The lines actually bulge outward and their density

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Unformatted text preview: 0 i r2 = 0 ir : B (r) = 2a2 r 2a2 Use s dB = 2rB = 0 i enclosed , or B = 0 i enclosed =2r. (a) r < c: 0 i encl = 0 i r2 = 0 ir : B (r) = 2r 2r c2 2c2 (b) c < r < b: (c) b < r < a: (d) r > a: encl 0 i B (r) = 0 ir = 2r : 2 47PH (r2 2 0 i 2 2 encl 0 i B (r) = 0 ir = 2r 1 (a2 b2) = 2r((a 2 r 2) : 2 b) a b) encl B (r) = 0 ir = 0 : 2 (e) For example, check what happens if b = c. In this case the expressions in (a), (b) and (c) above should yield the same result at r = b = c. This is indeed the case. (f) B (mT) 7.6 1.7 0 0.40 1.8 2.0 r (cm) 818 CHAPTER 30 AMPERE'S LAW 48P For r < a, 0 i enclosed = 0 Z J (r)2r dr = 0 Z J r 2r dr = 0 J0 r2 : B (r) = 2r 2r 0 2 0 0 a 3a r r 49P The eld at the center of the pipe (point C ) is due to the wire alone, with a magnitude of C 0 iwire 0 iwire B = 2(3R) = 6R : For the wire we have B wire > B wire , so for B = B = B wire ; iwire must be into the page: 0 iwire B = B wire B pipe = 2R 20 iR) : (2 Let B = B to obtain iwire = 3i=8 . P; C; P C C; P P; P; C P (a) Take the magnetic eld at a point within the hole to be the sum of the elds due to two current distributions. The rst is the solid cylinder obtained by lling the hole and has a current density that is the same as that in the original cylinder with the hole. The second is the solid cylinder that lls the hole. It has a current density with the same magnitude as that of the original cylinder but it is in the opposite direction. Notice that if these two situations are superposed the total current in the region of the hole is zero. Recall that a solid cylinder carrying current i, uniformly distributed over a cross section, produces a magnetic eld with magnitude B = 0 ir=2R2 a distance r from its axis, inside the cylinder. Here R is the radius of the cylinder. For the cylinder of this problem the current density is 50P i J = A = (a2 i b2 ) ; where A [= (a2 b2 )] is the cross-sectional area of the cylinder with the hole. The current in the cylinder without the hole is 2 I1 = JA = Ja2 = a2ia b2 and the magnetic eld it produces at a point inside, a distance r1 from its axis, has magnitude 2 I1 B1 = 20ar1 = 2a20(ir21 a b2 ) = 2(0 ir1 b2 ) : 2 a a2 CHAPTER 30 AMPERE'S LAW 819 The current in the cylinder that lls the hole is I2 = Jb2 = a2ib b2 and the eld it produces at a point inside, a distance r2 from the its axis, has magnitude ir 2 0 I2 B2 = 2br2 = 2b0a22 b b2 ) = 2(0 ir2 b2 ) : 2 2( a2 At the center of the hole this eld is zero and the eld there is exactly the same as it would be if the hole were lled. Place r1 = d in the expression for B1 and obtain 2 0 B = 2(a2 id b2 ) : for the eld at the center of the hole. The eld points upward in the diagram if the current is out of the page. (b) If b = 0 the formula for the eld becomes This correctly gives the eld of a solid cylinder carrying a uniform current i, at a point inside the cylinder a distance d from the axis. If d = 0 the formula gives B = 0. This is correct for the eld on the axis of a cylindrical shell carrying a uniform current. (c) The diagram shows the situat...
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