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1 CHAPTER l x B is in the downward direction, /Frol = IIlIlIIBI, and I = VIR = 12 V/ 4 Q = 3 A. Therefore, IBI = IFrol = =410 ( T) IIlIll 3 x 0.1 m. Problem 5.4 The rectangular loop shown in Fig (P5A) consists of 20 closely wrap turns and is hinged along the -axis. The plane of the loop makes an angle of 30 with the y-axis, and the current in the windings is 0.5 A. What is the magnitude of the tc!rque exerted on the loop in the p~sence of a unifonn field B = Y1.2 T? When viewed from above, is the expected direction of rotation clockwise or counterclockwise? 20 turns y x Figure P5A: Hinged rectangular loop of Problem 504. Solution: The magnetic torque on a loop is given by T = m x B (Eq. (5.20», where m =iinia (Eq. (5.19». For this problem, it is given that! = 0.5 A, N = 20 turns, and A = 0.2 m xoa m =0.08 m2 From the figure, n= -xcos 30 +Ysin30. Therefor, m = no.8 (A m2) and T =fio.s(a m2) x Yl.2 T = -O.83 (N m).. As the torqje is negative, the direction of rotation is clockwise, looking from above. Problem 5.s In a cylindrical coordinate system, a 2-m-long straight wire canying a current of 5 A in the positive -direction is located at r = 4 em, cp= Te12, and -1 m < < 1m. (a) IfB = ro.2coscp (T). what is the magnetic force acting on the wire?

2 CHAPTER when -n/2< 4> ~ n/2 and negative over the second half of the circle. Thus, work is provided by the force between cp = nl2 and cp = -n!2 (when rotated in the -+-direction), and work is supplied for the second half of the rotation, resulting in a net work of ero. (c) The force F is maximum when coscp = 1, or cp= O. (f)robjem s..) A 20-turn rectangular coij with side I = 15 cm and w = 5 cm is placed in the y- plane as shown in Fig (P5.6). x Figure P5.6: Rectangular loop of Problem 5.6. (a) If the coil, which carnes a current / = 10 A, is in the presence of a magnetic flux density B = 2 x 10-2 (x+y2) (T), determine the torque acting on the coil. (b) At what angle cp is the torque ero? (c) At what angle cp is the torque maximum? Determine its value. Solution: (a) The magnetic field is in direction (x + y2), which makes an angle CPo= tan-i f = The magnetic moment of the loop is m = fin/a = fi20 x 10x (15 x 5) x 10-4 = fi 1.5 (A m2),

3 208 CHAPTER 5 2 y x/ Figure P5.6: (a) Direction of B. where it is the surface normal in accordance with the right-hand rule. Wben the loop is in the negative-y of the y- plane, II is equal to i, but when the plane of the loop is moved to an angle $, fi becomes or Ii = icos$+ysin$, (b) The torque is ero when T=mxD=n1.5x2x 1O-2(i+y2) = (icos$+ysincp) 1.5 x 2 x 1O-2(i+y2} = i3 x W-2[2coscp-sin<PJ (N m). tancp = 2, 2cos</>- sincp = 0, cp = or Thus, when fi is parallel to B, T = O. (c) The torque is a maximum when fi is perpendicular cp=63.43 ± 90 = or to D, which occurs at Mathematically, we can obtain the same result by taking the derivative of T and equating it to ero to find the values of cpat which ITI is a maximum. Thus, at a.,. acp = acp(3 x 1O--(2cosCP - sm$» = 0

4 CHAP1ER or -2sinlj!+cos<p = 0, which gives tancjl= -~. or cp = or Section 5-2: Biot-Savart Law C Problem si) An 8 cm x 12 em rectangular loop of wire is situated in the x-y 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 25 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 individuai fieids or each of the four wire segments: B = B1 + B2 + B3 +B4. An expression for the magnetic field from a wire segment is given by Eg. (5.29). y x Figure P5.7: Problem 5.7. For au 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 - direction at the origin. With r = 6 em and 1 = 8 em, B} = - I'J/ 21try' 4, = - 41t X 10-7 X 25 x t x 0.06 x v4 X = x 10-5 (T).

5 210 CHAPTER 5 For segment 2, r = 4 em and I = 12 em, Similarly. 82 = - f111 21trN+-]2 = - 41t X 10-7 X 25 x t x 0.04 x v4 x = -loao x 10-5 (T). B3= X 10-5 (T), B4 = X 10-5 (T). The total field is then B = BI + B2+B3 +84 = -ZO.30 (rot). Problem 5.8. Use the approach outlined in Example 5-2 to develop an expression for the magnetic field H at an arbitrary point P due to the linear conductor defined by the geometry shown in Fig (P5.8). If the conductor extends between 1 = 3 m and Z = 7 m and carries a current I = 5 A. find H at P(2, $, 0). 92 P2(Z2)t" J " "," I " "", " " "" "" --- " _:::.~ r-- Per, <P. ) Figure P5.8: Current-carrying linear conductor of Problem 5.8. Solution: The solution follows Example 5-2 up through Eq. (5.27), but the expressions for the cosines of the angles should be generalied to read as

6 CHAFTER 5 21J instead of the expressions in Eq. (5.28), which are specialied to a wire centered at the origin. Plugging these expressions back into Eq. (5.27), the magnetic field is given as H =+ 47tr 1 (-, Jr2+ (Z-Zl)2 - Jr2+(Z-Z2)2 Z-,). For the specific geometry of Fig. P5.8, CProbJem 5.?) The loop shown in Fig (p5.9) consists of radiaj lines and segments of circles whose centers are at point P. Determine the magnetic field H at P in tenus of a, b, e, and J. Figure P5.9: Configuration of Problem 5.9. Solution: From the solution to Example 5-3, if we denote the -axis as passing out of the page through point P, the magnetic field pointing out of the page at P due to the current flowing in the outer arc is Houler = -ZI9/41tb and the field pointing out of the page at P due to the current flowing in the inner arc is HinDer = ZI9/41ta. The other wire segments do not contribute to the magnetic field at P. Therefore, the totaj field flowing directly out of the page at P is H=Houter+Hinner=ZI9 41t (!_!) a b =/9(b-a) --' Problem 5.10 An infinitely long, thin conducting sheet defined over the space o ~ x ~ w and -00 < Y =::; 00 is carrying a current with a uniform surface current

7 CHAPTER 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 12 f.lt when the current is flowing through the cable and 20 p.t when the current is ero, what is the magnitude of J? 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 which gives I = 320 A. B = (20-12),uT = 8f.lT. 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 8 T=8X 10-6= J.lQI = 41t x 10-1/ f-l 21td 21t x 8 ' ~blem s--:iu Two parallel, circular loops carrying a current of 20 A each are arranged as shown in Fig (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 = 2 m. If the two loops have the same radius a = 3 m, determine the magnetic field at: (a) = 0, (b) = 1m. (c) = 2 m. Solution: The magnetic field due to a circular loop is given by (5.34) for a loop in the x-y plane canying a current / in the +~irection. Considering that the bottom loop in Fig. P5.14 is in the x-y plane, but the current direction is along -+, H A It? I = - 2(a )3/2 where is the observation point along the -axis. For the second loop, which is at a height of 2 m, we can use the same expression but should be replaced with (1. - 2). Hence, 1a2 H2 =- 2[a2 + (1._ 2)2]3/2. '

8 216 CHAPTER 5 y x Figure P5.14: Parallel circular loops of Problem The total field is H=HI + H2 = -T ~[a2 [I(02 + 2)3/2 + [a2 + ( _ 1] 2)2]3/2 Afm. (a) At = 0, and with a = 3 m and 1= 20 A, H=-Z-2- A20X9[ (9+4)3/2 1]_ =-5.25AIm. (b) At = 1 m (midway between the loops): H = --2- A 20 x 9 [1(9+ 1)3/2 + (9+ 1] 1)3/2 = - A Af m. (c) At = 2 m, H should be the same as at = O. Thus, H = Afm. Section 5-3: Forces between Currents ~oblem s.ld The long, straight conductor shown in Fig (p5.15) lies in the plane of the rectangular loop at a distance d = 0.1 ffi. The loop has dimensions

9 CHAPTER Ib=05m d=o.lm a = O.2m Figure P5.15: Current loop next to a conducting wire (Problem 5.15). a = 0.2 m and b = 0.5 m. and the currents are II = 10 A and /2 = 15 A. Determine the net magnetic force acting on the loop. Solution: The net magnetic force on the loop is due to the magnetic field surrounding the wire canying current l). The magnetic forces on the loop as a whole due to the current in the loop itself are canceled out by symmetry. Consider the wire carrying II to coincide with the -axis, and the loop to lie in the +x side of the x- plane. Assuming the wire and the loop are surrounded by free space or other nonmagnetic material, Eg. (5.30) gives B =.uoh. 21tr In the plane of the loop, this magnetic field is B=Y~~. Then, from Eq. (5.12), the force on the side of the loop nearest the wire is The force on the side of the loop farthest from the wire is

10 218 CHAPTER 5 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 F = Fml +Fm2,.,/lOh lb,., p.o1ihb =-x---+x--- 21td 21t(a +d),., /loll lab =-x td(a+d) _-x xl - -AU _,.,41tx 10-7 X lox 15xO.2xO.5 _,.,0-4 (N)- => 1 (mn) 21t X 0.1 x 0.3 The force is pulling the loop toward the wire. Problem 5.16 In the ammgement shown in Fig (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 0.3 glcm. Due to the repulsive force acting on the conductors, the angle a between the supporting strings is 10. Determine the magnitude ofl and the relative directions of the currents in the two conductors. x too (a) y (c) " ~1F' d m (b) :'2 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 horiontal 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' = 0.3 g/cm. The

11 ClLA..PTER 5 side 1 is attractive. That is, F = ~JlQltha = A 4n x 10-7 X 5 x 10 x 2 = A 2 x 10-5 N I Y 2n(aj2) y 21t X 1 Y. h and / are in opposite directions for side 3. Hence, the force on side 3 is repulsive, which means it is also along y. That is, F3 = F]. The net forces on sides 2 and 4 are ero. Total net force on the loop is F=2F, =y4x 10-5 N. Section 5-4: Gauss's Law for Magnetism and Ampere's Law CProblem 5.2V Current I flows along the positive -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 a ~ r ~ b, b ~ r ~ c. and r ~ c. of the following regions: 0:5 r :5 a, (b) Plot the magnitude of H as a function of r over the range from r = 0 to r = 10 em, given that I = loa, a = 2 em, b = 4 em, and c = 5 em. Solution: (a) Following the solution to Example 5-5. the magnetic field in the region r < a, and in the region a < r < b, A ri H = If>2m2 ' A I H=If>-. 2nr The total area of the outer conductor is A = n( c2 - 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 n(? _b2) _? _b2 rc( c2 - b2) - c2 - b2. The total current enclosed by a contour of radius r is therefore

12 CHAPTER and the resulting magnetic field is H = _len_cl_osed_ = +_1 (_t?_-_,2_) 21tr 21tr \.c2 - JJ2 / For r > c, the total enclosed current is ero: 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 =O. (b) See Fig. P "0 0.6 ::I: (jj '';: -'c::j 03 I;:: c 1. ~ (,) 4) "0 E~ 0.1 '-' O. G) 0.0 5' Radial distance r (em) Figure P5.20: Problem 5.20(b). ~blem 5~ A long cylindrical conductor whose axis is coincident with the -axis has a radius a and carries a current characteried by a current density J = Vo/ r, where Jo is a constant and r is the radial distance from the cylinder's expression for the magnetic field Hfor (a) 0 5r 5a and (b) r> a. Solution: This problem is very similar to Example 5-5. (a) For () 5 ri =:; a, the total current flowing within the contour CI is h= L I J ds= $:0iT!(Vo) r:0 - r.(rdrdcp)=21t r:0 Jodr=21tr]Jo. if i21[ ir' axis. Obtain an

13 CHAPTER 5 Therefore, since /1 = 2nr, HI, HI = Jo within the wire and HI =~Jo. (b) For r 2: a, the total current flowing within the contour is the total current flowing within the wire: Therefore, since 1 = 21trH2. H2 =loa/r within the wire and H2 =~lo(a/r). Problem 5.22 Repeat Problem 5.21 for a current density J = Voe-r s Figure P5.22: Cylindrical current. Solution:

March 11. Physics 272. Spring Prof. Philip von Doetinchem

March 11. Physics 272. Spring Prof. Philip von Doetinchem Physics 272 March 11 Spring 2014 http://www.phys.hawaii.edu/~philipvd/pvd_14_spring_272_uhm.html Prof. Philip von Doetinchem philipvd@hawaii.edu Phys272 - Spring 14 - von Doetinchem - 32 Summary Magnetic