This preview shows page 1. Sign up to view the full content.
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 412, (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 42), 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 airﬁlled parallelplate capacitor with separation of 1 cm and rectangular plates each 10 cm 2 in area Fig. 433 (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 434a (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. 434a (P4.52(a)) is equivalent to two capacitances in parallel, as illustrated in Fig. 434b (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. P452(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. 435(a) (P4.53(a)), (b) conducting plates are on front and back faces of structure in Fig. 435(a) (P4.53(a)), (c) conducting plates are on top and bottom faces of the cylindrical structure in Fig. 435(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 412: Image Method
Problem 4.56 With reference to Fig. 437 (P4.56), charge Q is located at a distance d above a grounded halfplane located in the x–y plane and at a distance d from another grounded halfplane 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 yaxis is shielded from the region of interest, there might as well be a conducting halfplane 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 xz 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. 438 (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. 439 (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 20turn 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 20turn rectangular coil with side l placed in the y–z plane as shown in Fig. 534 (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 righthand rule. When the loop ˆ ˆ is in the negativey 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 52: 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 xaxis. 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 righthand 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 25A current in the positive xdirection is placed along the xaxis in the vicinity of a 20turn circular loop located in the x–y plane as shown in Fig. 537 (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 6A currents in opposite directions. Determine the magnetic ﬂux density at point P in Fig. 538 (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, EastWest 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 magneticﬁeld 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 righthand 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. 539 (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 zaxis. 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 53: Forces between Currents
Problem 5.15 The long, straight conductor shown in Fig. 540 (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. 541 (P5.16), each of the two long, parallel conductors carries a current I , is supported by 8cmlong 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 53, 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 54: Gauss’s Law for Magnetism and Amp` re’s Law e
Problem 5.20 Current I ﬂows along the positive zdirection 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 55, 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 zaxis 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 55. (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 55: Magnetic Potential
Problem 5.24 With reference to Fig. 510, (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 ...
View
Full
Document
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.
 Spring '08
 Johnson,J

Click to edit the document details