Title 22. Electromagnetic Induction Hard Ans.pdf ✅

1. (a) By using 2 2 0 N r L  = ; 25 10 H 25mH 2 (3.14) 4 (3.14) 10 (500) 5 10 L 3 7 2 2          = − − − 2. (d) By using l N N A M  0 1 2 = and dt di | e | = M ; M H 3 3.01 10 − =   0.25 {2 ( 2)} 3.01 10 3 − − =   − e ; e = 48 mV . 3. (b) We know for air cored solenoid l N A L 2  0 = In case of soft of iron core it’s self inductance l N A L r 2 0 '   = ; L' =  r L . So here L = 900 × 0.18 = 162 mH Note : The self-inductance of a solenoid may be increased by inserting a soft iron core. The function of the core is to improve the flux linkage between the turns of the coil. 4. (b) At t = 1 sec, i = 2 + 3  1 = 5A and dt di | e | = L  9  10–6 (2 3t) dt d = L  +  L = 3  10–3 H So energy 2 3 2 (3 10 ) (5) 2 1 2 1 = =   − U Li = 37.5 mJ. 5. (c) Flux linkage = N2 2 = Mi1 = 24  2 = 48 mwb 6. (b)  2 L  N  2 2 2 2 1 2 1 500 108 600        =         = N L N L L ; L2 = 75mH 7. (a) volt dt dI e L 1 0.01 1.0 10 10 3 = − = −  = − −  | e | = 1volt 8. (a) [3 2 ] 2 t t dt d dt dI e = −L = − + [6 2] 10 10 [6 2] 3 = − + = −  + − L t t ( ) 10 10 (6 2 2) 3 at 2 = −   + − t= e 10 10 (14) 0.14 volt 3 = −  = − − ; | e | = 0.14 volt 9. (b) U = 1KWH = 3.6  106 J. By using 2 2 1 U = Li  3.6  106 2 (200 ) 2 1 =  L   L = 180 H 10. (d) 2 L  N 11. (a) By using | | ; 1 2 dt di e = M dt di R M R e i 1 2 2 2 2 = =  dt di1 5 0.5 0.4 =  ; A sec dt di 4 / 1 = 12. (b) By using | | ; 1 dt di e = L L L 0.2H 0.05 (2 0) 8  = − =  13. (a)  L N r 2  ; 2 1 2 2 1 2 1 r r N N L L          =  2 1 2 / 2 1 2 2  =             = r r L L ; L2 = 2L 14. (c) When contact is broken induced current flows in the same direction of main current. So bulb suddenly glows more brightly. 15. (c) U Li 25 J 2 10 2 2 1 2 1 2 2  =      = =  16. (b) By using         = − − L Rt i i 1 e 0 ; At t = , i = i0 and at t = 1 sec         = −  − 5 10 1 i i0 1 e ;         − = − = − 2 2 0 2 0 1 (1 ) e e i i e i ; 1 2 2 0 − = e e i i 17. (b) e   when  doubles, ‘e’ gets doubled. 18. (c)
e e sint = 0 = e0 sin  = 60 sin30o = 30 volts 19. (a) Zero, because transformer works on ac only. 20. (d) We know that, the transformation of current or voltage from primary to secondary or vice-versa in an ideal transformer takes place according to transformation ratio. Since, it is a step-down transformer, the turns in secondary are smaller in number. Hence current in secondary must be larger. Therefore the secndary current must be 0.1 1 times the primary current. Hence I s = 10 10 mA = 100 mA = 0.1 amp 21. (a) V 220 V,V 110 V,V I 550 W, p = s = s s = Now p p s s V I = V I or A V V I I p s s p 2.5 220 550 = = = 22. (d) V 200 V, p = Vs = 6V  out s s P = V i  30 = 6  is  is = 5A From s p p s i i V V =  200 5 6 p i =  ip = 0.15 A 23. (a) Vp = 220 V, Vs = 22 V, Rs = 220  secondary current amp R V i s s s 10 1 220 22 = = = So by using the relation i A i i V V p p s s p = , = 0.01 24. (a) 2 C 1 XC  =  2 X (C) 1  C  = 6 5 10 1000 1 2 1 −     = . 100 MHz  = 25. (a) Peak value of voltage C i V i X C 2 0 0 = 0 =  50V 2 3.14 50 100 10 1.57 6 =     − Hence if equation of current i i sin t = 0 then in capacitive circuit voltage is       = − 2 0 sin  V V  t          =  −       =   − 2 50sin 100 t 2 V 50 sin 2 50t 26. (a) Resistance of the bulb =   = 360 10 60 60 R . For maximum illumination, voltage across the bulb VBulb = VR = 60V By using 2 2 V = VR + VL  2 2 2 (100 ) (60) = + VL  VL = 80V Current through the inductance (L) = Current through the bulb A 6 1 60 10 = = Also V iX i(2 L) L = L =   1.28 . 6 1 2 3.14 60 80 (2 ) H i V L L =    = =   27. (a) Current is maximum i.e. the given circuit is in resonance, and at resonance LC 1 0 =  500 / . 2 10 1 0.5 8 10 1 3 6 0 = rad sec  =   = − −  28. (c) =  =     =    − X 2 L 2 3.14 50 95.5 10 29.98 30 3 L Impedance = + = (40) + (30) = 50  2 2 2 2 Z R XL 29. (c) 2 6 2 2 2 1000 2 10 1 (300) 1000 0.9 C 1 Z R L         = +  −          = +  − −  = (300 ) + (400 ) = 500  2 2 Z . 30. (c) By using  LC  2 1 0 =  C L 1   C C C C L L ' 4 ' = =  4 ' L L = . 31. (b) Current will be maximum in resonance i.e. XL = XC  6 100 2 10 1 100 −    =   L  . 50 2 L Henry  = 100V, 60Hz L 60V, 10W
32. (c) VL = VC; This is the condition of resonance and in resonance V = VR = 220 V. In the condition of resonance current through the circuit 2.2 . 100 220 A R V i rms = = = 33. (d) X L =  L = 2000  5  10–3 = 10  and =    = − 10 2000 50 10 1 C 6 X Total impedance of the circuit = 6 + ( ) + ( − ) = 6 + (4) + 0 = 10 2 2 2 R XL XC Ammeter reads r.m.s. current so it's value A V i rms rms 2 1.41 10 20 / 2 Total impedance = = = = Since XL = XC ; this is the condition of resonance and in this condition V = VR = iR = 1.4  4 = 5.6 V. 34. (a) iL 10 A 24 240 = = iC 5 A 48 240 = = Hence i = iL − iC = 5A 35. (d) The induced emf =  = − dt d  = − (B.A.) dt d = − A dt dB = − (r 2 ) t 2 t (e ) r e dt d − − =   o =  r 2 t 0 t e = − =  r 2  The electrical power developed in the resistor just at the instant of closing the key = P = R 10 r R r R 2 2 4 4 0   =  .Hence (D) is correct 36. (a) According to given problem, I = V Z V = / [R2 + (1/C 2 )]1/2 . . . (1) and 2 2 2 [R (3/C ) ] V 2 I +  = ............(2) substituting the value of I from equation (1) in (2), 4 2 2 2 2 2 2 C 9 R C 1 R   = +       + , i.e., 2 2 2 R 5 3 C 1 =  so that 5 3 R [(3/5)R ] R (1/C ) R X 2 1/ 2 = =  = Hence (A) is correct. 37. (d) The induced emf =  = − dt d  = − (B.A.) dt d = − A dt dB = − (r 2 ) t 2 t (e ) r e dt d − − =   o =  r 2 t 0 t e = − =  r 2  The electrical power developed in the resistor just at the instant of closing the key = P = R 10 r R r R 2 2 4 4 0   =  . Hence (D) is correct. 38. (b) = p 5 p 4 p 3 p 2 p Bp B1 B B B B         +         +         +         +         = → → → → → → Where p B1        → = ^ 0 B = B i → (semi-infinite wire) p B2        → = (k) 2 3a 4 i ^ o        p B3        → = 0 p B4        → = ( k) 2 a 4 i ^ o −        =        → p B5 ( j) 2 a 4 i ^ o −          ^ 0 ^ 0 ^ 0 ^ 0 k 2a i k 6a i j 2 a i j 6 a i B  −  +   −   = − →  ^ 0 ^ 0 j a i 3 1 i 3 a 2 i  −   −  4 3 a i | B | 0 2  +   = → 39. (b) The net electric field 0 E    = XL =24 240 V XC = 48 iC iL i
The net force acting on the electron is zero because it moves with constant velocity.  = + =  → → → F net F e F m 0 F e F m → → =  eE = evB  v= B B E 0   = The time of motion inside the capacitor = t =   = B v l 0  40. (a) The gravitational torque must be counter balanced by the magnetic torque about O, for equilibrium of the sphere. The gravitational force = gr  =mg X r sin   gr  = mgr sin The magnetic torque m = → →  X B Where the magnetic moment of the coil =  = (ir 2 )  m = ir2 Sin  ir2 B Sin = mgr Sin  B= ir mg  41. (c) (A),(B),(D), in going from P to Q increase in kinetic energy = 2 1 m ( ) − 2 2V 2 1 m 2 V = 2 1 m ( ) 2 3v = work done by electric field. or 2 3 m 2 v = Eq x 2a or E=         qa mv 4 3 2 The rate of work done by E at P = force due to E x velocity. = (qE)v = qv                 qa mv 4 3 2 =         a mv 4 3 3 At q, V is perpendicular to E and B , and no work is done by either field 42. (b) The net magnetic force on the conducting wire  = F = 2dFcos  F = ( )           cos 2 R di i 2 0 0  F =     di cos R i 0 0 When    =  = id x Rd R i di  ( )       = id cos R i F 0 0        =   = / 2 0 2 0 0 2 0 o R i i cos d R i i F 43. (b) Magnetic field due to one side =   = 0 0 1 2sin 30 4 (a 3/ 2) i B 2 3 a i 0   Resultant magnetic field B = 3B1 cos       −   2 = 3                       2 3a 2 3 a 2 3 a i 0 = 2 3 a i 0   44. (d) From the equation B = 0 i0n = ( )( )         − 1.23m 5 850 turns 4 10 T.m / A 5.57A 7 = 2.42 10 T 24.2mT 2  = − 45. (b) The work done = Initial Final .B .B        −           → → → → = B(cos0 cos180 ) 2 B 0 0  − =  = 2N a iB2  = 2 100 3.14 (0.05) 0.1 1.5 2      =0.236 Joule di d     dF dF P dFcos dFcos mg O r 