the two sides of length b cancel out. For the remaining two sides i 1 1 i F = 0 i212 L a a + d = 2a(1 i2 b b) a+ 6 2 A)(20 = (1:26 10 T m/A...
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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.

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