ECE312_W09_Homework_2_Solutions

ECE312_W09_Homework_2_Solutions - CHAPTER 5 251 I2 Figure...

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Unformatted text preview: CHAPTER 5 251 I2 Figure P5.11: (b) Direction of I2 . µ0 I1 2πd ˆ where z is out of the page. Since the net field is zero at the center of the loop, I 2 must be clockwise, as seen from above, in order to oppose I1 . The field due to I2 is, from Eq. (5.35), µ0 NI2 ˆ z B µ0 H 2a Equating the magnitudes of the two fields, we obtain the result NI2 I ECE 312 - Homework1 2 Solutions 2a or 2πd 2aI1 2πNd 1 25 20 2 the wire is Problem 5.12 Two infinitely long, parallel wires carry 6-A currents in opposite directions. Determine the magnetic flux density at point P in Fig. 5-38 (P5.12). I1 = 6A P 0.5m 2m Figure P5.12: Arrangement for Problem 5.12. Solution:  £ ¦¢ © £ ¢ ©¢ £ B ˆ µ0 I1 φ 2π 0 5 ˆ µ0 I2 φ 2π 1 5 ˆ µ0 6 φ π   π I2 = 6A 2 ˆ 8µ0 φ π   £ £ £ I2  ¡  ¨ £ £ £ B1 ˆ z £ 02A (T) 258 1 1 2 ln ¦ CHAPTER 5 x2 h2 w2 x w2 At P in Fig. P5.17, the field is pointing to the left. The z-component could have been assumed zero with a symmetry argument. An alternative solution is to employ Eq. (5.24a) directly. (b) From Eq. (5.9), a differential force is of the form d F m I d l B or, assuming ˆ d l a d , the force per unit length is given by The force is repulsive; the wire is experiencing a force pushing it up. Problem 5.18 Three long, parallel wires are arranged as shown in Fig. 5-43 (P5.18(a)). Determine the force per unit length acting on the wire carrying I 3 . I1 = 10A 2m 2m I3 = 10A 2m I2 = 10A Figure P5.18: (a) Three parallel wires of Problem 5.18. Solution: Since I1 and I2 are equal in magnitude and opposite in direction, and equidistant from I3 , our intuitive answer might be that the net force on I3 is zero. As    £   ¨  £  £ £ Fm ∂Fm ∂ ˆ Ia B ˆ Iy ˆ x µ0 I tan πw 1 w 2h ˆ z µ0 I 2 tan πw   £   §  1 (A/m) 1 I ˆ x tan πw w 2h  ¡ x w2 w 2h ¡ (N) ¡ § ¢ © ©    ˆ xh ˆ z  I 2πw 1 tan h x h w2 § ¨ ¨ £ £ £ CHAPTER 5 z 259 I1 x 2m into the page (y) R1 B2 I3 R2 x x 2m B1 R2 = R1 I2 out of the page (-y) R1 Figure P5.18: (b) B fields due to I1 and I2 at location of I3 . I1 x F' 31 ' F32 θ x I3 x I2 Figure P5.18: (c) Forces acting on I3 . ˆ we will see, that’s not the correct answer. The field due to I1 (which is along y) at location of I3 is ˆ µ0 I1 B1 b1 2πR1 ˆ where b1 is the unit vector in the direction of B1 shown in the figure, which is £ 260 ˆ perpendicular to R1 . The force per unit length exerted on I3 is CHAPTER 5 Similarly, the force per unit length excited on I3 by the field due to I2 (which is ˆ along y) is µII ˆ 023 F32 R2 2πR2 ˆ ˆ The two forces have opposite components along x and equal components along z. 1 1 Hence, with R1 R2 8 m and θ sin 2 8 sin 1 2 45 , Problem 5.19 A square loop placed as shown in Fig. 5-44 (P5.19) has 2-m sides and carries a current I1 5 A. If a straight, long conductor carrying a current I2 10 A is introduced and placed just above the midpoints of two of the loop’s sides, determine the net force acting on the loop. z I2 4 1 a x 2 3 I1 a y Figure P5.19: Long wire carrying current I2 , just above a square loop carrying I1 (Problem 5.19). Solution: Since I2 is just barely above the loop, we can treat it as if it’s in the same plane as the loop. For side 1, I1 and I2 are in the same direction, hence the force on  £   £ ¡    ¡   ©  § £ £ £ © £ F3 F31 F32 µ0 I1 I3 µ0 I2 I3 sin θ 2πR1 2πR2 4π 10 7 10 20 ˆ z2 2π 8 ˆ z 1 2 ˆ z2 10 5 N/m ¥ £ ¢ ¡   £ ¤¢  ¨ ¡ £ ¢   £ £ £ F31 µ0 I1 I3 ˆ y 2πR1 ˆ b1 ˆ R1 µ0 I1 I3 2πR1 § ¡ £ £ ¨ CHAPTER 5 265 Solution: (a) From the text immediately following Eq. (5.65), that equation may take the form 2 µ I µ0 I ˆ A dl z dz 2 4π R 4π z 2 z r2 2 4r2 (b) From Eq. (5.53), which is the same as Eq. (5.29). Problem 5.25 In a given region of space, the vector magnetic potential is given by ˆ ˆ A x5 cos πy z 2 sin πx (Wb/m). © ¡ 2 4r2 2πr 2 4r2  ¢ © ¡ © ˆ µ0 I φ 4π 2 4r2 4r ˆ φ µ0 I (T) © ¡ ¡¢ © ¡ 2 2 ¢ © © © ¡ ¨ ©¨§ © ¨ ˆ µ0 I φ 4π ¡ 2 4r2 4r2 2 4r2 4r2 ¢ 2 4r2 4r 2 4r2 ¡ 2 ¢ © ¡ © ¨ ¢ © ¡ © © ¨ ¢ ¡ £ © ¢ © ¨ ¢ © ¨ © ¡ © ¢ 2 4r2 ∂ ∂r 2 4r2 2 4r2 ∂ ∂r ¡ 2 © © ¡ ¡ ©© © ¨ ˆ µ0 I φ 4π 2 4r2 4r2 ¡ © ¡ © ¨ ¡ 2 2 © © © © ¡ ¡ ©© ¨ ˆ µ0 I φ 4π ¡ 2 4r2 4r2 ∂ ∂r ¡ 2 © © ¡ ¡ © © ˆ µ0 I ∂ ln φ 4π ∂r 2 4r2 4r2 ¡¡ 2 © © ¡ © © ¨ ¡ ¨ © ¡ ©  ∇ ˆ z µ0 I ln 4π ¡   ¨ ¨ ¨ ¨ ¨ ¨ £ £ B ∇ A 2 4r2 4r2 2 4r2 4r2 ¡ © © ¡ © © ˆ z µ0 I ln 4π 2 4r2 © ¢ 2 2 © ¨ © ˆ z ¢ © ¡   µ0 I ln 4π 2 2 2 r2 2 r2 ¢£  ¡ z ¢ z2 r2 © ¢ © µ0 ˆ zI ln z 4π © ¢ ¨ ¨ ¡  ¡  £ £ £ £ £ 2 2 £ £ £ £ £ £  2 4r2 266 (a) Determine B. CHAPTER 5 (b) Use Eq. (5.66) to calculate the magnetic flux passing through a square loop with 0.25-m-long edges if the loop is in the x–y plane, its center is at the origin, and its edges are parallel to the x- and y-axes. (c) Calculate Φ again using Eq. (5.67). Solution: (a) From Eq. (5.53), B (b) From Eq. (5.66), sides of length 0.25 m centered at the origin. Thus, the integral can be written as C C where Sfront , Sback , Sleft , and Sright are the sides of the loop. y ¡ ¢ ¡ ¢  ¡ y 0 125  x 0 125 ¢ © © £ Sleft ˆ z2 ¢¢ ˆ x5 cos πy 0 125 ¢ sin πx ¨ y 0 125 ¢ ¡ ¢  ¡ x 0 125 ¨ ¡  £ ¢  ¢ 5x cos πy ¢ ¡ ¢  ¡ x 0 125 ¢ ¢  ¨ ¡ x 0 125 0 125 5 cos πy y 0 125 dx 5 π cos 4 8 0 125 ¨ y 0 125 ¢ ¡ ˆ y dy ¢ © ©  ¨ £ Sback ˆ z2 ¢¢ ˆ x5 cos πy sin πx 0 125 ¢ ¢  ¡ ¢  ¡ y 0 125 x 0 125 ˆ x dx ¡  £ ¨ £ ¢  ¢ 5x cos πy 0 125 5 cos 4 § ¢ ¡ ¢  ¡ x 0 125  ¢  ¢ ¡ x 0 125 0 125 0 125 ¢ 5 cos πy y 0 125 dx π 8 5 π cos 4 8 ¢  © ©  £ Sfront ˆ z2 ¢¢ ˆ x5 cos πy sin πx 0 125 ¢ ˆ x dx ¡ © © © £ £ Φ Ad £ (c) From Eq. (5.67), Φ A d , where C is the square loop in the x-y plane with Sfront Sback Sleft  £  ¨ cos ¨  5 π cos 4 8 π 8 0 ¢  ¡ ¢ § ¡ x 0 125 y ¢  ¢  cos πy 5πx π ¢  ¡ ¢ § ¨  § ¡ y 0 125 m x 0 125 m 0 125 0 125 0 125 Sright ¢ ¨ ¨£ £ B ds ¦¢ ¢ ¢ £ ¦ £ £ £ £ £ Φ ˆ z5π sin πy ˆ yπ cos πx 0 125 m 0 125 m ¨ £  ∇ A ˆ z5π sin πy ˆ yπ cos πx. £ ˆ z dx dy CHAPTER 5 0 125 267 0x ¡ Thus, c Problem 5.26 A uniform current density given by gives rise to a vector magnetic potential (a) Apply the vector Poisson’s equation to confirm the above statement. (b) Use the expression for A to find H. (c) Use the expression for J in conjunction with Amp` re’s law to find H. Compare e your result with that obtained in part (b). Solution: (a) ¤ ¥ ¤ ¢ ¨ © © ¨ § ¨ ¨ ¢  © ©  ¨ ¨ ¨§ © £ § ¨ £ £ £ £  £ H 1 ∇ µ0 A ∂Ay ∂Ax 1 ∂Az ∂Ay ∂Ax ∂Az ˆ ˆ ˆ x y z µ0 ∂y ∂z ∂z ∂x ∂x ∂y 1 ∂Az ∂Az ˆ ˆ x y µ0 ∂y ∂x J0 2 J0 2 1 ∂ ∂ ˆ ˆ µ0 x y2 µ0 x y2 x y µ0 ∂y 4 ∂x 4 J0 y J0 x ˆ ˆ (A/m) x y 2 2 § § § £ ¨ £ Hence, ∇2 A (b) µ0 J is verified. ¤¢ ©  ¨  ¨ £ ¤¢ © © © ¨ £ £ © © £ ∇2 A ˆ x ∇2 A x ˆ y ∇2 A y ˆ z ∇2 A z ˆ z J0 2 ∂2 ∂2 ∂2 µ0 x ∂x2 ∂y2 ∂z2 4 J0 ˆ ˆ z µ0 z µ0 J0 22 4 £  ¢ © § ¨ £ A ˆ z µ0 J0 2 x 4 y2 (Wb/m) ¡ £ J ˆ zJ0 (A/m2 ) y2  £ © ©  ¨   £ © © © £ £ Φ Ad Sfront ¢  ¡ y 0 125 Sback Sleft Sright  £ ¢ ¡ 0x 0 125 dy 0 5 π cos 4 8 ¢ ¡ ¢  ¢ ¡ y 0 125 0 125 x 0 125 5 π cos 4 8 ¢ © © £ Sright ˆ z2 ¢¢ ˆ x5 cos πy sin πx 0 125 y 0 125 ¡ ˆ y dy 0 0 0 £ ¢  ¢  ¢ ¡ ¢ 0 125 dy 0 ¨ £ £ 268 z CHAPTER 5 r J0 Figure P5.26: Current cylinder of Problem 5.26. (c) C S We need to convert the expression from cylindrical to Cartesian coordinates. From Table 3-2,  © £ r x2 y2 © ¢ © x2 y2 x2 y2 ¡ © ¢ ¨ £ © ¢ ¨ £ ˆ φ ˆ x sin φ ˆ y cos φ ˆ x y  £ £ H ˆ φ Hφ ¡ ˆ y x ¦ £ ˆ ˆ φ Hφ φ2πr J0 πr2 ˆr φ J0 2 ¡ £ £ H dl I J ds ¦ ¦ CHAPTER 5 Hence 269 which is identical with the result of part (b). Problem 5.27 A thin current element extending between z L 2 and z L 2 ˆ carries a current I along z through a circular cross section of radius a. (a) Find A at a point P located very far from the origin (assume R is so much larger than L that point P may be considered to be at approximately the same distance from every point along the current element). (b) Determine the corresponding H. z L/2 R θ P I -L/2 Cross-section πa2 Solution: (a) Since R L, we can assume that P is approximately equidistant from all segments of the current element. Hence, with R treated as constant, (5.65) gives L2  £  £ ¢ £ £ A ˆ z µ0 J dV 4π V R I µ0 ˆ z πa2 dz 4πR V πa2 µ0 I 4πR L2 dz ˆ z Figure P5.27: Current element of length L observed at distance R L. µ0 IL 4πR  £  © ¨ £ ¨ © ¡ © ¢ © x2 y2 x2 y2 ¡ £ © © ¢ ¨ £ H ˆ x ˆ y ¢¦ y x J0 2 x2 y2 ˆ x yJ0 2 ˆ y xJ0 2 270 (b) CHAPTER 5 Section 5-6: Magnetic Properties of Materials Problem 5.28 In the model of the hydrogen atom proposed by Bohr in 1913, the electron moves around the nucleus at a speed of 2 10 6 m/s in a circular orbit of radius 5 10 11 m. What is the magnitude of the magnetic moment generated by the electron’s motion? Solution: From Eq. (5.69), the magnitude of the orbital magnetic moment of an electron is Problem 5.29 Iron contains 8 5 1028 atoms/m3 . At saturation, the alignment of the electrons’ spin magnetic moments in iron can contribute 1.5 T to the total magnetic flux density B. If the spin magnetic moment of a single electron is 9 27 10 24 (A m2 ), how many electrons per atom contribute to the saturated field? Solution: From the first paragraph of Section 5-6.2, the magnetic flux density of a magnetized material is Bm µ0 M, where M is the vector sum of the microscopic magnetic dipoles within the material: M Ne ms , where ms is the magnitude of the spin magnetic moment of an electron in the direction of the mean magnetization, and Ne is net number of electrons per unit volume contributing to the bulk magnetization. If the number of electrons per atom contributing to the bulk magnetization is n e , then Ne ne Natoms where Natoms 8 5 1028 atoms/m 3 is the number density of atoms for iron. Therefore,         Natoms  £ £ £ £ £ ne 7  Ne M ms Natoms B µ0 ms Natoms 15 4π 10 9 27 10 1 5 (electrons/atom) 24 85 1028    £    £       £ £  £ ¨  £ m0 16 10 2 5 10 8 10 ¦ 1 2 eur 1 2 19 106  ¢ x2 2  ¤ © © ¨ ©  IL 4π ˆ xy y2 ˆ yx z2 3 11 24 (A m2 ) ¡ © © ¢ ¨ ¡ ¡ x2 y2 z2 x2 y2 z2 ¡ ¢¡ © © ¢ £ 1 µ0 ˆ x ∂ ∂y µ0 IL 4π 1 ˆ y ∂ ∂x µ0 IL 4π ¤ ¨  £   £ £ £ £ H 1 ∇A µ0 ∂Az 1 ˆ x µ0 ∂y ˆ y ∂Az ∂x 1 £  CHAPTER 5 271 Section 5-7: Magnetic Boundary Conditions Problem 5.30 The x–y plane separates two magnetic media with magnetic permeabilities µ1 and µ2 , as shown in Fig. 5-45 (P5.30). If there is no surface current at the interface and the magnetic field in medium 1 is (b) θ1 and θ2 , and z θ1 H1 µ1 x-y plane µ2 Figure P5.30: Adjacent magnetic media (Problem 5.30). Solution: (a) From (5.80), and in the absence of surface currents at the interface, (5.85) states ¡ £ H1y H2y ¡ £ H1x H2x ¡ £ £ In this case, H1z H1n , and H1x and H1y are tangential fields. Hence, µ1 H1z µ2 H2z  £ H1t H2t ¡ £ µ1 H1n µ2 H2n £ £ £ £ (c) evaluate H2 , θ1 , and θ2 for H1x and µ2 4µ0 . £ ¡ find: (a) H2 2 (A/m), H1y ¡ 0, H1z 4 (A/m), µ1 µ0 , © © £ H1 ˆ xH1x ˆ yH1y ˆ zH1z 272 and µ1 H1z µ2 CHAPTER 5 (b) H1z 2 2 H1x H1y µ1 H1z µ2 (c) ˆ Problem 5.31 Given that a current sheet with surface current density J s x 8 (A/m) ˆ exists at y 0, the interface between two magnetic media, and H 1 z 11 (A/m) in medium 1 y 0 , determine H2 in medium 2 y 0 . Solution: or ¡ £  ˆ y H2 ˆ x3 ¡   ˆ x 11 ˆ y H2 ˆ x8 ¡ ¨ ¨  ˆ y ˆ z 11 H2 ˆ x8 ¡ £ £¢ £ ¢ ¨   ˆ n2 H1 H2 Js £ ˆ H1 is tangential to the boundary, and therefore H 2 is also. With n2 (5.84), we have  £ H1 ˆ z 11 A/m ¡ £ Js ˆ x 8 A/m ˆ y, from Eq. £ £ ¢ ¥ ¡  £  £ θ2 tan 1 63 44 ¡¥  £ §  £ θ1 tan 1 26 56 ¡ © £ §© £ H2 ˆ x2 ˆ z 1 4 4 2 4 2 1 ˆ x2 ˆ z (A/m)  £ © ¦ £ £ tan θ2 H2t H2z ¡ © £ £ tan θ1 H1t H1z 2 H1x ¡ © £ H1t 2 H1x 2 H1y 2 H1y µ2 tan θ1 µ1  © © £ H2 ˆ xH1x ˆ yH1y ˆ z ¢ £ CHAPTER 5 y H1 n2 Js x H2 273 Figure P5.31: Adjacent magnetic media with J s on boundary. which implies that H2 does not have an x-component. Also, since µ 1 H1y µ2 H2y and H1 does not have a y-component, it follows that H 2 does not have a y-component either. Consequently, we conclude that Problem 5.32 In Fig. 5-46 (P5.32), the plane defined by x y 1 separates medium 1 of permeability µ1 from medium 2 of permeability µ2 . If no surface current exists on the boundary and ˆ ˆ B1 x2 y3 (T) find B2 and then evaluate your result for µ1 5µ2 . Hint: Start out by deriving the equation for the unit vector normal to the given plane. ˆ Solution: We need to find n2 . To do so, we start by finding any two vectors in the plane x y 1, and to do that, we need three non-collinear points in that plane. We choose 0 1 0 , 1 0 0 , and 1 0 1 . Vector A1 is from 0 1 0 to 1 0 0 : Vector A2 is from 1 0 0 to 1 0 1 : Hence, if we take the cross product A2 A1 , we end up in a direction normal to the given plane, from medium 2 to medium 1, ¡ ¨ ¡ © ¡   A2 A1  £ ¨ £ © £ £ ˆ n2 ¢    A2 A2 A1 A1 ˆ z1 ˆ x1 ˆ y1  £ A2 ˆ z1  ©  £ A1 ˆ x1 ˆ y1 ˆ ˆ y1 x1 11 ˆ y 2 ˆ x 2 £ ¨ ¡  £ £ © H2 ˆ z3 £ ¢¡¡ ¢ ¡ ¡ ¢¡¡ £ ¨ ¢¡ ¡ ¢ ¡ ¡ ¢¡¡ ¢¡ £ ¨ ¡¨ 274 y CHAPTER 5 n2 Medium 1 µ1 n2 (1, 0) x Medium 2 µ2 (0, -1) In medium 1, normal component is Tangential component is Boundary conditions:  £ £ For µ1 5µ2 , B2 ˆ y (T) ¢  ©  © ¨ £ © £ B2 B2n B2t § Finally, ˆ y 2 ˆ x 2 µ2 ˆ x2 5 µ1 ¢  ©  £ £ B2t Hence, µ2 B1t µ1 µ2 ˆ x2 5 µ1 ˆ y2 5  £ ¡ £ H1t H2t or B2t µ2 ¡ ¨ £ ¡ £ B1n B2n or B2n ˆˆ yx 22 B1t µ1 ˆ y2 5  ©  £ ¨ ¨ ¢ © £ ¨ £ B1t B1 B1n ˆ x2 ˆ y3 ˆ y 2 ˆ x 2  ¨ £ §  ¡ ¨ ¡ £ £ B1n ˆ n2 B1n 1 2 ˆ y 2 ˆ x 2 ˆ x2 5 ˆ y2 5 ¡ ¡ £ ¡ ¨ ¡ £ ¤¢ © ¡¦  ¡ ¨ ¡ § £ £ B1n ˆ n2 B1 ¦ ˆ y 2 ˆ y 2 ˆ x 2 ˆ x 2 ˆ x2 ˆ y3 3 2 £ Figure P5.32: Magnetic media separated by the plane x y 1 (Problem 5.32). ¨ § 2 2 1 2 ¦ CHAPTER 5 275 Problem 5.33 The plane boundary defined by z 0 separates air from a block of ˆ ˆ ˆ iron. If B1 x4 y6 z8 in air (z 0), find B2 in iron (z 0), given that µ 5000µ0 for iron. Solution: From Eq. (5.2), or Problem 5.34 Show that if no surface current densities exist at the parallel interfaces shown in Fig. 5-47 (P5.34), the relationship between θ 4 and θ1 is independent of µ2 . B3 θ4 B2 θ2 θ3 µ3 µ2 µ1 B1 θ1 Figure P5.34: Three magnetic media with parallel interfaces (Problem 5.34).  © ¨ £ £ £ where µ2 µ1 µr 5000. Therefore, B2 ˆ x20000 ˆ y30000 ˆ z8 ¡ ¨ £ £ ¡ £ £ B2x µ2 H2x µ2 4 µ1 B2y µ2 H2y µ2 6 µ1 ¡ ¨ £ £ ¡ £ £ H2x H1x 1 4 µ1 H2y H1y 1 6 µ1 £ The z component is the normal component to the boundary at z Eq. (5.79), B2z B1z 8 while, from Eq. (5.85), ¢ © ¨ £ £ H1 B1 µ1 1 ˆ x4 µ1 ˆ y6 ˆ z8 0. Therefore, from £ £ £ © ¨ £ £  276 Solution: CHAPTER 5 and which is independent of µ2 . Sections 5-8 and 5-9: Inductance and Magnetic Energy Problem 5.35 Obtain an expression for the self-inductance per unit length for the parallel wire transmission line of Fig. 5-27(a) in terms of a, d , and µ, where a is the radius of the wires, d is the axis-to-axis distance between the wires, and µ is the permeability of the medium in which they reside. Solution: Let us place the two wires in the x–z plane and orient the current in one of them to be along the z-direction and the current in the other one to be along the z-direction, as shown in Fig. P5.35. From Eq. (5.30), the magnetic field at point P x 0 z due to wire 1 is where the permeability has been generalized from free space to any substance with ˆˆ permeability µ, and it has been recognized that in the x-z plane, φ y and r x as long as x 0. Given that the current in wire 2 is opposite that in wire 1, the magnetic field created by wire 2 at point P x 0 z is in the same direction as that created by wire 1, and it is given by  ¢ ¨ £ B2 ˆ y µI 2π d x £ £ ¡ £ £ B1 ˆ µI φ 2πr ˆ y µI 2πx ¡ £ £ £ ¢¡¡ © £ tan θ4 £ We note that θ2 θ3 and µ3 tan θ3 µ2 µ3 tan θ2 µ2 µ3 µ2 tan θ1 µ2 µ1 µ3 tan θ1 µ1  £ £ £ £ But B2n B1n and B2t µ2 B1t . Hence, µ1 tan θ2 B1t µ2 B1n µ1 µ2 tan θ1 µ1  £ tan θ2 B2t B2n ¡ £ tan θ1 B1t B1n ¢¡¡ ¨ ...
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This note was uploaded on 10/20/2010 for the course ECE 312 taught by Professor Johnson,j during the Spring '08 term at Ohio State.

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