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Unformatted text preview: T m/A)(10 A) 1 1 + 2 e = (1:0 10 3 T) e : = (4 10:0 10 3 m) 2(5 a a (b) Now we only need to consider the two straight wires:
Bb = 20 ie = 2(4 10 7 T m/A)(10 A)e = (8:0 10 4 T) e : 2R 2(5:0 10 3 m) 806 CHAPTER 30 AMPERE'S LAW Here the factor of 2 in the numerator is due to the fact that two wires are involved. Sum the elds of the two straight wires and the circular arc. Look at the derivation of the expression for the eld of a long straight wire, leading to Eq. 30{6. Since the wires we are considering are innite in only one direction the eld of either of them is half the eld of an innite wire. That is, the magnitude is 0 i=4r, where r is the distance from the end of the wire to the center of the arc. It is the radius of the arc. The elds of both wires are out of the page at the center of the arc. Now nd an expression for the eld of the arc, at its center. Divide the arc into innitesimal segments. Each segment produces a eld in the same direction. If ds is the length of a segment the magnitude of the eld it produces at the arc center is (0 i=4r2 ) ds. If is the angle subtended by the arc in radians, then r is the length of the arc and the total eld of the arc is 0 i=4r. For the arc of the diagram the eld is into the page. The total eld at the center, due to the wires and arc together, is For this to vanish must be 2 radians.
17P 16P 0 i 0 i B = 4r + 4r 0 i = 0 i (2 ) : 4r 4r p p and r are functions of x. Replace r with x2 + R2 and sin with R=r = R= x2 + R2 , then integrate from x = L=2 to x = L=2. The total eld is 2 0 iR Z 2 dx x 0 iR 1 B = 4 = 4 R2 2 2 1 2 = 20 i p 2 L 2 : (x + R ) 2 R L + 4R 2 (x2 + R2 )3 2
L= L= L= = = L= Put the x axis along the wire with the origin at the midpoint and the current in the positive x direction. All segments of the wire produce magnetic elds at P that are into the page so we simply divide the wire into innitesimal segments and sum the elds due to all the segments. The diagram shows one innitesimal segment, with width dx. According to the BiotSavart law the magnitude of the eld it produces at P is given by dB = 0 i sin2 dx : 4 r i 0 x dx R P r If L R then R2 in the denominator can be ignored and B = 20 i R CHAPTER 30 AMPERE'S LAW 807 is obtained. This is the eld of a long straight wire. For points close to a nite wire the eld is quite similar to that of an innitely long wire.
18P Follow the same steps as in the solution of 17P above but replace R with D, change the lower limit of integration to L, and change the upper limit to 0. The magnitude of the total eld is 0 dx x 0 iD 1 0 iD Z 0 = 4 D2 2 = 40 i p 2L 2 : B = 4 D L + D (x2 + D2 )3 2 (x + D2 )1 2
L = = L 19P You can easily check that each of the four sides produces the same magnetic eld B1 at the center of the square. Apply the result of 17P for B1 and let R = a=2 and L = a to obtain 2 4 B = 4B1 = 2(0 i2) (a2 +aa2 )1 2 = 2 a 0 i : a=
= p 20P The Beld produced by the four sides of the rectangle have the same direction. For each of the two longer sides (see 17P) B = 2(0 i 2) (L2 + L 2 )1 2 : W= W
L = Similarly for each of the shorter sides
0i B = 2L=2)...
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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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