ECE312_W09_Homework_1_Solutions_012309

ECE312_W09_Homework_1_Solutions_012309 - ECE 312 - Homework...

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Unformatted text preview: ECE 312 - Homework 1 Solutions 212 Problem 4.48 Solution: From Eq. (4.131), CHAPTER 4 Problem 4.49 Dielectric breakdown occurs in a material whenever the magnitude of the field E exceeds the dielectric strength anywhere in that material. In the coaxial capacitor of Example 4-12, (a) At what value of r is E maximum? (b) What is the breakdown voltage if a 1 cm, b 2 cm, and the dielectric material is mica with εr 6? Solution: ˆ (a) From Eq. (4.114), E rρl 2πεr for a r b. Thus, it is evident that E is maximum at r a. (b) The dielectric breaks down when E 200 (MV/m) (see Table 4-2), or which gives ρl 200 MV/m 2π 6 8 854 10 12 0 01 667 6 µC/m). From Eq. (4.115), we can find the voltage corresponding to that charge density, Problem 4.50 An electron with charge Q e 1 6 10 19 C and mass 31 kg is injected at a point adjacent to the negatively charged plate in me 9 1 10 the region between the plates of an air-filled parallel-plate capacitor with separation of 1 cm and rectangular plates each 10 cm 2 in area Fig. 4-33 (P4.50). If the voltage across the capacitor is 10 V, find (a) the force acting on the electron, (b) the acceleration of the electron, and (c) the time it takes the electron to reach the positively charged plate, assuming that it starts from rest. Solution: (a) The electric force acting on a charge Q e is given by Eq. (4.14) and the electric field in a capacitor is given by Eq. (4.112). Combining these two relations, we have    ¨ £    ¨ £ £ £ F Qe E Qe V d 16 10 19 10 0 01 16 10 16   ¨ £    £ Thus, V 1 39 (MV) is the breakdown voltage for this capacitor. (N)   £ ¤¢ ¢    £ £ V ¢  ρl b ln 2πε a 667 6 µC/m ln 2 12π 8 854 10 12 F/m 1 39  £ ¢  ¢  ¢ 2π 6ε0 10 2 ¡ £  ¢  ¢ £ ¢ £ E ρl 2πεr ρl 200 (MV/m) (MV)    £ ¨ £ ¡  ¡  £ ¢   £ ¨ £ ¨ £ § £ £ ¨ £ F ˆ z ˆ z 2ε0 5 εA E 2 2 10 4 § 50 0 02 2 ˆ z 55 3 10 9 (N) £ 214 Solution: Electrostatic potential energy is given by Eq. (4.124), z0y0x 1 ¡ CHAPTER 4 Problem 4.52 Figure 4-34a (P4.52(a)) depicts a capacitor consisting of two parallel, conducting plates separated by a distance d . The space between the plates A1 ε1 ε2 A2 + - V d (a) + C1 C2 V (b) Figure P4.52: (a) Capacitor with parallel dielectric section, and (b) equivalent circuit. contains two adjacent dielectrics, one with permittivity ε 1 and surface area A1 and another with ε2 and A2 . The objective of this problem is to show that the capacitance C of the configuration shown in Fig. 4-34a (P4.52(a)) is equivalent to two capacitances in parallel, as illustrated in Fig. 4-34b (P4.52(b)), with ¡ © £ C C1 C2    4 62 10 9 (J)  £ 4ε0 2 1304 5 (4.132) ¡ ¡  ¡ x 1 y0 ¢ © © © © 4ε0 2 25 x yz 5 223 zxy 3 43 z xy 3 1 y 12 1 2 zx 4 ¢£ ¢ ¢£ © © © ¢ ©  ¡ ¡ ¡ £ § ¡ ¡ § £ £ £ We x2 2z 2 x4 1 ε E 2 dV 2V ε 2 3 2 1 y z 2 dx dy dz 3 z0 CHAPTER 4 where 215 To this end, you are asked to proceed as follows: (a) Find the electric fields E1 and E2 in the two dielectric layers. (b) Calculate the energy stored in each section and use the result to calculate C 1 and C2 . (c) Use the total energy stored in the capacitor to obtain an expression for C. Show that Eq. (4.132) is indeed a valid result. Solution: ε1 E1 E2 ε2 d + V - (c) Figure P4.52: (c) Electric field inside of capacitor. (a) Within each dielectric section, E will point from the plate with positive voltage to the plate with negative voltage, as shown in Fig. P4-52(c). From V Ed , (b) But, from Eq. (4.121),  £ ¢ © £ © £ We We1 We2 £ A1 Hence C1 ε1 . Similarly, C2 d (c) Total energy is ε2 A2 . d ε2 A 2 1 CV 2 2 1 V2 ε1 A 1 2d  £ We1 1 C1V 2 2  £ £ £ We1 ¦ 1 2 ε1 E 1 V 2 1 V2 ε1 A1 d 2 d2  £ £ E1 E2 V d 1 A1 ε1V 2 2 d £  £ C2 ¡ £ C1 ε1 A 1 d ε2 A 2 d (4.133) (4.134) ¦ £ 216 Hence, ε1 A 1 d ε2 A 2 d CHAPTER 4 Problem 4.53 Use the result of Problem 4.52 to determine the capacitance for each of the following configurations: (a) conducting plates are on top and bottom faces of rectangular structure in Fig. 4-35(a) (P4.53(a)), (b) conducting plates are on front and back faces of structure in Fig. 4-35(a) (P4.53(a)), (c) conducting plates are on top and bottom faces of the cylindrical structure in Fig. 4-35(b) (P4.53(b)). Solution: (a) The two capacitors share the same voltage; hence they are in parallel. (b) 4 (c)     £ © © £ C ¡    £ ¡ ¢  ¨ ¢   £  ¨  £ £ C3 32 0 12 10 12 F ¡    £ ¡ ¢  ¨ ¢   £ ¢  ¨  £ £ C2 32 ¡    £¢   ¢ £   £ £ C1 2 A1 πr1 4πε0 8ε0 2 10 3 2 0 04 10 2 d 2 10 10 2 2 2 A2 π r2 r1 2πε0 ε2 4ε0 4 10 3 2 2 10 d 2 10 2 10 2 2 2 π r3 r2 A3 πε0 ε3 2ε0 8 10 3 2 4 10 2 d 2 10 10 2 C1 C2 C3 0 22 10 12 F ε1    £ © ¢ £ C 12 ¡  £   £ £ C2 24 ε0 5 10 ¡   £    ¦¢   £ £ C1 ε1  A1 2 1 10 2ε0 d 5 10 2 A2 3 2 10 ε2 4ε0 d 5 10 2 C1 C2 0 5 10 12 F 4 0 8 ε0 10 2 2 F 0 06    £  £  ¦¢  ¤¢  © £ © £ C ¡  £   £ £ C2 ¡  £    ¤¢   £ £ C1 A1 5 1 10 4 2ε0 5ε0 10 2 d 2 10 2 A2 5 3 10 4 ε2 4ε0 30ε0 10 2 d 2 10 2 C1 C2 5ε0 30ε0 10 2 0 35ε0 3 1 ε1  10 12 © £  ©  ¤¢  £ C C1 C2 F 10 12 F CHAPTER 4 1 cm 5 cm 2 cm 3 cm 221 εr = 2 εr = 4 Figure P4.55: Dielectric section for Problem 4.55. Section 4-12: Image Method Problem 4.56 With reference to Fig. 4-37 (P4.56), charge Q is located at a distance d above a grounded half-plane located in the x–y plane and at a distance d from another grounded half-plane in the x–z plane. Use the image method to (a) establish the magnitudes, polarities, and locations of the images of charge Q with respect to each of the two ground planes (as if each is infinite in extent), and (b) then find the electric potential and electric field at an arbitrary point P 0 y z . z P(0, y, z) Q(0, d, d) y d d Figure P4.56: Charge Q next to two perpendicular, grounded, conducting half planes. Solution: (a) The original charge has magnitude and polarity Q at location 0 d d . Since the negative y-axis is shielded from the region of interest, there might as well be a conducting half-plane extending in the y direction as well as the y direction. This ground plane gives rise to an image charge of magnitude and polarity Q at location ¢¡¡ ¢¡¡ ¨ © © ¨ 222 z CHAPTER 4 -Q d d Q P(y, z) y -d Q -d -Q Figure P4.56: (a) Image charges. 0 d d . In addition, since charges exist on the conducting half plane in the z direction, an image of this conducting half plane also appears in the z direction. This ground plane in the x-z plane gives rise to the image charges of Q at 0 d d and Q at 0 d d . (b) Using Eq. (4.47) with N 4, 1 1 x2 x2 x2 y2 y2 y2 2yd 1 2yd 2yd z2 z2 1 z2 2zd 2zd 2zd 2d 2 2d 2 ¡ 2d 2  © © © © ¨ © ¨ © © © © © ¨ ¨ © © © © © © ¢ Q 4πε 1 z2 x2 y2 2yd 1 2zd 2d 2 (V) ¢ © © ¢ ¨ © © © ¢ y d z d x2 ¨¢ 2 2 x2 y d 2 z d 2 ¢ ¢£ ¨ © ¢ © © ¨ © ¢ x2 y d z d ¨¢ 2 2 x2 y d 2 ¨ © © © Q 4πε 1 1 © © ¢ ¨ © © ¢ ˆ xx ˆ zz d ˆ xx ˆ zz d z ¢ ¨ © ¢ © © © ¢ ˆ xx ˆ yy 1 ˆ yy d ˆ zz d ˆ xx ˆ yy 1 ˆ yy d ˆ zz d d ¢ © ¨ ¢ ¨ ¢ ¨ ¨ © © © ¡ ¢ ¢ ¢ ¨ © © ¨ © £ £ £ ¢ ¡ ¡ V xyz Q 4πε 1 d 1 d 2 ¢¡ © ¨ ¡ ¨ ¨ £ ¢ § ¨ ¡ ¨ ¡ ¢ ¨ ¡¡ © CHAPTER 4 From Eq. (4.51), 223 Problem 4.57 Conducting wires above a conducting plane carry currents I 1 and I2 in the directions shown in Fig. 4-38 (P4.57). Keeping in mind that the direction I2 I1 (a) (b) Figure P4.57: Currents above a conducting plane (Problem 4.57). of a current is defined in terms of the movement of positive charges, what are the directions of the image currents corresponding to I1 and I2 ? £ Solution: (a) In the image current, movement of negative charges downward positive charges upward. Hence, image of I1 is same as I1 . ¢¢ © © ¢ ¨ © ¢¢ © © ¢ x2 y d z d x2 y d z d ¡ 2 232 2 232 movement of  ¢ © © ¢ ¨ © ¨ ¢ © © ¢ ˆ xx ˆ yy d ˆ zz d ˆ xx ˆ yy d ˆ zz d ¢¢ ¨ © ¢ © © ¢¢ ¨ © ¢ x2 y d 2 z d 232 x2 y d 2 z d 232 ¢ ¨ © ¢ © © ¨ ¢ ¨ © ¢ ¨ ¨ © © © © © Q 4πε ˆ xx ˆ yy d ˆ zz d ˆ xx ˆ yy d ˆ zz ¢ © © ¢ ¨ © © © ¢ x2 y d z d ¨¢ 2 2 x2 y d 2 © ∇ 1 ∇ 1 z d 2 d ¢ ¢£ ¨ © ¢ © © ¨ © ¢ y d z d x2 ¨¢ 2 2 x2 y d 2 ¨ © © © Q 4πε ¡ © © ¨ £ £ £ E ∇V ∇ 1 ∇ 1 z d 2 (V/m) 224 + q @ t=t1 I1 + q @ t=0 CHAPTER 4 I1 (image) - q @ t=0 - q @ t=t1 Figure P4.57: (a) Solution for part (a). I1 @t=0 +q + q @ t=t1 -q @t=0 I1 - q @ t=t1 (image) Figure P4.57: (b) Solution for part (b). Problem 4.58 Use the image method to find the capacitance per unit length of an infinitely long conducting cylinder of radius a situated at a distance d from a parallel conducting plane, as shown in Fig. 4-39 (P4.58). Solution: Let us distribute charge ρ l (C/m) on the conducting cylinder. Its image cylinder at z d will have charge density ρ l . For the line at z d , the electric field at any point z (at a distance of d z from the center of the cylinder) is, from Eq. (4.33), ¢ ¨ ¨ £ E1 ˆ z ρl 2πε0 d z ¨ £ (b) In the image current, movement of negative charges to right positive charges to left. movement of ¨ £¨ £ 244 CHAPTER 5 when π 2 φ π 2 and negative over the second half of the circle. Thus, work is provided by the force between φ π 2 and φ π 2 (when rotated in the ˆ φ -direction), and work is supplied for the second half of the rotation, resulting in a net work of zero. (c) The force F is maximum when cos φ 1, or φ 0. z 20-turn coil w l I y φn ^ x Figure P5.6: Rectangular loop of Problem 5.6. determine the torque acting on the coil. (b) At what angle φ is the torque zero? (c) At what angle φ is the torque maximum? Determine its value. ¡ £  ¤¢    £ £ m ˆ nNIA ˆ n 20 10 30 10 10 ˆ n6 ¦ 4 ¢ © Solution: (a) The magnetic field is in direction φ0 tan 1 2 63 43 . 1 The magnetic moment of the loop is ˆ x ˆ y 2 , which makes an angle ¡ ¢ 2 © £  (a) If the coil, which carries a current I flux density B 2 10 10 A, is in the presence of a magnetic ˆ x ˆ y2 (T) (A m2 ) £ £ Problem 5.6 A 20-turn rectangular coil with side l placed in the y–z plane as shown in Fig. 5-34 (P5.6). 20 cm and w  ¨ £ £ £  £  £ ¥   £   ¨ £ ¨ 10 cm is CHAPTER 5 z 245 2 y 1 φ0 = 63.43o B x Figure P5.6: (a) Direction of B. ˆ where n is the surface normal in accordance with the right-hand rule. When the loop ˆ ˆ is in the negative-y of the y–z plane, n is equal to x, but when the plane of the loop is ˆ moved to an angle φ, n becomes (b) The torque is zero when or ˆ Thus, when n is parallel to B, T 0. ˆ (c) The torque is a maximum when n is perpendicular to B, which occurs at Mathematically, we can obtain the same result by taking the derivative of T and equating it to zero to find the values of φ at which T is a maximum. Thus, £ ¤¢ ¢ ¨  £ ∂T ∂φ ∂ 0 12 2 cos φ ∂φ sin φ 0 ¥  © ¥  ¨ £¥ £  £ φ 63 43 90 26 57 or 153 43 ¥  ¨ ¥  £ £ ¡ £ tan φ 2 φ 63 43 or ¡ £ ¨ 2 cos φ sin φ 0 116 57  (N m) ¦ ¡ ¨ ˆ z 0 12 2 cos φ sin φ ¢ ©   ¢ © ˆ x cos φ ˆ y sin φ 6 2 10 2 ˆ x ¢ ©     £   £ £ £ T m B ˆ n6 2 10 2 ¡ © £ ˆ n ˆ x cos φ ˆ y sin φ ˆ x ˆ y2 ˆ y2 246 or CHAPTER 5 Section 5-2: Biot–Savart Law 12 cm rectangular loop of wire is situated in the x–y Problem 5.7 An 8 cm plane with the center of the loop at the origin and its long sides parallel to the x-axis. The loop has a current of 50 A flowing with clockwise direction (when viewed from above). Determine the magnetic field at the center of the loop. Solution: The total magnetic field is the vector sum of the individual fields of each of the four wire segments: B B1 B2 B3 B4 . An expression for the magnetic field from a wire segment is given by Eq. (5.29). z -6 cm 4 2 3 -4 cm 4 cm 1 6 cm x Figure P5.7: Problem 5.7. For all segments shown in Fig. P5.7, the combination of the direction of the current and the right-hand rule gives the direction of the magnetic field as z direction at the origin. With r 6 cm and l 8 cm,     ¨ £  ©      ¡  ©   ¡  ¨ ¨ £ B1 µIl 2πr 4r2 l 2 4π 10 7 50 0 08 ˆ z 2π 0 06 4 0 062 0 082 ˆ z ˆ z 9 24 10 ¨ © © © £ £   £ £ £ at which T ˆ z 0 27 (N m). I y ¡¥  ¥  ¨ £ ¦ ¨ £ which gives tan φ 1 2, or φ 26 57 or 153 43 ¡ 5 £ © 2 sin φ cos φ 0 ¨ (T) 250 Therefore, the field d H due to the current Ix is 2 CHAPTER 5 and the total field is An alternative approach is to employ Eq. (5.24a) directly. Problem 5.11 An infinitely long wire carrying a 25-A current in the positive x-direction is placed along the x-axis in the vicinity of a 20-turn circular loop located in the x–y plane as shown in Fig. 5-37 (P5.11(a)). If the magnetic field at the center of the loop is zero, what is the direction and magnitude of the current flowing in the loop? 1m d = 2m x I1 Figure P5.11: (a) Circular loop next to a linear current (Problem 5.11). Solution: From Eq. (5.30), the magnetic flux density at the center of the loop due to ¡  £ £  ¤¢¢ ¡ § ¢ © © ¤ ¥ ¨ ¢  © ¦ © © © § ¡ ¡ ¢  © © © © © §© ¢ § §  ¢   © ¡ § © ¡ £ £ § § ¡ £ £ £ £ £ £ ¢ ¡ ¡ H00z Js dx 2π x2 z2 x0 w Js dx ˆ ˆ xz zx 2 2π x 0 x z2 w w dx x dx Js ˆ ˆ xz z 2 2 2 2π z z2 x 0x x 0x w Js 1 x w ˆ ˆ2 tan 1 xz z 1 ln x2 z2 x 0 2π z z x0 5 w ˆ ˆ2 for z 0 x2π tan 1 z 1 ln w2 z2 ln 0 z2 2π z 5 w w2 z2 ˆ ˆ2 (A/m) for z 0 x2π tan 1 z 1 ln 2π z z2 w ˆ xz ˆ zx ¡ ¢ © © £ ¢ © © £ dH ¢ ˆ ˆ xz zx x2 z2 1 Ix 2πR ˆ ˆ xz zx Js dx 2π x2 z2 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 the wire is or 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  2aI1 2πNd 1 25 20 2 ¡ £ NI2 2a I1 2πd  ¨ £ £ B1 ˆ z £ 02A (T) 252 CHAPTER 5 Problem 5.13 A long, East-West oriented power cable carrying an unknown current I is at a height of 8 m above the Earth’s surface. If the magnetic flux density recorded by a magnetic-field meter placed at the surface is 15 µT when the current is flowing through the cable and 20 µT when the current is zero, what is the magnitude of I ? Solution: The power cable is producing a magnetic flux density that opposes Earth’s, own magnetic field. An East–West cable would produce a field whose direction at the surface is along North–South. The flux density due to the cable is As a magnet, the Earth’s field lines are directed from the South Pole to the North Pole inside the Earth and the opposite on the surface. Thus the lines at the surface are from North to South, which means that the field created by the cable is from South to North. Hence, by the right-hand rule, the current direction is toward the East. Its magnitude is obtained from Problem 5.14 Two parallel, circular loops carrying a current of 40 A each are arranged as shown in Fig. 5-39 (P5.14). The first loop is situated in the x–y plane with its center at the origin and the second loop’s center is at z 2 m. If the two loops have the same radius a 3 m, determine the magnetic field at: (a) z 0, (b) z 1 m, (c) z 2 m. Solution: The magnetic field due to a circular loop is given by (5.34) for a loop in ˆ the x–y plane carrying a current I in the φ -direction. Considering that the bottom ˆ loop in Fig. P5.14 is in the x–y plane, but the current direction is along φ , 32 where z is the observation point along the z-axis. For the second loop, which is at a height of 2 m, we can use the same expression but z should be replaced with z 2 . Hence, Ia2 ˆ z2 H2 2a z 2232 ¢ ¨  ¡ ¡ ¢ ¢ ¨ © © ¨ £ H1 ˆ z Ia2 2 a2 z2 ¨ £ © ¨ £ £ £ which gives I 200 A. ¡  £ £  £ 5 µT 5 10 6   µ0 I 2πd 4π 10 7 I 2π 8  £ ¢ ¨ £ B 20 15 µT 5µT £ £ £ CHAPTER 5 z a z = 2m I 253 a 0 I x y Figure P5.14: Parallel circular loops of Problem 5.14. The total field is a2 32 Section 5-3: Forces between Currents Problem 5.15 The long, straight conductor shown in Fig. 5-40 (P5.15) lies in the plane of the rectangular loop at a distance d 0 1 m. The loop has dimensions   £  ¨ £ H ˆ z 10 5 A/m £ £ (c) At z 2 m, H should be the same as at z 0. Thus, ¨ ¢ © © ¢ 2 32 32   £ ¤ © ¨ £ H ˆ z  40 9 £ £ (b) At z 1 m (midway between the loops): 1 91 1 91 ¨ ¢ © © 2 32 ˆ z 11 38 A/m   £ ¤ ¨ £ H ˆ z  40 9 1 33 £ £ £ £ (a) At z 0, and with a 3 m and I 40 A, 1 94 ˆ z 10 5 A/m ¡ ¢ a2 z 2 232  ¤ ¨ © © ¢ © ¨ £ © £ H H1 H2 ˆ z Ia2 2 £ 1 z2 1 A/m CHAPTER 5 255 The other two sides do not contribute any net forces to the loop because they are equal in magnitude and opposite in direction. Therefore, the total force on the loop is The force is pulling the loop toward the wire. Problem 5.16 In the arrangement shown in Fig. 5-41 (P5.16), each of the two long, parallel conductors carries a current I , is supported by 8-cm-long strings, and has a mass per unit length of 1.2 g/cm. Due to the repulsive force acting on the conductors, the angle θ between the supporting strings is 10 . Determine the magnitude of I and the relative directions of the currents in the two conductors. z z 8 cm θ = 10° 1 x 2 y x F' θ F'v 2 F'h (b) (a) Figure P5.16: Parallel conductors supported by strings (Problem 5.16). Solution: While the vertical component of the tension in the strings is counteracting the force of gravity on the wires, the horizontal component of the tension in the strings is counteracting the magnetic force, which is pushing the wires apart. According to Section 5-3, the magnetic force is repulsive when the currents are in opposite directions. Figure P5.16(b) shows forces on wire 1 of part (a). The quantity F is the tension force per unit length of wire due to the mass per unit length m 1 2 g/cm 0 12  £   £  ¨ £   ¥   ¢  ©    ¢ © ©  © ¨ ¨ ¨ £ £ £ £ F Fm1 Fm2 µ0 I1 I2 b µ0 I1 I2 b ˆ ˆ x x 2πd 2π a d µ0 I1 I2 ab ˆ x 2πd a d 4π 10 7 20 30 0 2 ˆ x 2π 0 1 0 3 05 ˆ x0 4 (mN) θ 2 d (c) CHAPTER 5 side 1 is attractive. That is, 261 I1 and I2 are in opposite directions for side 3. Hence, the force on side 3 is repulsive, ˆ which means it is also along y. That is, F3 F1 . The net forces on sides 2 and 4 are zero. Total net force on the loop is Section 5-4: Gauss’s Law for Magnetism and Amp` re’s Law e Problem 5.20 Current I flows along the positive z-direction in the inner conductor of a long coaxial cable and returns through the outer conductor. The inner conductor has radius a, and the inner and outer radii of the outer conductor are b and c, respectively. (a) Determine the magnetic field in each of the following regions: 0 r a, a r b, b r c, and r c. (b) Plot the magnitude of H as a function of r over the range from r 0 to r 10 cm, given that I 10 A, a 2 cm, b 4 cm, and c 5 cm. Solution: (a) Following the solution to Example 5-5, the magnetic field in the region r The total area of the outer conductor is A π c 2 b2 and the fraction of the area of the outer conductor enclosed by a circular contour centered at r 0 in the region b r c is The total current enclosed by a contour of radius r is therefore ¡ ¨ ¨ £ ¨ ¨ ¨ £ Ienclosed I1 r2 c2 b2 b2 I  ¨ ¨ £ ¢ ¢ ¨ ¨ π r2 π c2 b2 b2 r2 c2 b2 b2 c2 c2 r2 b2 £ ¢ ¨  £ £ § ¡ ¡ and in the region a r b, H ˆI φ 2πr ¡ £ H ˆ rI φ 2πa2 ¡ £ £  £   £ £ £ F 2F1 ˆ y4 10 5 N  a,   £  £ £ £ ¢ £ F1 ˆ y ˆ y     µ0 I1 I2 a 2π a 2 4π 10 5 10 2π 1 7 2 ˆ y2 10 5 N £ ¡ ¡ 262 and the resulting magnetic field is CHAPTER 5 For r c, the total enclosed current is zero: the total current flowing on the inner conductor is equal to the total current flowing on the outer conductor, but they are flowing in opposite directions. Therefore, H 0. (b) See Fig. P5.20. 0.8 Magnetic field magnitude H (A/cm) 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Radial distance r (cm) Figure P5.20: Problem 5.20(b). Problem 5.21 A long cylindrical conductor whose axis is coincident with the z-axis ˆ has a radius a and carries a current characterized by a current density J zJ0 r, where J0 is a constant and r is the radial distance from the cylinder’s axis. Obtain an expression for the magnetic field H for (a) 0 r a and (b) r a. Solution: This problem is very similar to Example 5-5. (a) For 0 r1 a, the total current flowing within the contour C1 is φ0r0 ¡ ¡ r0 ¡  £ £ ¤¢  £ £ I1 J ds ˆ zr dr d φ 2π ¦ 2π r1 ˆ zJ0 r r1 J0 dr 2πr1 J0  £  ¨ ¨ £ £ § £ H ˆ Ienclosed φ 2πr ˆI φ 2πr § c2 c2 r2 b2 ¦ 264 CHAPTER 5 a, Amp` re’s law is e (a) For r c S r 0 0 r Hence, Problem 5.23 In a certain conducting region, the magnetic field is given in cylindrical coordinates by Find the current density J. Solution: Section 5-5: Magnetic Potential Problem 5.24 With reference to Fig. 5-10, (a) derive an expression for the vector magnetic potential A at a point P located at a distance r from the wire in the x–y plane, and then (b) derive B from A. Show that your result is identical with the expression given by Eq. (5.29), which was derived by applying the Biot–Savart law.   ¡ £ ¡ ¢  © ¨ ¢ ¨ ©  £ £   £ J H ˆ z ¦ ∇ 4 1∂ r 1 1 3r e 3r r ∂r r 1 ˆ ˆ 36 z 12e 2r 1 2r 12e 2r z 24e r ¡ ¢ © ¨ § £ H ˆ4 φ1 r 1 3r e 3r 3r  ¡ ¡¢ ©  ¨ £ £ H ˆ φH e a ˆ J0 1 φ r a 1 r a ¡ ¢¢ ¡ ©  ¨ £ ¡ ¢ ¢ ©  ¨ £ 2πrH 2πJ0 e r r 1 a 0 2πJ0 1 e a (b) For r a, a 1  ¡ ¢¢ ¡ ©  ¨ £ £ H ˆ φH e r ˆ J0 1 φ r r 1 for r a A/m2  ¢¢ ¡ ©  ¨ £ ¡ ¢ ¢ ©  ¨ 2πJ0 e r r 1 r 0 2πJ0 1 e r 0  £ £ 2πrH 2πJ0 r re dr ¡  £ J ds ˆ zJ0 e ¦ ¦ £ ˆ ˆ φ H φ 2πr ¡ £ £ H dl I J ds ¦ ¦ ¦ r r ˆ z 2πr dr r 1 ...
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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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