CHAPTER 32. Answer to Checkpoint Questions

Transcription

1 CHAPTER 3 MAGNETISM AND MATTER 865 CHAPTER 3 Answer to Checkpoint Questions 1. (), (b), (c), (a) (zero). (a) ; (b) 1 3. (a) away; (b) away; (c) less 4. (a) towar; (b) towar; (c) less 5. a, c, b, (zero) 6. tie of b, c, an, then a Answer to Questions 1. (a) a, c, f; (b) bar gh. (a) an (c) 3. supplie 4. b 5. (a) all own; (b) 1 up, own, 3 zero 6. (a) increase; (b) increase 7. (a) 1 up, up, 3 own; (b) 1 own, up, 3 zero 8. (a) 1 own, own, 3 up; (b) 1 up, up, 3 zero 9. (a) rightwar; (b) leftwar 10. counterclockwise 11. (a) ecreasing; (b) ecreasing 1. (a) counterclockwise; (b) clockwise; (c) no irection (no inuce B) 13. (a) tie of a an b, then c, ; (b) none (plate lacks circular symmetry, so B is not tangent to a circular loop); (c) none 14. (a) rightwar; (b) leftwar; (c) into 15. 1=4

2 866 CHAPTER 3 MAGNETISM AND MATTER 16. 1, a;, b; 3, c an Solutions to Exercises & Problems 1E (a) (b) The sign of BA for every A on the sie of the paper cyliner is negative. (c) No, because Gauss's law for magnetism applies to an enclose surface only. In fact if we inclue the top an bottom of the cyliner to form an enclose surface S then H s BA = 0 will be vali, as the ux through the open en of the cyliner near the magnet is positive. E Use P 6 n=1 Bn = 0 to obtain B6 = 5X n=1 Bn = ( 1 Wb + Wb 3 Wb + 4 Wb 5 Wb) = +3 Wb : H B A = 0. Write H B A = C, where 3P Use Gauss' law for magnetism: 1 is the magnetic ux through the rst en mentione, is the magnetic ux through the secon en mentione, an C is the magnetic ux through the curve surface. Over the rst en the magnetic el is inwar, so the ux is 1 = 5:0 Wb. Over the secon en the magnetic el is uniform, normal to the surface, an outwar, so the ux is = AB = r B, where A is the area of the en an r is the raius of the cyliner. It value is = (0:10 m) (1: T) = +7: Wb = +7:4 Wb :

3 CHAPTER 3 MAGNETISM AND MATTER 867 Since the three uxes must sum to zero, C = 1 = 5:0 Wb 7:4 Wb = 47:4 Wb : The minus sign inicates that the ux is inwar through the curve surface. 4P* From Gauss' law for magnetism the ux through S 1 is equal to that through S, the portion of the xz plane that lies within the cyliner. Here the normal irection of S is in the positive y-axis. So S 1 B (S 1 ) = B (S ) = = = Z r r Z r r Z r B(x)L x r B left (x)l x 0 i = 0iL ln 3 : 1 r x L x S x 5E The horizontal component of the Earth's magnetic el is given by B h = B cos i, where B is the magnitue of the el an i is the inclination angle. Thus B = B h cos i = 16 T = 55 T : cos 73 6E The ux through Arizona is = B r A = ( T)(95; 000 km )(10 3 m/km) = 1: Wb ; inwar. By Gauss' law this is equal to the negative value of the ux 0 through the rest of the surface of the Earth. So 0 = 1: Wb, outwar. 7E (a) From = ia = ir e we get i = R e = 8:0 10 J/T (6: m) = 6:3 108 A :

4 868 CHAPTER 3 MAGNETISM AND MATTER (b) Yes, because far away from the Earth the els of both the Earth itself an the current loop are ipole els. If these two ipoles cancel out then the net el will be zero. (c) No, because the el of the current loop is not that of a magnetic ipole in the region close to the loop. 8P (a) q s Bh + 0 B v = 4r cos 0 3 m + r sin 3 m q q = 0 cos 4r 3 m + 4 sin m = sin 4r 3 m ; B = where cos m + sin m = 1 was use. (b) tan i = B v B h = ( 0=r 3 ) sin m ( 0 =4r 3 ) cos m = tan m : 9P (a) At the magnetic equator m = 0, so B = 0 4r 3 = ( Tm/A)(8:00 10 Am ) 4(6: m) 3 an i = tan 1 ( tan m ) = tan 1 (0) = 0: (b) At m = 60 : q B = sin 4r 3 m = (3: ) an i = tan 1 ( tan 60 ) = 74. (c) At the north magnetic pole m = 90:0, so p = 3: T sin 60 = 5: T B = 0 4r 3 q sin m = (3: ) p 1 + 3(1:00) = 6: T an i = tan 1 ( tan 90 ) = P (a) At a istance r from the center of the Earth, the magnitue of the magnetic el is given by q B = sin 4r 3 m ;

5 CHAPTER 3 MAGNETISM AND MATTER 869 where is the Earth's ipole moment an m is the magnetic latitue. The ratio of the el magnitues for two ierent istances at the same latitue is B = r3 1 B 1 r 3 : Take B 1 to be at the surface an B to be half of B 1. Set r 1 equal to the raius R e of the Earth an r equal to R e + h, where h is altitue at which B is half its value at the surface. Thus 1 = Re 3 (R e + h) : 3 Take the cube root of both sies an solve for h. You shoul get h = 1=3 1 R e = 1=3 1 (6370 km) = 1660 km : (b) Use the expression for B obtaine in 8P, part (a). For maximum B set sin m = 1. Also, r = 6370 km 900 km = 3470 km. Thus B max = 0 4r 3 q1 + 3 sin m = ( Tm/A)(8:00 10 Am ) 4(3: m) 3 p 1 + 3(1) = 3: T : (c) The angle between the magnetic axis an the rotational axis of the Earth is 11:5, so m = 90:0 11:5 = 78:5 at the Earth's north geographic pole. Also r = R e = 6370 km. Thus q B = sin 4RE 3 m = ( Tm/A)(8:0 10 J/T) p sin 78:5 4(6: m) 3 = 6: T ; an i = tan 1 ( tan 78:5 ) = 84:. A plausible explaination to the iscrepancy between the calculate an measure values of the Earth's magnetic el is that the formulas we obtaine in 8P are base on ipole approximation, which oes not accurately represent the Earth's actual magnetic el istribution on or near its surface. (Incientally, the ipole approximation oes get better when we calculate the Earth's magnetic el far from its center.) 11E Use Eq. 3-7 to obtain U = ( s;z B) = B s;z, where s;z = eh=4m e = B (see Eqs. 3-4 an 3-5). Thus U = B[ B ( B )] = B B = (9: J/T)(0:5 T) = 4: J :

6 870 CHAPTER 3 MAGNETISM AND MATTER 1E Use Eq. 3-11: orb;z = m l B. (a) For m l = 1, orb;z = (1)(9: J/T) = 9: J/T. (b) For m l =, orb;z = ( )(9: J/T) = 1: J/T. 13E (a) E = (b) Use Eq. 30-9: e 4 0 r = (1: C)(8: Nm = C ) = 5: N/C : (5: m) B = 0 p r 3 (c) From Eq. 3-10, = ( Tm/A)(1: J/T) (5: m) 3 = :0 10 T: orb p = eh=4m e p = B p = 9: J/T 1: J/T = 6:6 10 : 14E (a) Since m l = 0, L orb;z = m l h= = 0. (b) Since m l = 0, orb;z = m l B = 0. (c) Since m l = 0, from Eq. 3-1 U = orb;z B ext = m l B B ext = 0. () Regarless of the value of m l, for the spin part U = s;z B = B B = (9: J/T)(35 mt) = 3: 10 5 J : () Now m l = 3 so L orb;z = m lh = ( 3)(6: Js) = 3: Js an orb;z = m l B = ( 3)(9: J/T) = : J/T : The potential energy associate with the electron's orbital magnetic moment is now U = orb;z B ext = (: J/T)( T) = 9: J ; while the potential energy associate with the electron spin, being inepenent of m l, remains the same: 3: 10 5 J.

7 CHAPTER 3 MAGNETISM AND MATTER E For a given value of l, m l varies from l to +l. Thus in our case l = 3, an the number of ierent m l 's is l + 1 = (3) + 1 = 7. (a) Since L orb;z / m l, there are a total of seven ierent values of L orb;z. (b) Similarly, since orb;z / m l, there are also a total of seven ierent values of orb;z. (c) Since L orb;z = m l h=, the greatest allowe value of L orb;z is given by jm l j max h= = 3h=; while the least allowe value is given by jm l j min h= = 0. () Similar to part (c), since orb;z = m l B, the greatest allowe value of orb;z is given by jm l j max B = 3eh=4m e ; while the least allowe value is given by jm l j min B = 0. (e) From Eqs. 3-3 an 3-9 the z component of the net angular momentum of the electron is given by L net;z = L orb;z + L s;z = m lh + m sh : For the maximum value of L net;z let m l = [m l ] max = 3 an m s = 1. Thus [L net;z ] max = h = 3:5h : (f) Since the maximum value of L net;z is given by [m J ] max h= with [m J ] max = 3:5 (see the last part above), the number of allowe values for the z component of L net;z is given by [m J ] max + 1 = (3:5) + 1 = 8. 16P (a) an (b) The potential energy of the atom in association with the presence an external magnetic el B ext is given by Eqs an 3-1: U = orb B ext = orb;z B ext = m l B B ext : For level E 1 there is no change in energy as a result of the introuction of B ext so U / m l = 0, meaning that m l = 0 for this level. For level E the single level splits into a triplet (i.e., three separate ones) in the presence of B ext, meaning that there are three ierent values of m l. The mile one in the triplet is unshifte from the original value of E so its m l must be equal to 0. The other two in the triplet then correspon to m l = 1 an m 1 = +1, respectively. (c) For any pair of ajacent levels in the triplet jm l j = 1. Thus the spacing is given by U = j( which is equivalent to 4: J. m l B B)j = jm l j B B = B B = (5: ev/t)(0:50 T) = : ev ; 17P (a) The perio of rotation is T = =! an in this time all the charge passes any xe point near the ring. The average current is i = q=t = q!= an the magnitue of the magnetic ipole moment is = ia = q! r = 1 q!r :

8 87 CHAPTER 3 MAGNETISM AND MATTER (b) Curl the ngers of your right han in the irection of rotation. Since the charge is positive your thumb points in the irection of the ipole moment. It is the same as the irection of the angular momentum of the ring. 18E (a) S S i B (b) Since the material is iamagnetic its magnetic moment is oppose to the irection of B, so points away from the bar magnet. (c) In orer to prouce a magnetic moment which points away from the bar magnet the current in the ring must be running clockwise when viewe along the symmetry axis from the location of the bar magnet. () Consier the magnetic force F B exerte by the bar magnet on an innitesimal segment L of the ring. Use F B = il B to convince yourself that the net magnetic force on the ring points away from the magnet. 19P An electric el with circular el lines is inuce as the magnetic el is turne on. Suppose the magnetic el increases linearly from zero to B in time t. Accoring to Eq. 3-4 the magnitue of the electric el at the orbit is given by r B r B E = = t ; where r is the raius of the orbit. The inuce electric el is tangent to the orbit an changes the spee of the electron, the change in spee being given by v = at = ee m e t = e m e r B t t = erb m e : The average current associate with the circulating electron is i = ev=r an the ipole moment is = 1 evr : = Ai = r ev r

9 CHAPTER 3 MAGNETISM AND MATTER 873 The change in the ipole moment is = 1 er v = 1 er erb m e = e r B 4m e : 0E Let E = 3 kt = jb ( B)j = B to obtain T = 4B 3k = 4(1: J/T)(0:50 T) 3(1: J/K) = 0:48 K : 1E The magnetization is the ipole moment per unit volume, so the ipole moment is given by = MV, where M is the magnetization an V is the volume of the cyliner. Use V = r L, where r is the raius of the cyliner an L is its length. Thus = Mr L = (5: A/m)(0: m) (5:00 10 m) = :08 10 J/T : E This problem is analogous to 18E. (a) S S i B (b) Since the material is paramagnetic its magnetic moment is in the same irection with B, so points towar the bar magnet. (c) In orer to prouce a magnetic moment which points towar the bar magnet the current in the ring must be running counterclockwise when viewe along the axis from the location of the bar magnet.

10 874 CHAPTER 3 MAGNETISM AND MATTER () Consier the magnetic force F B exerte by the bar magnet on an innitesimal segment L of the ring. Use F B = il B to convince yourself that the net magnetic force on the ring points towar the magnet. 3P For the measurements carrie out the largest ratio of the magnetic el to the temperature is (0:50 T)=(10 K) = 0:050 T/K. Look at Fig to see if this is in the region where the magnetization is a linear function of the ratio. It is quite close to the origin, so we conclue that the magnetization obeys Curie's law. 4P (a) From Fig we see that B=T = 0:50 when M=M max = 50%. So B = 0:50 T = (0:50)(300 K) = 150 K. (b) Similarly, now B=T = :0 so B = (:0)(300) = 600 T. (c) Not yet. 5P (a) Again from Fig. 3-10, for M=M max = 50% we have B=T = 0:50. So T = B=0:50 = =0:50 = 4 K. Now B=T = :0 so T = =:0 = 1 K. 6P (a) A charge e traveling with uniform spee v aroun a circular path of raius r takes time T = r=v to complete one orbit, so the average current is i = e T = ev r : The magnitue of the ipole moment is this times the area of the orbit: = ev r r = evr : Since the magnetic force, with magnitue evb, is centripetal, Newton's law yiels evb = m e v =r, so r = m ev eb : Thus = 1 (ev) me v = eb 1 B 1 m ev = K e B : The magnetic force ev B must point towar the center of the circular path. If the magnetic el is into the page, for example, the electron will travel clockwise aroun the circle. Since the electron is negative, the current is in the opposite irection, counterclockwise an, by the right-han rule for ipole moments, the ipole moment is out of the page. That is, the ipole moment is irecte opposite to the magnetic el vector.

11 CHAPTER 3 MAGNETISM AND MATTER 875 (b) Notice that the charge cancele in the erivation of = K e =B. Thus the relation = K i =B hols for a positive ion. If the magnetic el is into the page, the ion travels counterclockwise aroun a circular orbit an the current is in the same irection. Thus the ipole moment is again out of the page, opposite to the magnetic el. (c) The magnetization is given by M = e n e + i n i, where e is the ipole moment of an electron, n e is the electron concentration, i is the ipole moment of an ion, an n i is the ion concentration. Since n e = n i, we may write n for both concentrations. Substitute e = K e =B an i = K i =B to obtain M = n B (K e + K i ) = 5:3 101 m 3 1: T 6: 10 0 J + 7: J = 310 A/m : 7P (a) M = NP () NP ( ) P () + P ( ) = N eb=kt e B=KT e B=KT + e B=KT B = N tanh kt : (b) For B kt (i.e., B=kT 1) we have e B=kT 1 B=kT, so B N[(1 + B=kT ) (1 B=kT )] M = N tanh = N B : kt (1 + B=kT ) + (1 B=kT ) kt (c) For B kt we have tanh(b=kt ) 1, so B M = N tanh kt N : () The following is a plot of tanh(x) as a function of x for = 1. Notice the resemblence between this plot an Fig By ajusting properly you can t the curve in Fig with a tanh function. 1 tanh(x) 0.5 ) x

12 876 CHAPTER 3 MAGNETISM AND MATTER 8E The Curie temperature for iron is 770 C. If x is the epth at which the temperature has this value, then 10 C + (30 C/km)x = 770 C or x = 770 C 10 C 30 C/km = 5 km : 9E (a) The el of a ipole along its axis is given by Eq. 30{9: B = 0 z 3 ; where is the ipole moment an z is the istance from the ipole. Thus B = ( Tm=A)(1: J/T) ( m) = 3: T : (b) The energy of a magnetic ipole in a magnetic el B is given by U = B = B cos, where is the angle between the ipole moment an the el. The energy require to turn it en-for-en (from = 0 to = 180 ) is U = B = (1: J/T)(3: T) = 9: J = 5: ev : The mean kinetic energy of translation at room temperature is about 0:04 ev. Thus if ipole-ipole interactions were responsible for aligning ipoles, collisions woul easily ranomize the irections of the moments an they woul not remain aligne. 30E The saturation magnetization correspons to complete alignment of all atomic ipoles an is given by M sat = n, where n is the number of atoms per unit volume an is the magnetic ipole moment of an atom. The number of nickel atoms per unit volume is n = =m, where is the ensity of nickel an m is the mass of a single nickel atom, calculate using m = M=N A, where M is the atomic mass of nickel an N A is the Avogaro constant. Thus n = N A M = (8:90 g/cm3 )(6: atoms/mol) 58:71 g/mol = 9:16 10 atoms/cm 3 = 9: atoms/m 3 : The ipole moment of a single atom of nickel is = M sat n = 4: A/m 9: m 3 = 5: Am :

13 CHAPTER 3 MAGNETISM AND MATTER E (a) The number of iron atoms in the iron bar is N = Thus the ipole moment of the iron bar is (7:9 g/cm 3 )(5:0 cm)(1:0 cm ) (55:847 g/mol)=(6: =mol) = 4:3 103 : = (: J/T)(4: ) = 8:9 Am : (b) = B sin 90 = (8:9 Am )(1:57 T) = 13 Nm : 3P (a) If the magnetization of the sphere is saturate the total ipole moment is total = N, where N is the number of iron atoms in the sphere an is the ipole moment of an iron atom. We wish to n the raius of an iron sphere with N iron atoms. The mass of such a sphere is Nm, where m is the mass of an iron atom. It is also given by 4R 3 =3, where is the ensity of iron an R is the raius of the sphere. Thus Nm = 4R 3 =3 an Substitute this into total = N to obtain N = 4R3 3m : total = 4R3 3m : Solve for R an obtain The mass of an iron atom is 1=3 3mtotal R = : 4 m = 56 u = (56 u)(1: kg/u) = 9: kg : So R = " 3(9: kg)(8:0 10 J/T) 4( kg/m 3 )(: J/T) # 1=3 = 1: m : (b) The volume of the sphere is an the volume of the Earth is V s = 4 3 R3 = 4 3 (1:8 105 m) 3 = : m 3 V e = 4 3 (6: m) 3 = 1: m 3 ;

14 878 CHAPTER 3 MAGNETISM AND MATTER so the fraction of the Earth's volume that is occupie by the sphere is : m 3 1: m 3 = : : 33P From the way the wire is woun it is clear that P is the magnetic north pole while P 1 is the south pole. (a) The eection will be towar P 1 (away from the magnetic north pole). (b) As the electromagnet is turne on the magnetic us through the aluminum changes abruptly, causing a strong inuce current which prouces a magnetic el opposite to that of the electromagnet. As a result the aluminum sample will be pushe towar P 1, away from the magnetic north pole of the bar magnet. As reaches a constant value, however, the inuce current isappears an the aluminum sample, being paramagnetic, will move slightly towar P, the magnetic north pole of the electromagnet. (c) A magnetic north pole will now be inuce on the sie of the sample closer to P 1, an a magnetic south pole will appear on the other sie. If the el of the electromagnet is stonger near P 1 then the sample will move towar P 1. 34P The interacting potenial energy between the magnetic ipole of the compass an the Earth's magnetic el is U = B e = B e cos, where is the angle between an B e. For small angle U() = B e cos B e 1 = 1 k B e ; where k = B e. Conservation of energy for the compass then gives 1 I + 1 k = const. : This is to be compare with the following expression for the mechanical energy of a springmass system: 1 x m + 1 kx = const. ; which yiels! = p k=m. So by analogy, in our case which gives! = r r k I = Be I = s B e ml =1 ; = ml! = (0:050 kg)(4:0 10 m) (45 ra/s) 1B e 1( T) = 8:4 10 J/T :

15 CHAPTER 3 MAGNETISM AND MATTER P (a) The magnitue of the toroial el is given by B 0 = 0 ni p, where n is the number of turns per unit length of toroi an i p is the current require to prouce the el (in the absence of the ferromagnetic material). Use the average raius (r = 5:5 cm) to calculate n: Thus n = N r = i p = B 0 0 n = 400 turns (5:5 10 m) = 1: turns/m : 0: T ( Tm/A)(1: m 1 ) = 0:14 A : (b) If is the magnetic ux through the seconary coil then the magnitue of the emf inuce in that coil is E = N(=) an the current in the seconary is i s = E=R, where R is the resistance of the coil. Thus i s = N R : The charge that passes through the seconary when the primary current is turne on is Z Z q = i s = N Z R = N = N R R : The magnetic el through the seconary coil has magnitue B = B 0 + B M = 801B 0, where B M is the el of the magnetic ipoles in the magnetic material. The total el is perpenicular to the plane of the seconary coil, so the magnetic ux is = AB, where A is the area of the Rowlan ring (the el is insie the ring, not in the region between the ring an coil). If r is the raius of the ring's cross section then A = r. Thus = 801r B 0 : 0 The raius r is (6:0 cm 5:0 cm)= = 0:50 cm an = 801(0:50 10 m) (0: T) = 1: Wb : Thus q = 50(1: Wb) 8:0 = 7: C : 36E Let R be the raius of a capacitor plate an r be the istance from axis of the capacitor. For points with r R the magnitue of the magnetic el is given by B = 0 0 r

16 880 CHAPTER 3 MAGNETISM AND MATTER an for r R it is B = 0 0 R r : The maximum magnetic el occurs at points for which r = R an its value is given by either of the formulas above: B max = 0 0 R : There are two values of r for which B = B max =: one less than R an one greater. To n the one that is less than R, solve 0 0 r = 0 0 R 4 for r. The result is r = R= = (55:0 mm)= = 7:5 mm. To n the one that is greater than R, solve 0 0 R r = 0 0 R 4 for r. The result is r = R = (55:0 mm) = 110 mm. 37E Use the result of part (b) in Sample Problem 3-4: to solve for =: B = 0 0 R r (for r R) = = Br 0 0 R (: T )(6: m) ( Tm/A)(8: C =Nm )(3: m) = : V/ms : 38P (a) Use the result of part (a) in Sample Problem 3-4: where r = 0:80R an B = 0 0 r = V = 1 (for r R) ; V 0 e t= = V 0 e t= :

17 CHAPTER 3 MAGNETISM AND MATTER 881 Here V 0 = 100 V. Thus 0 0 r B(t) = V0 e t= = 0 0 V 0 r e t= = ( Tm/A)(8: C =Nm )(100 V)(0:80)(16 mm) ( s)(5:0 mm) = (1: T) e t=1 ms : e t=1 ms The minus sign here is insignicant. (b) At time t = 3, B(t) = (1: T)e 3= = 5: T. 39P Apply Eq. 3-8 to a circular loop of raius R centere at the center of the plates: Thus I Bs = RB = 0 0 E = 0 0 R = 0 0 (R E) = 0 0 R [V m cos(!t + )] = 0 0 R V m! sin(!t + ) : B max = 0 0 R V m! [sin(!t + )] max = 0 0 RV m f R = ( m)(150 V)(60 Hz) (3: m/s) (5: m) = 1: T : = RV mf c (b) For r R B(r) / r, an for r R B(r) / r 1 (see Eqs an 3-39). The plot is shown below. B max (r) /B max (R) 1 1/ 1/ r (mm)

18 88 CHAPTER 3 MAGNETISM AND MATTER 40E The isplacement current is given by i = 0 A ; where A is the area of a plate an E is the magnitue of the electric el between the plates. The el between the plates is uniform, so E = V=, where V is the potential ierence across the plates an is the plate separation. Thus i = 0A V : Now 0 A= is the capacitance C of a parallel-plate capacitor without a ielectric, so i = C V : 41E Let the area plate be A an the plate separation be. Use Eq. 3-3: or i = 0 E V = i 0 A = i C = = 0 (AE) = 0A V = 0A V 1:5 A : F = 7:5 105 V/s : So we nee to change the voltage ierence across the capacitor at the rate of 7:510 5 V/s. ; 4E Consier an area A, normal to a uniform electric el E. The isplacement current ensity is uniform an normal to the area. Its magnitue is given by J = i =A. For this situation so i = 0 A ; J = 1 A 0A = 0 : 43E Use Eq. 3-36: i = 0 A :

19 CHAPTER 3 MAGNETISM AND MATTER 883 Note that here A is the area over which a changing electric el is present. In this case r > R so A = R. Thus = i 0 A = i 0 R = :0 A (8: C =Nm )(0:10 m) = 7: 101 V/ms : 44E Let the area of each circular plate be A an that of the central circular section be a, then A a = R (R=) = 4 : Thus from Eqs an 3-37 the total ischarge current is given by i = i = 4(:0 A) = 8:0 A. 45P From Eq. 3-3 E i = 0 = 0 A = 0A (4: ) = 0 A(6: V/ms) (6: t) = (8: C =Nm )(4:0 10 m )(6: V/ms) = : A : (b) If one raws a counterclockwise circular loop s aroun the plates then accoring to Eq. 3-3 I s B s = 0 i < 0 ; which means that Bs < 0. Thus B must be clockwise. 46P (a) From Sample Problem 3-4 we know that B / r for r R an B / r 1 for r R. So the maximum value of B occurs at r = R, an there are two possible values of r at which the magnetic el is 75% of B max. Denote these two values as r 1 an r, where r 1 < R an r > R. Then 0:75B max =B max = r 1 =R, or r 1 = 0:75R; an 0:75B max =B max = (r =R) 1, or r = R=0:75 = 1:3R. (b) From Eqs an 3-43 B max = 0i R = 0i R = ( Tm/A)(6:0 A) (0:040 m) = 3: T :

20 884 CHAPTER 3 MAGNETISM AND MATTER 47P (a) Use H Bs = 0 I enclose to n B = 0I enclose r = 0(J r ) r = 1 0J r = 1 (1: H/m)(0 A/m )( m) = 6: T : (b) From we get = J 0 = i = J r = 0 E = 0 r 0 A/m 8: F/m = :3 101 V/ms : 48P (a) ji j = 0 E = 0A = (8: F/m)(1:6 m ) 4:5 105 N/C 6: N/C 4: s = 0:71 A: (b) i / = = 0: (c) ji j = 0 A = (8: F/m)(1:6 m ) 4: N/C s s = 1:1 A : 49P (a) At any instant the isplacement current i in the gap between the plates equals the conuction current i in the wires. Thus i = i = :0 A. (b) = 1 0 A 0 E = i 0 A = :0 A (8: F/m)(1:0 m) = : V/m s : (c) i 0 = i area enclose by the path = (:0 A) area of each plate 0:50 m = 0:50 A : 1:0 m

21 CHAPTER 3 MAGNETISM AND MATTER 885 () I Bs = 0 i 0 = (1: H/m)(0:50 A) = 6: Tm : 50P (a) (b) (c) E = J i = 0 E = J = i A = (1: m)(100 A) 5: m = 0:34 V/m : = 0 A = 0A i = 0 i A = (8: F)(1: )(000 A/s) = : A : B( ue to i ) B( ue to i) = 0i =r 0 i=r = i i = : A 100 A = : : 51P (a) At any instant the isplacement current i in the gap between the plates equals the conuction current i in the wires. Thus i max = i max = 7:60 A. (b) Since i = 0 ( E =), (c) Accoring to 40E max = i max 0 = 7: A 8: F/m = 8: Vm/s : i = C V = 0A V : Now the potential ierence across the capacitor is the same in magnitue as the emf of the generator, so V = E m sin!t an V= =!E m cos!t. Thus an This means i = 0A!E m cos!t i max = 0A!E m = 0A!E m = (8: F/m)(0:180 m) (130 ra/s)(0 V) i max 7: A = 3: m ; where A = R was use. :

22 886 CHAPTER 3 MAGNETISM AND MATTER H () Use the Ampere-Maxwell law in the form B s = 0 I, where the path of integration is a circle of raius r between the plates an parallel to them. I is the isplacement current through the area boune by the path of integration. Since the isplacement current ensity is uniform between the plates I = (r =R )i, where i is the total isplacement current between the plates an R is the plate raius. The el lines are circles centere on the axis of the plates, H so B is parallel to s. The el has constant magnitue aroun the circular path, so Bs = rb. Thus an The maximum magnetic el is given by rb = 0 r B = 0i r R : R i B max = 0i max r R = ( Tm/A)(7: A)(0:110 m) (0:180 m) = 5: T :