Tiêu đề 20. Moving Charges and Magnetism Medium Ans.pdf ✅

Hence 2 2 1 2 1 R R M M         = 19. (b) m 2E mv v 2 1 E 2 =  = Be 2mE m 2E Be m Be mv r = = = 3.58 10 m 1.6 10 0.4 2 1800 1.6 10 9.1 10 r 4 19 19 31 − − − − =         = 20. (b) As r mv qvB 2 = 2 3 V V r r r v or qB mv r B A B A  =   = = 21. (c) E  0,B= 0 22. (d) As frequency of revolution in a cyclotron 2 m Bq C   = is independent of r. So the radius of path in the dees will remain unchanged When the frequency is changed. 23. (b) The orbit of the electron spirals down inside the cube in a plane parallel to x-y plane. therefore B must be along +z direction and E should be along x-axis i.e. iwhereE 0 E E ˆ Kand B B ˆ = 0 = 0 0  24. (b) 7 1 v 3.2 10 ms− =  B 5 10 T −4 =  The frequency of electron is 31 19 4 2 3.14 9.1 10 1.6 10 5 10 2 m qB − − −       =   = 1.4 10 Hz 14MHz 7  =  = 25. (a)In a cyclotron the centripetal force is balanced by magnetic force then v m qBr r mv qvB 2 =  = Where c v = 2r 2 m qB 2 r m qBr c c  =    = 26. (d) The charged particles and ions in cyclotron can move along circular path only. 27. (d) Here, 11MHz 11 10 Hz 6 c = =  B 1T,R 55cm 55 10 m, −2 = = =  e 1.6 10 C;m 1.67 10 kg 27 p −19 − =  =  27 2 2 2 19 2 2 2 2 1.67 10 (1.6 10 ) (1) (55 10 ) 2m q B R K.E. − − −        = = 23.19 10 J −13 =  14.49 10 eV 14.49MeV 1.6 10 23.19 10 6 19 13 =  =   = − − 28. (b) Here, B 5.4 10 T −4 =  r 32cm 32 10 m −2 = =  2.5MHz 2.5 10 Hz 6  = =  The speed of electron on circular path v = r22 6 = 3210 23.142.510 − 4 6 1 502.4 10 5.024 10 ms− =  =  29. (d) In the cyclotron charge particle is accelerated while point from one dee to another dee due to electric field at all time 30. (a) The time period of revolution of a charge particle in a magnetic filed is Bq 2 m T  = For proton, mp = m,qp = q Bq 2 m Tp   = Now, for −particle m = 4m,q = 2q 2 1 T T Bq 2 m 2 B(2q) 2 (4m) T p  =          =   =   31. (c) Bq 2me r = q m r  Hence,    = q m : q m : q m r :r :r d d p p p d 1: 2 :1 2e 4m : e 2m : e m = = 32. (d) 2m q B r E 2 2 2 = 1.22 10 J 2 1.67 10 (1.6 10 ) (0.7) (1.8) 11 27 19 2 2 2 − − − =       = = 76 MeV 33. (c) In arrangements (a), (b) and (d), the field at the centre due to the currents flowing in the square frames is zero the field
here in option (c) due to the current flowing in and out of the frame is non zero. 34. (a) The resultant magnetic field induction at O due to current through all the five wires will be zero. 35. (c) 2 0 r Idlsin 4 dB    = Here, dl = x = 0.05mI =10A,r =1m sin = sin90o =1 36. (d) As B = Bxy + Byz + Bzx ........ (i) Where j ˆ 4 R I ˆ i,B 4 R I k ˆ ,B 4 R I B 0 zx 0 yz 0 xy     =    =   = Substituting these values in equation (i) we get, k) ˆ j ˆ i ˆ ( 4 R I B 0  + +   = Here ,I 4A 2 =   = and R 5cm 5 10 m −2 = =  k) ˆ j ˆ i ˆ ( 5 10 2 4 4 a B 2 0 + +        = − =100 ( ˆ i + ˆ j+ k ˆ )T 37. (c) Here, N = 150 R 12cm 12 10 m −2 = =  I = 2A 1.57 10 T 12 10 2 10 150 2 2R NI B 3 2 7 0 − − − =      =   = 1.57 10 T 15.7 10 T 15.7G 3 4 =  =  = − − 38. (a) Here, N = 90 R 15cm 15 10 m,B 4 10 T −2 −4 = =  =  2R NI B 0  = 1.06A 4 [ 10 90 2 15 10 4 10 N 2RB I 7 2 4 0 =        =   = − − 39. (a) 2 2 3/ 2 2 0 2(R x ) IR B +  = Here, I 12.5A,R 3cm 3 10 m −2 = = =  X= 4 cm = 4 10 m −2  5.65 10 T 2[(3 10 ) (4 10 ) ] 4 10 12.5 (3 10 ) B 5 2 2 2 2 3/ 2 7 2 2 − − − − − =   +       = 40. (c) Here N = 100 R = 9 cm = 9 10 m −2  and I = 0.4A Now, 2 7 0 9 10 2 10 100 0.4 2R NI B − −     =  = 3 10 9 2 3.14 0.4 −    = 0.279 10 T 2.79 10 T −3 −4 =  =  41. (d) Magnetic field at the centre M due to current through the curved portion DA is           = 2 3 4 R I B 0 1 into the plane of the paper 8R 3 I 0 = into the plane of the paper Magnetic field at the centre M due to current through the straight portion AB is B2 = 0 (  point M lies on the axis of the straight line portion AB) Magnetic field at centre M due to current through the curbed portion BC is 4 (2R) 2 I B 0 3     = into the plane of the paper 16R I 0 = into the plane of the paper Magnetic field at the centre M due to current through the straight portion CD is B4 = 0 (  point M lies on the axis of the straight portion CD) The resultant magnetic field at M is B= B1 +B2 +B3 +B4 16R I 8R 3 I 0 16R I 0 8R 3 I 0 0 0 0 +  + =  + +  = 16R 7 I 0 = into the plane of the paper 42. (b) The value of  B.dl is independent of sense of C. 43. (b) The magnetic field from the centre of wire of radius R is given by r (r R B r) 2R I B 2 0           = And ) r 1 (r R B 2 r I B 0      = From the above descriptions we can say that the graph (b) is a correct representation. 44. (b) 0 enc B.dl =  I  45. (a) I= 40 A
r 15cm 15 10 m −2 = =  46. (c ) 10 5.34 10 T 15 80 15 10 2 10 40 2 r I B 5 5 2 7 0 − − − − =  =     =    = (c) From Ampere circuital law 0 enc B.dl =  I  0 enc B2r =  I 5 10 T 3 75 2 10 2 r I B 0 enc −7 −6 =   =    = The direction of field at the given point will be vertical up determined by the screw rule or right hand rule. 47. (a) Here,I = 2.5A,l = 50cm= 0.50m and 1 200m 0.50 100 n − = = 3.14 10 T 2 4 10 200 2.5 2 nI B 4 7 0 − − =     =   = 48. (c) Here 1000turns/m,I A 0.6 600 n = = =  l = 0.6m,r = 0.02m 30i.e.1 r r 1  =  Hence we can use long solenoid formula then 7 3 4 B 0nI 4 10 10 4 50.24 10 − −  =  =    =  5.024 10 T −3 =  49. (c) The magnetic field inside the core of toroid is constant. 50. (a) The number of turns per unit length for the given toroid av 2 r N n  = The average radius of toroid 28.5cm 28.5 10 m 2 28 29 r 2 av − = =  + = 2067.27 2067 2 3.14 28.5 10 3700 n 2 =      = − Now, B nI 4 10 2067 10 7 = 0 =    − 259615.2 10 T 2.60 10 T −7 −2 =  =  51. (c) For six layers of windings the total number of turns = 6450= 2700 Now number of turns per unit length 3000 90 10 2700 l N n 2 =  = = − Then the field inside the solenoid near the centre B nI 4 10 3000 6 72 10 T 7 4 0 − − =  =    =  = 72G 52. (a) Here I I 30A, l 1m, 1 = 2 = = m 3g 3 10 kga −3 = =  In equilibrium position h 2I I l 4 mg 0 1 2   = 3 10 10 2 30 30 1 10 mg 2I I l 4 h 3 0 2 2 7       =    = − − 0.6cm 6 10 m −3 = =  53. (b) Parallel currents attract and antiparallel currents repel. 54. (c) Given I1 = 2A,I2 = 5A,r = 2m 0 1 2 7 6 1 1 10 Nm 2 2 2 5 10 r 2I I 4 f − − − =    =     = 55. (a) As ,HereN 100, I 3.2A, 2R NI B 0 = =  = R 10cm 10 10 m −2 = =  2.01 10 T 2 0.1 4 10 100 3.2 B 3 7 − − =       = 56. (d) Magnetic moment M = NIA = 2 2 NIr i.e. M  r 57. (b) The magnetic moment is given by 2 m = NIA = NIr 2 2 200 4 3.14 (15 10 ) − =     4 2 = 20043.14151510 = 56.5Am − 58. (a) Shape of loop 59. (b) N= 70 r 5cm 5 10 m,I 8A 2 = =  = − B=1.5T, = 30o The counter torque to prevent the coil from turning will be equal and opposite to the torque acting on the coil NIABsin NI r Bsin30o 2  =  =  3.297Nm 2 1 70 8 3.14 (5 10 ) 1.5 2 2 =       = − = 3.3Nm 60. (a) The torque acting on the coil  = mB = mBsin here the circular coil is placed normal to the direction of magnetic field then the angle between the direction of magnetic moment (m) and magnetic field (B) is zero, then  = mBsin = mBsin0 = 0  = 0