Title 21.Magnetism and Matter Med Ans.pdf ✅

1. (a): Atomic currents and intrinsic spins of electrons are responsible for origin of magnetism. 2. (b): As equatorial magnetic field due to bar magnet is 3 0 E 4 r m B   = ...(i) and axial magnetic field due to bar magnet is 3 0 A r 2m 4 B   = ...(ii) From (i) and (ii), we get, A BE B = 2 3. (d): The magnetic field lines do not intersect. 4. (b): magnetic monopole does not exist. 5. (d): Torque,  = mB = mBsin    and magnetic potential energy Um = m.B = mBcos   at o  = 0 the dipole will be in most stable position  = msin = mBsin0 = 0 and, Um = −mBcos = −mBcos0 = −mB 6. (c): Here, n = 900 turns, 2 s 4 2 A = 210 m ,m = 0.6Am − The magnetic moment of solenoid, ms = NIA The current flowing through the solenoid is, 3.33A 900 2 10 0.6 NA m I 4 s =   = = − 7. (c): Since the most stable position is at  = 0 and the most unstable position is at 180 , o  = then the work done is given by  = = =    o 180 0 W ( )d  =   = −  o o 180 0 180 0 mBsin d mB[cos ] mB[cos180 cos0] mB[ 1 1] o = − − = − − − = −mB[−2] = 2mB Here, 2 m = 0.5Am and B = 0.09T  W = 20.500.09 = 0.09J 8. (a): Here, 2 2 m 5.0 10 Am − =  7 2 T 0.65s,I 8.5 10 Kgm − = =  The magnetic field, 2 2 mT 4 I B  = 2 2 2 7 5 10 (0.65) 4 (3.14) 8.5 10      = − − 158.69 10 T 0.0016T 5 =  = − 9. (b): Magnetic field of equatorial point is, 3 0 E 4 r m B   = Here, 10 ,r 1m 4 0 7 = =   − and 2 m = 0.60Am 6.0 10 T (1) 10 0.60 B 8 3 7 E − − =    = 10. (c): Here, m 0.6Am ,r 75cm 0.75m 2 = = = 3 7 3 0 (0.75) 2 0.6 10 2 4  =       = − r m Baxial   2.13 10 0.213 . 7 Baxial =  T = T − 11. (c): Here, 30 ,B 0.35T o  = = 4.5 10 J −2  =  then, o 2 0.35 sin30 4.5 10 Bsin m   =   = − 0.35 2 4.5 10 2 1 0.35 4.5 10 m −2 −2   =   = 1 m 0.26JT − = 12. (d): Here, n = 750 turns, A 5 10 m ,I 3A 4 2 =  = − Then, magnetic moment, 4 4 m nIA 750 3 5 10 11250 10 − − = =    =  1 1.125JT − = 1 1.13JT− = along the axis of solenoid. 13. (d): Here, n = 25 turns, r = 12 cm, B = 0.5T Since the coil is placed in uniform magnetic field normal to the plane of the coil. Hence the angle between magnetic moment and magnetic field direction is zero
(i.e.,  = 0 )  = mBsin = mBsin   = 0 14. (d): On the axis of the magnet, 3 0 d 2m . 4 B   = Here, 0 7 2 10 A m 4 − − =   m 0.48JT ,d 10cm 0.1m 1 = = = − Then, 3 7 (0.1) 10 2 0.48 B   = − 0.96 10 T, −4 =  along S –N direction. 15. (c): Here, n 1000,A 1.4 10 m ,I 3A 4 2 = =  = − 4 m nIA 1000 3 1.4 10− = =    1 0.42JT − = 16. (c): Here, 1 o M − 0.65JT ,B = 0.25T, = 30 −   = MBsin o = 0.650.25sin300.08125Nm 2 1 = 0.65 0.25 =  0.081Nm 17. (a): Here, m 0.39JT ,d 20cm 0.2m 1 = = = − On the equatorial line of magnet, 3 0 d m . 4 B   = 4 3 3 7 3 7 10 8 0.39 2 10 0.39 10 (0.2) 0.39 10 − − − − =    =  = 4 0.049 10− =  T, N –S direction = 0.049 G; N – S direction 18. (b): Option (a), (c) and (d) all are correct about electrostatic shielding and magnetostatic shielding. 19. (d): On the axis of a magnetic dipoles, magnetic induction is, 3 0 r 2m 4 B    = Here, 10 26 2 o r 1A 10 m,m 1.4 10 Am − − = = =  and 7 2 10 NA 4 − − =   10 3 7 26 (10 ) 10 2 1.4 10 B − − −     = 2.8 10 T 2.8mT 10 10 2.8 10 3 30 7 26 =  =   = − − − − 20. (d): According of fleming’s left hand rule, the direction of the magnetic force on the wire is . 21. (b): Here, N 300,I 15A,r 7cm 7 10 m −2 = = = =  2 M = NIA = NIr 2 2 300 15 3.14 (7 10 ) − =     1 69.2JT− = 22. (a): On axial line, , 2 2 2 0 (d l ) 2md 4 B  −  = Given, , 2l =12cm = 0.12m Pole strength , = 20Am,d =10cm = 0.1m Magnetic moment , 2 m = 200.12Am [(0.1) (0.06) ] 2(20) (0.12) 0.1 B 10 2 2 7 −    =  − 1.17 10 T −3 =  23. (b): Potential energy, , U = −M.B = −MBcos   (1) U MBcos180 MB o = − = (2) U MBcos90 0 o = − = (3) U = −MBcos (4) U = +MBcos 24. (a): In solenoid, H field, H = nI and B field B = nI 0r Now the core is heated beyond the curie temperature due to that r decrease and hence B will also decrease. 25. (a): The moment of inertia of the bar magnet of mass m, length l about an axis passing through its center and perpendicular to its length. TC
12 ml I 2 = and magnetic dipole moment is M. when the magnet is cut into two equal pieces perpendicular to length then moment of inertia of each piece of magnet about an axis perpendicular to length passing through its centre 8 I 8 1 12 ml 2 (l/ 2) 2 m I' 2 = =  = and magnetic dipole moment, 2 M M' = Now, time period of oscillation, MB I T = 2 and time period of oscillation on each piece of magnet B 2 M 8 I 2 M'B I' T'= 2 =  2 T MB I 2 2 T' =  = 26. (b): For circular coil, 2 R L C ,n l l  = and magnetic moment, 1 1 1 M = n IA 2 LIR .I. R 2 R L 2  =  = For square coil C2 4a L n2 = and magnetic moment, 2 2 2 M = n IA 4 LIa .I a 4a L 2 =  = Now moment of inertia about an axis through the diameter of circular coil 2 MR I 2 1 = and moment of inertia of square coil about an axis passing through its centere and parallel to edge of square 12 Ma I 2 2 = Then, frequency of oscillation of 1 1 1 1 I M B C . = and frequency of oscillation of 2 2 2 2 I M B C . = As, 2 2 1 2 1 2 2 1 I M I M  =   = a 3R 12 Ma 4 LIa 2 MR 2 LIR 2 2 = =  = 27. (c): Here, , 2 m = 30 ˆ jAm  and j)T. ˆ i 5 ˆ B = (2 +  Since , m B     =  j ˆ j ˆ i 150 ˆ j ˆ j) 60 ˆ i 5 ˆ j (2 ˆ = 30  + =  +  k) 150 0 ˆ = 60(− +  = −60k ˆ Nm [ k ˆ i ˆ j ˆ   = − and j 0 ˆ j ˆ  = ] 28. (d): Here, , N 3000,A 2 10 m ,I6A −4 2 = =  4 1 M NIA 3000 6 2 10 3.6JT − −  = =    = 29. (a): Torque, ,  = MBsin Here, , 1 2 o M =1.28JT ,B = 7.510 T, = 30 − − 2 o 1.28 7.5 10 sin30 −  =   2 1 1.28 7.5 10 2 =    − 0.64 7.5 10 4.8 10 Nm −2 −2 =   =  30. (d): Here, , 1 M 0.355JT − = B 5.0 10 T −2 =   = 2Hz As I MB 4 1 I MB 2 1 2 2   =   = 157.75 1775 10 4 (3.14) 2 0.355 5 10 4 MB I 2 2 2 2 2 2 − −  =     =    = 4 2 1.13 10 Kgm − =  31. (c): Here, , 60 ,B 1.2 10 T 2 1 o −  = =  o 1  = 30 and o o o 2 = 60 −30 = 30 in stable equilibrium, torques due to two fields must be balanced i.e., 1 2  =  1 1 2 2  MB sin = MB sin or 2 1 2 1 sin sin B B   =
o 1 o 2 B sin30 sin30 = B 1.2 10 T −2 =  32. (a): Magnetic induction at point E due to magnet at F (axial point) is 3 0 1 d 2m 4 B   = It acts along EF. Magnetic induction at point E due to magnet at D (equatorial point) is 3 0 2 d m 4 B   = It acts along FE. Resultant magnetic induction at point E is 3 0 1 2 4 d m B B B   = − = 33. (d): The magnetic field lines due to a bar magnet are closed continuous curves directed from N to S outside the magnet and directed from S to N inside the magnet. Hence option (d) is correct. 34. (b): South to North 35. (d): The strength of the earths magnetic field is not constant. it varies from one place to other place on the surface of earth. Its value being of the order of 5 10− T. 36. (a): Here, , b 0.4G 0.4 10 T 4 E − = =  r 6.4 10 m 6 =  As (pw¡fd), 3 0 E 4 r B   =    = / 4 B r m 0 3 E 7 4 6 3 10 0.4 10 (6.4 10 ) − −    = 23 2 =1.0510 Am 37. (a): Here, , HE = 0.25G and E E B H cos =  The magnetic field of earth at the given location is 0.50G 1/ 2 0.25 cos60 H B o E E = = = 38. (a) As Britain is located close to North pole hence dip angle in Britain is greater and its approximate value is o  = 70 39. (a): Here, , 4 HE 0.16G 0.16 10− = =  T, dip angle, o () = 60 Then magnitude of earth’s field. T cos60 0.16 10 cos H B o 4 E E −  =  = 1/ 2 0.16 10 B 4 E −  = 0.32 10 T 0.32G 4 =  = − 40. (d): In stable equilibrium, a compass needle points along magnetic north and experiences no torque. When it is turned through declination , it points along geographic north and experiences torque,  = mBsin 2 1 60 40 10 1.2 10 mB sin 6 3 =    =    = − − or , o  = 30 41. (b): The angle of dip at a pint on the geographical equator is positive, negative or zero depending upon the situation. 42. (c): As, , V BH B = 3 Also, 3 B 3B B B tan H H H V  = = = Or , 1 o  = tan ( 3) = 60 −  Angle of dip, o  = 60 43. (b): Refer to the adjacent figure. the point P is at the intersection of geographical equator (GE) and magnetic equator (ME) then at this point dip = 0 and declination o =11.3