Magnetics-08-ANSWERSHEET

1. (c) 2. (b) 3. (c) 4. (c) 5. (d) 6. (a) 7. (a) 8. (a) 9. (b) 10. (a) 11. (a) 12. (b) 13. (c) 14. (a) 15. (d) 1. (b) 2. (a) 3. (d) 4. (b), (c), (d) 5. (a) 6. (c) 7. (b) 8. (a) 9. (a), (b) 10. (c) 11. (c), (d) 12. (a) 13. (b) 14. (a) 15. (d) SUBJECTIVE LEVEL - I (C. B. S.E.) 1. 104 T □ 3.1104 T 2. 4 106 3. (a) 1.9 104 T normal to the plane of the paper going out of it → (b) same magnitude of B but opposite in direction to that in (a) 4. 8.1102 N; direction of force given by Fleming’s left-hand rule 5. 2 105 N; attractive force normal to Atowards B 6. 0.96 N m 7. 5104 T  1.6 103 T towards west. 8. (b) In a small region of lengh 2d about the mid-point between the coils.  IR 2 N  R 2 3 / 2  R 2 3 / 2  B  0    d   R 2     d   R 2   2  2    2       IR 2 N  5R 2 3 / 2  4d 3 / 2  4d 3 / 2  □ 0     1    1    2  4   5R   5R    IR 2 N  4 3 / 2  6d 6d  □ 0     1   1   2  5   5R 5R  where in the second and third steps above, terms containing d2 / R 2 and higher powers of d/R are neglected since d  1 . The terms linear in d/R cancle giving a R uniform field B in a small region:  4 3 / 2  IN  IN B      0 □ 0.72 0 R R 9. (a) Cicular trajectory of radius 1.0 mm normal to → (b) helical trajectory of radius 0.5 mm with velocity component 2.3 107 ms1 along → B 10. Deuterium ions or deuterons; the answer is not unique because only the ratio of charge of mass is determined. Other possible answers are He++, Li+++ etc. 11. (a) A horizontal magnetic field of magnitude 0.26 T normal to the conductor in such a direction that Fleming’s left-hand rule gives a magnetic force upwards. (b) 1.176 N. 12. (a) 2.1 N vertically dowarwards (b) 2.1 N vertically dowards (true for any angle between current and direction and B since sin l  remains fixed, equal to 20 cm) (c) 1.68 N vertically dowards. 13. Use →  → →  and → →  Il  (a) 1.8  102 N m along - y direction (b) same as in (a) (c) 1.8  102 N m along -x direction (d) 1.8  102 N m at an angle of 240º with the +x direction (e) zero (f) zero Force is zero in each case. Case (e) corresponds to stable, and case (f) corresponds to unstable equilibrium. 14. (a) zero (b) zero (c) force on each electron is evB  IB /(nA)  5 1025 N . Note Answer (d) denotes only the magnetic force. 15. 7.8  104 N , towards the long conductor if the current in the closer side of the loop is in the same direction as the current in the long conductor; repulsive otherwise. Not that the forces on the two smaller sides of the loop cancel each other. SUBJECTIVE UNSOLVED LEVEL - II 1. F1  1.2ˆiN , F2  1.2ˆjN , F3  (1.2ˆi  1.2ˆj)N , F4  1.2ˆjN . mv0 sin  2. qE   2 tan1  mv  3.  qBR  4. (a) 1.13N (b) 1.25 N 5. (a) mv qB (b)   2 (c) m (  2) mv ,   2 m (  2) qB qB 6. 0Qrni 2m 7. 0.115 T 8. Proton will move in circular path of radius 50 and Neutron will move tangentially be- cause neutron does not have any charge to be in circular path. 9. 4.4 Amp 10.   2 rad SUBJECTIVE UNSOLVED LEVEL - III 2. 3. (a) 2  R (b) 4 R 4 3 0 3 4. (a) qBd (b) d , 3d (c) 2m 2 2 m 6qB (d) the particles stick together and the combined mass moves with constant speed vm along the straight line drawn upward in the plane of figure through the point of collision. 5. x(t)  R cos  qB   R sin   qB t   0 m   0 m      0 is initigl angular position of the particle. 6. 2 ai Bsin  , out of the page 7. 1.63 A 3mv2  qB  mv2m 1 qE 2 8. x(t)  z(t)    R sin  t  , qE     cos qB t  . y(t)   t 2qE 2 m R 1  m     9. (a) 0Ia2 2c(b2  a2 ) (b) 0Ic 2(b2  a2 ) 10. 0I I0 sin  (b  a)  PROBLEMS 1. (a) B  6.54 105 T (b) Force acing on the wire at the centre is zero Force on arc AC = 0 force on segment CD is 8.1106 N (inwards) 2. (i) x   (ii) f  i 2d 3. B min  2mv cos   4.73 103 T q(GS) →  I2  L2  a2  ˆ 4. F   2 ln  a2  k , zero.  