Homework #1 P.6-21 A very large slab of material of thickness d lies perpendicularly to a uniform magnetic field of intensity H 0 az H 0 . Ignoring edge effect, determine the magnetic field intensity in the slab: a) if the slab material has a permeability , b) if the slab is a permanent magnet having a magnetization vector M i az M i . Solution: a) From constitute relation we have B2 2 H 2 and B.C. requires B2 z B1z . Thus, H 2 0 H 0 H 2 az H 2 az 0 H 0 . b) Given B2 0 H 2 M i , and B.C. requires B2 z B1z . Thus, 0 H 2 M i 0 H 0 H 2 az H 0 M i . P.6-24 Do the following by using Eq. (6-224): a) Determine the scalar magnetic potential at a point on the axis of a circular loop having a radius b and carrying a current I. b) Obtain the magnetic flux density B from 0Vm , and compare the result with Eq. (6-38). Solution: a) From Eq. (6-224) in P.6-23, we have I ds aR I Vm . 2 4 4 R Referring the right figure, ds aR (cos ) d d z z 2 2 d d . R z2 2 . Vm I 4 2 b 0 0 z z 2 2 3/2 I 2 d d 1 . z 2 2 z Vm 0 Ib 2 az b) B 0Vm az 0 , which is the same as Eq. (6-38). 3/2 z 2 z 2 b2 P.6-26 A ferromagnetic sphere of radius b is magnetized uniformly with a magnetization M = azM0. a) Determine the equivalent magnetization current densities Jm and Jms. b) Determine the magnetic flux density at the center of the sphere. Solution: z b M aˆ z M 0 a) J m M 0; J ms M aˆ n aˆ R cos aˆ sin M 0 aˆ R aˆ M 0 sin . b) Apply Eq. (6-38) to a loop of radius bsin carrying a current Jmsbd ( J bd )(b sin ) 2 M aˆ z 0 0 sin 3 d dB aˆ z 0 ms 2 3 / 2 2 2(b ) M 2 2 B dB aˆ z 0 0 sin 3 d aˆ z 0 M 0 0 M . 2 0 3 3 P.6-27 A toroidal iron core of relative permeability 3000 has a mean radius R = 80 (mm) and a circular cross section with radius b = 25 (mm). An air gap g 3 (mm) exists, and a current I flows in a 500-turn winding to produce a magnetic flux of 10–5 (Wb). (See Fig. 6-44.) Neglecting flux leakage and using mean path length, find a) the reluctances of the air gap and of the iron core, b) Bg and Hg in the air gap, and Bc and Hc in the iron core, c) the required current I. Solution: a) g g 0 S 3 103 1.21106 (H 1 ), 7 2 4 10 ( 0.025) c 2 0.08 3 103 6.75 104 (H 1 ). 7 2 S 3000 (4 10 ) ( 0.025) 105 b) Bg Bc a a 5.09 103 (T), 2 0.025 1 5.09 103 a 4.05 103 (A/m), Hg Bg a 7 4 10 0 1 4.05 103 a 1.35 (A/m). Hc B a 0 r c 3000 c c) NI c g , I c g 0.0256 (A) = 25.6 (mA). N P.6-32 Consider a plane boundary ( y = 0) between air (region 1, r1 = 1) and iron (region 2, r2 = 5000). a) Assuming B1 = ax0.5 – ay10 (mT), find B2 and the angle that B2 makes with the interface. b) Assuming B2 = ax10 + ay0.5 (mT), find B1 and the angle that B1 makes with the normal to the interface. Solution: a) B1 ax 0.5 a y 10 (mT), B2 ax B2 x a y B2 y . B2 x 0.5 H1 x B2 x 2500 (mT). 50000 0 B2 y B1 y 10 (mT), thus B2 ax 2500 a y 10 (mT). From B.C.: H 2 x B 2 tan 1 5000 1x 250 2 tan 1 250 89.8o , 2 0.2o. B1 y 1 tan 2 b) B2 ax 10 a y 0.5 (mT), B1 ax B1x a y B1 y . From B.C.: H1x B1x 0 H2x 10 5000 0 B1x 0.002 (mT). B1 y B2 y 0.5 (mT), thus B1 ax 0.002 a y 0.5 (mT). 1 0.002 0.004 2 tan 1 0.004 0.23o. tan 2 2 0.5 tan 1 P.6-35 Determine the self-inductance of a toroidal coil of N turns of wire wound on an air frame with mean radius ro and a circular cross section of radius b. Obtain an approximate expression assuming b << ro. Solution: NI B a B a 0 , 2 r in which, r ro cos . 0 NI 2 b 2 0 0 d d 0 NI ro ro2 b 2 . ro cos N 0 N 2 ro ro2 b 2 . I NI If ro b , B 0 (constant). 2 ro L B S B b 2 0 Nb 2 I 2ro L 0 N 2b 2 2ro 註: 此處用到的積分請查看網頁留的說明。 . P.6-37 Calculate the mutual inductance per unit length between two parallel two-wire transmission lines A A and B B separated by a distance D, as shown in Fig. 6-47. Assume the wire radius to be much smaller than D and the wire spacing d. Solution: B at a distance r from an infinitely long line carrying a current I: I B aˆ 0 2r For a unit length the flux due to I in line A that links with the second line pair B - B' is: (We select the surface formed by sub-surfaces B – C and C - B'. Since the normal direction of surface C - B' is perpendicular to the field B, thus the surface integral of this part equals zero and we only need to consider the surface integral over the surface B – C.) dr 0 I AC ln B r 2 AB Similarly, that due to I in line A' is: A A 0 I 2 C 0 I AC ln 2 AB Total flux linkage per unit length A A 12 L12 0 I AC AC 0 I AB AB 0 I D 2 d 2 ln ln ln 2 AB AB 2 AB AB 2 D2 0 d 2 ln 1 2 D 2 P.6-40 Find the mutual inductance between two coplanar rectangular loops with parallel sides, as shown in Fig. 6-50. Assume that h1 >> h2 (h2 > w2 > d). Solution: Approximate the magnetic flux due to the long loop linking with the small loop by that due to two infinitely long wires carrying equal and opposite current I. 1 1 h I w d w1 d . dx 0 2 ln 2 2 w1 w2 d d d x w1 d x 12 0 h2 I 2 L12 12 0 h2 w2 d w1 d ln . 2 I d w1 w2 d w2 0 Fig. 6-50 P.6-42 Calculate the force per unit length on each of three equidistance, infinitely long, parallel wires 0.15 (m) apart, each carrying a current of 25 (A) in the same direction. Specify the direction of the force. Solution: I1 = I2 = I3 = 25 (A); d = 0.15 (m). 3 0 I . B2 aˆ x 2 B12 cos 30 o aˆ x 2d Force per unit length on wire 2: f 2 aˆ y IB2 aˆ y 3 0 I 2 aˆ y 1150 0 aˆ y 1.44 10 3 (N/m). 2d Forces on all three wires are of equal magnitude and toward the center of the triangle. P.6-46 The bar AA’ in Fig. 6-53 serves as a conducting path (such as the blade of a circuit breaker) for the current I in two very long parallel lines. The lines have a radius b and are spaced at a distance d apart. Find the direction and the magnitude of the magnetic force. Solution: 1 I1 , d aˆ y dy Bat y aˆ z 0 4 y d y 1 I2 1 dy dF Id B aˆ x 0 4 y d y 1 I 2 d b 1 I2 d dy aˆ x 0 ln 1. F aˆ x 0 4 b y d y 2 b P. 6-47 A d-c current I = 10 (A) flows in an equilateral triangle loop in the xy-plane as in Fig. 6.-54. Assuming a uniform magnetic flux density B = ay0.5 (T) in the region, find the forces and torque on the loop. The dimensions are in (cm). y B A (0, 20) I B C (10,0)O (10,0) x FIGURE 6-54 A triangular loop in a uniform Magnetic field (Problem P.6-47). Solution: Assign point names A, B, and C as above, then we can obtain FAB B I ( AB) (a y 0.5) 10(a x 0.1 a y 0.2) a z 0.5 (N) FCA B I (CA) (a y 0.5) 10(a x 0.1 a y 0.2) a z 0.5 (N) FBC B I ( BC ) (a y 0.5) 10(a x 0.2) a z 1.0 (N) Ftotal 0 (N) T m B (a z IS ) (a y 0.5) (a z 10 12 0.2 0.2) (a y 0.5) a x 0.1 (N m) P.6-53 A current I flows in a long solenoid with n closely wound coil-turns per unit length. The cross-sectional area of its iron core, which has permeability , is S. Determine the force acting on the core if it is withdrawn to the position shown in Fig. 6-55. Fig. 6-55 Solution: Wm 1 H 2 dv 2 Assume a virtual displacement, x, of the iron core. Wm ( x x) Wm ( x) ( FI ) x 1 1 ( 0 ) H 2 dv Wm ( x) 0 ( r 1)n 2 I 2 Sx. S x 2 2 Wm 0 ( r 1)n 2 I 2 S , in the direction of increasing x. x 2