Tài liệu Physics exercises_solution: Chapter 28 doc

28.1: For a charge with velocity , ˆ )sm108.00( 6 jv   the magnetic field produced at a position r away from the particle is . ˆ 4 2 0 r qμ rv B     So for the cases below: a) 4 1 2 0 , ˆ ˆˆ ˆ )m0.500(  rkrvir  . ˆˆ )T101.92( ˆ )m0.50( )sm108.0)(C10(6.0 4 ˆ 4 0 5 2 66 0 2 0 0 kkkkB B π μ r qv π μ      b) .00 ˆˆ ˆ )m0.500(  Brvjr  c) . 4 1 , ˆ ˆˆ ˆ )m0.500( 2 0  rirvkr  . ˆˆ 4 0 0 0 iiB 2 B r qv    d) 0 2 2 2 1 , ˆ ˆˆ ˆ )m0.500( ˆ )m0.500( rr  irvkjr  22 ˆ 2 ˆ 2 2 ˆ 4 000 iBB r qv  ii B 2   28.2:            22 0 total 4 d vq d qv BBB   page.theintoT,1038.4 )m120.0( )sm100.9)(C100.3( )m120.0( )sm105.4)(C100.8( 4 4 2 66 2 66 0                B μ B  28.3: 3 0 4 r q π μ rv B      a) ; ˆ , iriv rv     0,    B0rv   b) ; ˆ , ˆ jriv rv    m0.500, ˆ  rvrkrv   kB ˆ )T01.31(sonegative,is T101.31 )m0.500( )sm106.80)(C104.80)(CsN101 4 6 6 2 56227 2 0                q r vq π μ B c) ); ˆ ˆ )(m0.500(, ˆ jiriv    v m0.7071, ˆ )m0.500(  rvkrv     kB ˆ )T104.62(;T104.62 )m0.7071( )sm106.80)(m0.500(C)104.80)(CsN101 4 77 3 56227 3 0               B rrvq π μ B d) ; ˆ , ˆ kriv rv    m0.500, ˆ  rvrjrv   jB ˆ T)101.31(;T101.31 m)0.500( )sm106.80)(C104.80)(CsN101 4 66 2 56227 2 0              B r vq π μ B 28.4: a) Following Example 28.1 we can find the magnetic force between the charges: down).points charge lowertheonforcetheanduppointschargeuppertheonforce(theN101.69 )m0.240( m 109.00)(sm104.50)(C103.00)(C108.00( )AmT10( 4 3 2 6666 7 2 0         r vvqq π μ F B The Coulomb force between the charges is N3.75)CmN108.99( 2 212 m)(0.240 C100)(8.00)(3.0 229    2 21 r qq kF (the force on the upper charge points up and the force on the lower charge points down). The ratio of the Coulomb force to the magnetic force is 21 2 vv c    3 N101.69 N3.75 102.22 3 . b) The magnetic forces are reversed when the direction of only one velocity is reversed but the magnitude of the force is unchanged. 28.5: The magnetic field is into the page at the origin, and the magnitude is page.theintoT,101.64 m)0.400( )sm108.0)(C101.5( )m0.300( )sm102.0)(C104.0( 4 4 6 2 56 2 56 0 22 0                           B π B r vq r qv π μ BBB  28.6: a) 2 0 4 ; πd qvμ Bqq q   into the page; 2 0 4 πd vqμ B q    out of the page. (i) )2(42 2 0 dπ qvμ B v v   into the page. (ii) .0     Bvv (iii) 2 0 4 2 d qv Bvv     out of the page. b) 2 2 0 )2(4 dπ vvqμ q q     BvF  and is attractive. c) 25 0000 2 0 2 2 2 0 )sm103.00( )2(4 , )2(4      εμvvεμ F F d πε q F d π vvqμ F C B CB .101.00 6  28.7: a) . ˆ )0.500( ˆ )0.866( ˆ )150(sin ˆ )150(cos ˆ sin ˆ cos ˆ jijijir  θθ b) kkjiirl ˆ )m105.00( ˆ )0.500() ˆ )0.500( ˆ )0.866(() ˆ ( ˆ 3  dldld  c) kk rl B ˆ )m1.20( )m0.500)(m0.010)(A125( 4 ˆ m)0.500( 4 ˆ 4 2 0 2 0 2 0 π μ r dlI π μ r dI π μ d     . ˆ )T104.3( 8 kB   d 28.8: The magnetic field at the given points is: 2 0 2 0 2 0 6 2 0 2 0 6 2 0 2 0 6 2 0 2 0 sin 4 .0 )0(sin 4 sin 4 T.102.00 )m0.100( )m0.000100(A)200( 4 sin 4 T.100.705 )m0.100(2 45sin)m0.000100(A)(200 4 sin 4 T.102.00 )m0.100( )m0.000100(A)200( 4 sin 4 r θdlI π μ dB r dlI π μ r θdlI π μ dB π μ r θdlI π μ dB π μ r θdlI π μ dB π μ r θdlI π μ dB e d c b a             T.10545.0 3 2 )m100.0(3 )m00100.0(A)200( 4 6 2 0    e e dB dB   28.9: The wire carries current in the z-direction. The magnetic field of a small piece of wire 2 0 ˆ 4 r dI π μ d rl B    at different locations is therefore: a) jrlir ˆ ˆ ˆ ˆ )m00.2(   . ˆ 1000.5 m)00.2( 90sinm)105()A00.4( 4 ˆ sin 4 11 2 4 0 2 0 jjB T π μ r θdlI π μ d      b) . ˆ ˆ ˆ ˆ )m00.2( irljr   . ˆ T1000.5 ˆ )m00.2( )90(sinm)105()A00.4( 4 ˆ sin 4 11 2 4 0 2 0 i iiB        π μ r θdlI π μ d c) ) ˆˆ ( 2 1 ˆ ˆˆ )m00.2( ˆ )m00.2( ijrljir   ) ˆˆ T(1077.1 ) ˆˆ ( 2 1 m)2.00(m)00.2( m)10(5.0A)00.4( 4 ) ˆˆ ( 2 1sin 4 11 22 4 0 2 0 ij ijijB        π μ r θdlI π μ d d) 0 ˆ ˆ ˆ )m00.2(  rlkr  28.10: a) At , 3 4 3 8 223 1 2 1 2 : 2 000 πd Iμ dπ Iμ ddπ Iμ B d x                  in the j ˆ direction. b) The position 2 d x  is symmetrical with that of part (a), so the magnetic field there is d I B   3 4 0  , in the j ˆ direction. 28.11: a) At the point exactly midway between the wires, the two magnetic fields are in opposite directions and cancel. b) At a distance a above the top wire, the magnetic fields are in the same direction and add up: kkkkkB ˆ 3 2 ˆ )3(2 ˆ 2 ˆ 2 ˆ 2 000 2 0 1 0 a I a I a I r I r I            . c) At the same distance as part (b), but below the lower wire, yields the same magnitude magnetic field but in the opposite direction: kB ˆ 3 2 0 πa Iμ  . 28.12: The total magnetic field is the vector sum of the constant magnetic field and the wire’s magnetic field. So: a) At (0, 0, 1 m): . ˆ )T100.1( ˆ )m00.1(2 )A00.8( ˆ )T1050.1( ˆ 2 7 0 6 0 0 iiiiBB       r I b) At (1 m, 0, 0):     46.8atT,102.19 ˆ T)10(1.6 ˆ T)1050.1( ˆ )m00.1(2 )A00.8( ˆ )T1050.1( ˆ 2 666 0 6 0 0 θ π μ πr Iμ kiB kikBB from x to z. c) At (0, 0, – 0.25 m): iiiBB ˆ )m25.0(2 )A00.8( ˆ )T1050.1( ˆ 2 0 6 0 0 π μ πr I    . ˆ T)109.7( 6 i   28.13: . )( 2 4)(4)(4 2122 0 21222 0 2322 0 axx a π Iμ yxx y π Ixμ yx xdy π Iμ B a a a a          28.14: a) A.110 T)10(5.50m)040.0(22 2 0 4 0 00 0     μ π μ πrB I πr Iμ B b) T,1075.2 2 m)0.080(so, 2 4 00   B rB πr Iμ B T.10375.1 4 m)160.0( 4 0   B rB 28.15: a) ,T1090.2 m)(5.502 A)800( 2 5 00   π μ πr Iμ B to the east. b) Since the magnitude of the earth’s magnetic filed is 5 1000.5   T, to the north, the total magnetic field is now o 30 east of north with a magnitude of 5 1078.5   T. This could be a problem! 28.16: a) B = 0 since the fields are in opposite directions. b)          baba ba rrπ I πr Iμ πr Iμ BBB 11 222 000  T6.67T1067.6 m0.2 1 m0.3 1 2 )A(4.0)ATm014( 6 7                 π c) Note that aa rB and bb rB θB θBθBB a ba cos2 coscos    tan 22 m)(0.05m)20.0(:04.14 20 5  a rθθ  04.14cos )m(0.05m)02.0(2 )A(4.0)ATm104( 2 cos 2 2 22 7 0          θ r I B a ,T53.7T1053.7 6    to the left. 28.17: The only place where the magnetic fields of the two wires are in opposite directions is between the wires, in the plane of the wires. Consider a point a distance x from the wire carrying 2 I = 75.0 A. tot B will be zero where 21 BB  . A0.75A,0.25;)m400.0( 2)m400.0(2 2112 2010    IIxIxI πx Iμ xπ Iμ x = 0.300 m; 0 tot  B along a line 0.300 m from the wire carrying 75.0 A amd 0.100 m from the wire carrying current 25.0 A. b) Let the wire with 0.25 1 I A be 0.400 m above the wire with 2 I = 75.0 A. The magnetic fields of the two wires are in opposite directions in the plane of the wires and at points above both wires or below both wires. But to have 21 BB  must be closer to wire #1 since 1 I < 2 I , so can have 0 tot B only at points above both wires. Consider a point a distance x from the wire carrying 0.25 1 I A. tot B will be zero where . 21 BB  m200.0);m400.0( )m400.0(22 12 2010    xxIxI x π Iμ πx Iμ 0 tot B along a line 0.200 m from the wire carrying 25.0 A and 0.600 m from the wire carrying current 0.75 2 I A. 28.18: (a) and (b) B = 0 since the magnetic fields due to currents at opposite corners of the square cancel. (c) left. the toT,100.4 45cos m)210.0(2 A)(100)ATm104( 4 m20.10cm210cm)(10cm)(10 45cos 2 445cos4 45cos45cos45cos45cos 4 7 22 0                         π B r πr Iμ B BBBBB a dcba 28.19: 321 ,, BBB     ⊙ m200.0; 2 0  r r I B   for each wire T1000.2,T1080.0T,1000.1 5 3 5 2 5 1   BBB Let ⊙ be the positive z-direction. A0.20A,0.8A,0.10 321  III T100.2)( 0 T1000.2,T1080.0T,1000.1 6 3214 432z1 5 z3 5 z2 5 z1      zzzz zzz BBBB BBBB BBB To give 4 B in the  direction the current in wire 4 must be toward the bottom of the page. A0.2 )AmT10(2 T)100.2(m)200.0( )2( so 2 7 6 0 4 4 0 4       πμ rB I πr Iμ B 28.20: On the top wire: , 42 11 2 2 0 2 0 d I dd I L F             upward. On the middle wire, the magnetic fields cancel so the force is zero. On the bottom wire: , 42 11 2 2 0 2 0 d I dd I L F             downward. 28.21: We need the magnetic and gravitational forces to cancel: g I h h LI Lg λ22 λ 2 0 2 0      [...]... (16)(4.00 A) ˆ j 2π ˆ  3.72  10 6 Ti  1 .28  105 Tˆ j    ˆ j c) F  I l  B  I lB ˆ  I lB i    1  1 x 1 y ˆ  1.00 A 0.010 m [(3.72  106 T) ˆ  (1 .28  10 5 T) i ] j ˆ  3.72  108 Tˆ  1 .28  10 7 Ti ; F  1.33  10 7 N, 16.2 counterclockwise j from +x-axis 28. 55: a) If the magnetic field at point P is zero, then from Figure (28. 46) the current I 2 must be out of the page,... currents is B  80 B  80 (0.0267 T)  0.0263 T 28. 40: a) B  28. 41: a) If K m  1400  B  2πr B 2π (0.0290 m) (0.350 T) K m μ0 NI I    2πr K m μ0 N μ0 (1400)(500) 0.0725 A b) If K m  5200  I  1400 I part(a)  0.0195 A 5200 2πrB 2π (0.2500 m) (1.940 T) K m μ0 NI  Km    2021 2πr μ0 NI μ0 (500) ( 2.400 A) b) X m  K m  1  2020 28. 42: a) B  28. 43: a) The magnetic field from the solenoid... center, Bc   0 NI 2a At a distance x from the center, 28. 27: a) Bx  μ0 NI 2a , so I  Bx   a3  0 NIa 2   NI    0  2 2 2 32  ( x  a 2 )3 2   Bc  2( x  a )  2a    Bx  1 Bc means 2   a3  2  ( x  a 2 )3 2     a3 1  2 2 32 (x  a ) 2 ( x 2  a 2 )3  4a 6 , with a  0.024 m, so x  0.0184 m  NI  0 (600) (0.500 A) 28. 28: a) From Eq (29-17), Bcenter  0   9.42  10 3 T...  4 T 2 2 32 2 2 32 2( x  a ) 2((0.080 m)  (0.020 m) ) 28. 29: B( x)  μ0 NIa 2 2 B( x) ( x 2  a 2 )3 2 N 2( x 2  a 2 )3 2 μ0 Ia 2  2(6.39  104 T) (0.06 m) 2  (0.06 m) 2 μ0 (2.50 A) (0.06 m) 2 28. 30:    B  dl  μ I 0 encl  32  69  3.83  10 4 T  m  I encl  305 A   b)  3.83  10 4 since dl points opposite to B everywhere 28. 31: We will travel around the loops in the counterclockwise... ( I1  I 2 ) B μ0 ( I 1  I 2 ) 2πr 28. 34: Using the formula for the magnetic field of a solenoid: μ NI μ0 (600) (8.00 A) B  μ0 nI  0   0.0402 T (0.150 m) L μ0 NI BL (0.0270 T) (0.400 m) N   716 turns L μ0 I 0 (12.0 A) N 716 turns n   1790 turns m 0.400 m L b) The length of wire required is 2 rN  2 ( 0 0140 m ) (116 )  63 m 28. 35: a) B  28. 36: B  0 I N L BL 0 N (0.150 T) (1.40... ( N L) I , so I  BL μ0 N  237 A a) B  28. 37: 28. 38: Outside a toroidal solenoid there is no magnetic field and inside it the magnetic μ NI field is given by B  0 2πr a) r = 0.12 m, which is outside the toroid, so B = 0  NI  0 (250) (8.50 A) b) r = 0.16 m  B  0   2.66  10 3 T 2r 2 (0.160 m) c) r = 0.20 m, which is outside the toroid, so B = 0 28. 39: B   0 NI  0 (600) (0.650 A)  ... opposite directions b) Doubling the currents makes the force increase by a factor of four to F  2.40  10 5 N 28. 22: a) F  2 (0.0250 m) F  0 I1 I 2 F 2r   I2   (4.0  10 5 N m)  8.33 A L 2r L  0 I1  0 (0.60 A) b) The two wires repel so the currents are in opposite directions 28. 23: 28. 24: There is no magnetic field at the center of the loop from the straight sections The magnetic field from... (5200)(1.13  10 3 T)  5.88 T b)   2  J   N  m  C  m  2 28. 44:       Am  T   N  s C  m   s  28. 45: The material does obey Curie’s Law because we have a straight line for temperature against one over the magnetic susceptibility The Curie constant from the graph is 1 1 C   1.55  10 5 K  A T  m  0 (slope)  0 (5.13) 28. 46: The magnetic field of charge q  at the location of...  0   0 , 2 2  2R  4R into the page 28. 25: As in Exercise 28. 24, there is no contribution from the straight wires, and now we have two oppositely oriented contributions from the two semicircles: 1  B  ( B1  B2 )   0  I 1  I 2 , 2  2R  into the page Note that if the two currents are equal, the magnetic field goes to zero at the center of the loop 28. 26: a) The field still points along... I encl  I   B  dl  B 2r   0 I encl   0 I  B  0 2r 28. 69: a) I   JdA   αrrdrdθ  α 2π R r 2 dr    A   r2   r2  28. 70: a )r  a  I encl  I  r   I  2    B  dl  B 2πr   0 I encl  μ0 I  2  a  a  A       a μ Ir B 0 2 2πa μI When r  a, B  0 which is just what was found from Exercise 28. 32, part (a) 2πa A   r 2  b2  b) b  r  c  I encl  I  .     ckwisecounterclo2.16,N1033.1; ˆ T1 028. 1 ˆ T1072.3 ] ˆ T)1 028. 1( ˆ T)1072.3[(m010.0A00.1 778 56     Fij ij from +x-axis. 28. 55: a) If the magnetic. . ˆ T)109.7( 6 i   28. 13: . )( 2 4)(4)(4 2122 0 21222 0 2322 0 axx a π Iμ yxx y π Ixμ yx xdy π Iμ B a a a a          28. 14: a) A.110 T)10(5.50m)040.0(22 2 0 4 0 00 0     μ π μ πrB I πr Iμ B