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Unformatted text preview: (L2 +W 2 )1 2 : ( W
W = Thus
0 B = 2B + 2B = W (L22+iL 2 )1 2 + L(L20 W 2 )1 2 2+W W 2 + W2 1 2 = 20 i(LLW ) :
L W = = = For L W B ! 20 iL = 20 i ; LW W which is consistent with Eq. 3015 for W = 2d and x = 0. 808 CHAPTER 30 AMPERE'S LAW When the elds of the four sides are summed vectorially the horizontal components add to zero. The vertical components are all the same, so the total eld is given by B total = 4B cos = 42Ba = p 4Ba 2 : R 4x2 + a Thus 40p 2 ia : B total = (4x2 + a2 ) 4x2 + 2a2 For x = 0 the expression reduces to 40p 2 = 2 20 i ; B total = 2 ia a a 2a in agreement with the result of 19P.
22P 21P p The square has sides of length L=4. The magnetic eld at the center of the square is given by the result of 19P, with a = L=4 and x = 0. It is 2 Bsq = 8 L 0 i = 11:31 0 i : L p The radius of the circle is R = L=2. Use Eq. 30{28 of the text, with R = L=2 and z = 0. The eld is 2 Bcirc = L0 i = 9:87 0 i : L The square produces the larger magnetic eld.
23P Since ds is parallel to r,
BQ /
Z ids r = 0 : r3
a O i a P Let e be a unit vector unit vector pointing into the page. At point P 0 Z ids r = 0 ie Z sin ds B = 4 Z r3 4 Z r2 a dx 0i 0i = 4e ardx = 4e 3 2 + x2 )3 2 0 (a p 0 20 = 8aie :
P a a = r ds x CHAPTER 30 AMPERE'S LAW 809 24P Obviously, B 3 = B 6 = 0,
P P P P and B 1 = B 2 = 82a0 i ;
0 B 4 = B 5 = 82ai (2 )
P P p P 3 1 i 2 4 6 p (see 23P). Thus
BP = X
n 6 20 20 = 2 820 i 16ai e = 8aie ; a where e is a unit vector pointing into the page. =1p BP n = (BP 1 + BP 2 )e p p (B 4 + B 5 )e
P P 5 1 2 be a unit vector pointing into the page. Use the results of 18P and 23P to calculate B 1 through 8 B 8: p p P 20 i = 20 i ; B 1 = B 8 = 8(a=4) 2a p p 20 i = 20 i ; 7 B 4 = B 5 = 8(3a=4) 6a a= 0i B 2 = B 7 = 4a=4) = [(3a=4)23+ (4 4)2 ]1 2 ( a= 6 5 p30 i ; = 10a and i B 3 = B 6 = 4(30a=4) = [(a=4)2 +a=4a=4)2 ]1 2 = p 0 i : (3 3 10a Finally,
P P P P P P P P = P P = 25P Let e 3 i 4 BP =1 p p 0 i 2 + 2 + p3 + =2
n = X 8 B Pn e 1 p e a 2 6 10 3 10 7 T m/A)(10 A) p2 p2 1 3 + p e 2(4 10 p = (8:0 10 2 m) 2 + 6 + 10 3 10 = (2:0 10 4 T) e ; 810 CHAPTER 30 AMPERE'S LAW where e is a unit vector pointing into the page.
26P Consider a section of the ribbon of thickness dx located a distance x away from point P . The current it carries is di = i dx=w, and its contribution to B is
P Thus
P 0 idx dB = 0 di = xw : 2x 2
P and B points upward.
27E 0 i Z + dx = 0 i ln 1 + w ; B = d B = 2w x 2w d Z
d w P P d (a) If the currents are parallel, the two elds are in opposite directions in the region between the wires. Since the currents are the same the total eld is zero along the line that runs halfway between the wires. There is no possible current for which the eld does not vanish. (b) If the currents are antiparallel, the elds are in the same direction in the region between the wires. At a point halfway between they have the same magnitude, 0 i=2r. Thus the total eld at the midpoint has magnitude B = 0 i=r and 040 m)(300 10 6 i = rB = (0:4 10 7 Tm/A T) = 30 A : 0
28E The point P...
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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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