6.Solution Electodynamics

SOLUTIONS TO PART TEST-01 TOPIC : ELECTRODYNAMICS 1. Charge on outer surface of C = – charge on inner surface of C Hence potential at B due to charge on conductor C = 0 charge on outer surface of dielectric = – charge on inner surface of dielectric  Potential at B due to charge on dielectric = 0 Potential at B due to charge on A = Q 4 0 b  net potential at B = Q 4 0 b . 2. In parallel combination each branch should have equal potential but in AC circuit in a branch veff  V2 + V2 = V2 + V2  12 2 + 52 = 102 + V2  (V ) = R2 peak V 3. Potential difference across ‘C1’ C2 V V1 = C1  C2 = When dielectric is inserted, C2 will increase. So V1 will increase. 4. Current will have to increase in order to oppose the cause (decrease in flux). 5. Work done on 2C charge =  3 → = q Edr [ r for (1, 1, 0) = & r for (3,0, 0) = 3] = 2 × (area of E-r graph from r = m to r = 3) = 2 ×  1 (3  2 )20  = 20 (3  2 ) J. 6. The magnetic force F shall act on the negatively charged test particle in direction parallel to current carrying wire, if the velocity negative charge is as shown. → of the 3 KQ KQ KQ 1  4  3 R2 7. Vc – Vs = 2 R – R = 2R   R  8 R 3 6  0  0 0   0 ˆ ˆ 8. B = 4 2a = 8a (– i  k) 9. Four lines, perpendicular to lines of electric field and passing through A, B, C and D are drawn. These are equipotential lines. As potential decreases in the direction of electric field, therefore VA > VB > VD > VC x 10. Let r be the radius of new arc  r (2 – ) = L or r = L (2  )  Magnetic field at centre O is µ0i  2     µ i  2   2 B = 2 r   2  = 0  L   2  11. Electric field of a point charge is non-uniform hence net force can never be zero. 12. Magnetic field due to circular segment = 0i 3 . 0i 4 2a 0i Due to one straight wire segment = Net field 4b (sin 45° + sin 0°) = 4 2  b = 30i + 2 x 0i = 0i  3  8a 4 2 b 4  2a   q  13. Net charge enclosed inside the cube = 8  8  = q   q net electric flux through cube = 0  charges are symmetrically placed so the flux through each face = q . 6 0 14. The charge particle will experience magnetic as will as gravitational force due to wire. Gravitation force is  Gm1 m2   r 2  very small.  15. The distribution of charge on the outer surface, depends only on the charges outside, and it distributes itself such that the net, electric field inside the outer surface due to the charge on outer surface and all the outer charges is zero. Similarly the distribution of charge on the inner surface, depends only on the charges inside the inner surface, and it distributes itself such that the net, electric field outside the inner surface due to the charge on inner surface and all the inner charges is zero. Also the force on charge inside the cavity is due to the charge on the inner surface. Hence only induced charge on outer surface will redistribute 16. As the net force acting on the proton is zero, it will move on straight line. 17. From Gauss theorem surface charge density on C face is –  .  EP =  2 +  2 =  = 5.0 N/C 0 18. The force on the rod due to magnetic field and gravity is i 𝑙B – mg (upwards) i𝑙B  mg Hence the extension in the springs is 2k ( Note that effective spring constant is 2k) i𝑙B  mg Therefore the length of the spring is 𝑙0 + 2k 19. j = i and i = V  j = V using R = 1 . 𝑙  j = V (or as J =  E) A R RA  A 𝑙 using values j = 20 × 106 A/m2 20. The resultant magnetic dipole moment of toroid is zero. d→ of small parts of toroid turn along a circle and hence there resultant is zero.  Torque acting on it is zero. 21. = As C & D are at same potential by symmetry of circuit  It is balanced wheat-stone bridge Hence the circuit has equivalant resistance 7/3  22. No change in the flux occurs due to the described motion of the magnet. Hence no current will be induced in the coil. 23. Current through resistance will be from A to B if 20 –  > 2   < 18 and from B to A if 20 –  < 2   > 18 24. For any loop  = iR, For superconductor, R = 0 total = 0  d dt = 0   = constant 25. V = E – i r. Now, if we go from A  B, then resistance of the circuit increases, ' i ' decreases ∴ V increases 26. Here oscillating rod is an AC source because emf induced in it is (vB𝑙) ; which varies sinusoidally because v varies sinusoidally. Maximum current will flow through the circuit under resonance condition. Therefore time period of oscillation of rod is T = 2 = 2 sec. 27. Equivalent circuit is 28. E =  × 10–9 sint Also E = i × 2.  i =    × 10–9 sin t. 2 = 2 F 29. The potential difference across completely charged capacitor is V. As the switch is pushed to b, the initial current in the resistor R is V . Hence J is the correct curve. R 30. Current is decreasing and 17 L = 25kA = 680 H. SOLUTIONS TO PART TEST-02 TOPIC : ELECTRODYNAMICS 1. When shunted with 4  /5 Rg  5 g = 4  4 5 [Rg = 16  ] When further shunted with 2 ' × 16 = ( – ') × 4  ' =  3 13   8    Further reduction in current = – =   5 13 13  5  8 Hence further reduction in deflection = 13 of the deflection when shunted with 4 only. 2. The current leads in phase by ( XC > XL )  = 37°  i = 10 cos (100 t  37) Z = cos (100  t + 37°) The instantaneous potential difference across A B is = m (XC – XL) cos (100t + 37° – 90°) = 6 cos (100 t – 53°) The instantaneous potential difference across A B is half of source voltage.  6 cos (100  t – 53°) = 5 cos 100  t 24 solving we get cos 100 t = = 25  instantaneous potential difference = 5 × 24 = 24 volts 25 5 3. From given conditions, V = V and V = 0  V K (Q  q1) = + Kq2  kq1 = 0 A C B B 3 a 2a 2a  2Q + q + 3q2 = 0 .... (1) Using VC = VA K(Q  q2 ) + Kq2  K q1 = Kq1  K (Q  q1)  Kq2 3 a 3a 3a a 3a 2a  q = – q2 (2) 4 8 Using it in (1), q2 = – 1 Q 4. E.d𝑙 = d dt =  r 2 dB dt ; N = Ed𝑙 = – Rr2 dB dt Ndt = – Rr2 dB = Rr2 B =   r 2B q 2r = mr2   = qB 2m 5. Let the equivalent resistance of one infinite ladder be x. Then the complete network reduces to  R = x  3x  3 x B x C AB and RAC = x  3x 4 2x  2x 2x  2x = x x x A x D Hence RAB  3 RAC 4 6. A satellite is in a state of free fall & hence weightlessness. Thus only electric force is responsible for the tension Fe T = Fe = KQ2 (2L)2 7.  wheat stone bridge is in balanced condition 100 x 100 So 100 𝑙1 = 100  x 𝑙 2 𝑙1  𝑙 2 = 2 𝑙1 𝑙2  x = 100  8.  =  iR2 .r 2 2(R2  x2 )3 / 2 EA = – d dt = 0i 2 R2 r2 (–3/2) (R2 + x2)–5/2 . 2x EA is maximum when dEA = 0  dx d x dx (R2  x2 )5 / 2 = 0 or (R2 + x2)5/2 – 5x 2 (R2 + x2)3/2 2x = 0 or R2 + x2 – 5 x2 = 0 or x = R 2 9. V =  R = 10 × 10 = 100 volts (since,  = 10amp from figure) Also  =  e – t/RC Taking 𝑙n ; 𝑙n 0  = t  C = t   Rln (0 / ) At ; t = 2 sec,  = 2.5 A 2 C =  10  10 ln    2.5  2 C = 10 ln 4 1 2 = 10  2ln 2 1 = 10 ln 2 1 2 1  1  2 500 C = . Heat produced = CV =   (100 ) = joules. 10𝑙n2 Hence (C) is correct. P = I2 R 2 2  10𝑙n2  𝑙n2 RC 10 1 1 = I 2 R e–2t/RC = P e–2t/RC  time constant = 2 = 2 × 10𝑙n2 = 2𝑙n2 10. For given condition : Magnitude of Bsolenoid = Magnitude of Bloop  ni = 0  2R here n = Total no. of turn Total length 1300 = 0.65  1 i = 2R  n 8  0.65 = 2  0.02  1300 = 100 mA. For given condition : Total magnetic field at the centre of loop = |Bloop| + |Bsolenoid |  |Bloop| = |Bsolenoid| = 2|Bloop | = 2 × 0 2R 2  4 107  8 = 2  0.02 = 16  × 10–5 T. 11. Charge is distributed over the surface of conductor in such a way that net field due to this charge and outside charge q is zero inside the conductor. Field due to only q is non-zero. 12. Because both inductors are in parallel  L11 = L22 U1  ½L1 1 1 = 1  L2 U2 ½L2 2 2 2 L1 13. Statement  is true directly from Gauss Theorem. Statement 2 is false. Electric field at any point an Gaussian surface depends on all four charges. Statement-1 is True, Statement-2 is False 14. The current through solid metallic cylinder also produces magnetic field inside the cylinder. Hence statement- 1 is false 15. Statement-1 is obviously false if the current is sent in opposite direction given in figure of statement-2 → → 16. B  d𝑙 along any closed path within a uniform magnetic field is always zero. Hence the closed path can be chosen of any size, even very small size enclosing a very small area. Hence we can prove that net current through each area of infinitesimally small size within region of uniform magnetic field is zero. Hence we can say no current (rather than no net current) flows through region of uniform magnetic field. Hence statement -2 is correct explanation of statement-1. 17 - 19 i1 = 0.1A, E2 = 4V, i2 = 0 As ; 0.1 R1 + 0.1 R2 – E1 = 0 0.1 R2 – 4 V = 0 R = 40  Now ; i2 = 0.3A, i1 = 0.1 A, E2 = 8A Now ; 0.1 R1 +E1 – 8 = 0 0.1R1 + 4 – E1 = 0 40 0.2R1 – 4 = 0  R = 4 0.2 = 20  E1 = 2 + 4 = 6 V 20. Inside the cylinder B.2r =  .  R2 r 2 B = 0 .r 2R2 ..............(1) outside the cylinder B.2r =    B = 0 2r ..............(2) Inside cylinder B  r and outside B  1 r So at the surface equation of magnetic field changes. Hence clear from graph, wire 'c' has greatest radius. 21. Magnitude of magnetic field is maximum at the surface of wire 'a'. 22. Inside the wire B(r) = 0 .  .r = 0Jr  dB = 0J 2 R2 2 dr 2 i.e. slope  J  current density It can be seen that slope of curve for wire a is greater than wire C. J 23. Drift speed Vd = ne i = neA i = V R L where R = A V E = L and P = I2 R 24. (A) Because the magnetic field is parallel to x-axis, the force on wire parallel to x-axis is zero. The force on each wire parallel to y-axis is B i 𝑙. Hence net force on loop is B i𝑙. Torque of both the forces is canceling o 2 o each other. (B) Because the magnetic field is parallel to y-axis, the force on wire parallel to y-axis is zero. The force on each wire parallel to x-axis is B i 𝑙. Hence net force on loop is B i𝑙. Torque of both the forces is canceling o 2 o each other. (C) Since net displacement of current from entry point in the loop to exit point in the loop is along the diagonal of the loop. The direction of external uniform magnetic field is also along the same diagonal. Hence net force on the loop is zero. Torque of both the forces is canceling each other. (D) The net displacement of current from entry point in the loop to exit point in the loop is along the diagonal ( of length 2 𝑙.) of the loop. The direction of external uniform magnetic field is also perpendicular to the same diagonal. Hence magnitude of net force on the loop is Boi( 2𝑙 ). Torque of both the forces is canceling each other. SOLUTIONS TO PART TEST-03 (PT-03) TOPIC : ELECTRODYNAMICS 1. U = 1 C V2 ; C = k 0A  20 A 2 eq 1 d/ 2 (d/ 2) C2 = 0 A d / 2 ; Ceq = C1 C2 C1  C2  2 0 A   0 A    d/ 2  d/ 2 4 0 A Ceq = 3 0 A d/ 2 = 3 d 1  4 0 A  V 2 2  0 A  V2 U =   2  3 d  = 3  d  2. Charge on the differential element dx, dq = Q.dx 𝑙 equivalent current di = f dq  magnetic moment of this element d = (di) NA (N = 1) =   x 2  f Q dx   =  d   f Q 𝑙 x2 dx ;  = 1  fQ 𝑙2   𝑙 0 𝑙 0 3 3. tan = Fe Vg Fe / k = V(  )g  k =     2.4 = 2.4  0.8 = 1.5 → → → → 4. The magnetic field at point P is B1  B2 where B1 and B2 are magnetic field at P due to wire 1 and 2. → = 0 2  (cos ˆ + sin ˆj ) B1 4 r i → = 0 2  (– cos ˆ + sin ˆj ) B2 4 r i where r2 = x2 + (d/2)2  field is along +y direction at point P and its magnitude is 0 4  0 4  x B = 4 r sin = 4 [x2  (d/ 2)2 ] dB B is max when dx = 0 i.e. at x = d/2. 5. Let electric field at point . 'p' has both x and y component. So similar electric field will be, for other hemisphere (upper half). Now lets overlap both. (Enet)p = 2 Ex and it should be zero (as E inside a full shell = 0). So Ex = 0, So electric field at 'p' is purely in y direction. 6. When the rod rotates, there will be an induced current in the rod. The given situation can be treated as if a rod 'A' of length '3𝑙' rotating in the clockwise direction, while an other rod 'B' of length '2𝑙' rotating in the anticlockwise direction with same angular speed ''. As] e = 1 B𝑙2 2 For 'A' : eA = 1 B (3𝑙)2 2 For 'B' : eB = 1 B (– ) (2𝑙)2 2 Resultant induced emf will be : e = eA + eB = 1 B𝑙2 (9 – 4)  e = 2 5 B𝑙2 2 7. From symmetry, the current distribution in branches LP, MP, NP and OP are as shown in figure 1. Therefore junction at P can be broken as shown in figure 2  r 1.5r r   A B Hence equivalent resistance is 3.5 r. b b   8. Induced emf Bvdx  0 vdx 2x a a 0v 𝑙  b  E2  Induced emf = 2 n  a  Power dissipated =   R Also, power = F.V  F = E VR 1   V  b 2  F =  0 𝑙 n    VR  2  a  9. Net charge on both the capacitors is = C1V – C2V The effective capacitance of system is C1 + C2 because both are in parallel. Qeff Therefore p.d across the system is C1  C2 = (C1  C2 )V C1  C2 Initial energy= 1 (C  C )V2 2 1 2 1  (C  C )V 2 Final energy = (C 1  C 2 )  1 2  2  C1  C2   (C  C ) 2 Therefore ratio of final to initial energy is =  1 2   (C1  C2 )  10. Equivalent circuit :  2  Induced emf e =   ( Radius = a)     By nodal equation :  X  e   X  0  nodal %    r / 4 r  = 0  5X = 4e    4e 2Ba2  X 2Ba2  X = 5 = 5 and  = r = 5r also direction of current in ‘r’ will be towards negative terminal i.e. from rim to origin. Alternately; by equivalent of cells (figure (ii)) : e I = r  r 4 4 e = 5 r 11. after redrawing the circuit (a) 4 = 5A , (b), (c) From loop (1) – 8(3) + E1 – 4(3) = 0  E1 = 36 volt from loop (2) + 4(5) + 5(2) – E2 + 8(3) = 0 E2 = 54 volt (d) from loop (3) – 2R – E1 + E2 = 0 R = E2  E1 2 54  36 = 2 = 9  12. The x-component of velocity, being parallel to magnetic field, shall remain unchanged. The component of velocity perpendicular to x-axis will always have magnitude Voy, and at any time t it shall make an angle  = t with y-axis as shown.  y-component of velocity is Voy cost and z-component of velocity along negative z-direction at any time t is Voy sin t. Where  = qB m 13. Let E1 < E2 and a current i flows through the circuit. Then the potential difference across cell of emf E1 is E1 + ir1 which is positive, hence potential difference across this cell cannot be zero. Hence statement 1 is correct. For current in the circuit to be zero, emf of both the cells should be equal. But E1  E . Hence statement 2 is correct but it is not a correct explanation of statement 1. 14. The magnetic force on bob does not produce any restoring torque on bob about the hinge. Hence this force has no effect on time period of oscillation. Therefore both statements are correct and statement-2 is the correct explanation. 15-17. For t = 0 to to = RC seconds, the circuit is of charging type. The charging equation for this time is  t q  CE(1 e RC ) Therefore the charge on capacitor at time t0 = RC is qo  CE(1 1 ) e For t = RC to t = 2RC seconds, the circuit is of discharging type. The charge and current equation for this time are q  qoe  tto RC and i  qo e RC  tto RC Hence charge at t = 2 RC and current at t = 1.5 RC are q q  qoe  2RCRC RC  qo  e 1 CE(1 1) e e t and i  qo e RC 1.5RCRC RC  qo  eRC E (1 1 ) respectively e Since the capacitor gets more charged up from t = 2RC to t= 3RC than in the interval t=0 to t=RC, the graph representing the charge variation is as shown in figure 18. (A) Uniform electric field exerts constant force on the charged particle, hence the particle may move in straight line or a parabolic path. (B) Under action of uniform magnetic field, the charged particle may move in straight line when projected along or opposite to direction of magnetic field. The charged particle moves in circle when it is projected perpendicular to the magnetic field. If the initial velocity of the charged particle makes an angle between 0o and 180o (except 90o ) with magnetic field, the particle moves along a helical path of uniform pitch. (C) If charged particle is shot parallel to both fields it moves along a straight line. If the charged particle is shot at any angle with both the field (except 0° and 180°) , the particle moves along a helix with non-uniform pitch. (D) from results of A and B all the given paths are possible. 19. Electric field at P is Q x E = 4 0 (x 2  r 2 )3 / 2  2 ir 2  2 Qf r 2 Magnetic field at P is B = 0 4 (x2 0  r 2 )3 / 2 4 (x2  r 2 )3 / 2 f = frequency of revolution. Electric energy density = 1 0 E2 2 B2 ; Magnetic energy density 2 Electric energy density 1 0 E2 x2 0 x 2 c 2 9 magnetic energy density = 2 2 = 42 = 0 0 f 2 r 4 42 f 2 r 4 = 2 × 1010 = 9 × 109 20 20. When S1 is closed current in inductor remains, i =  2R    V1 +   V1 =    2  R 2 R 2 R  V 1      Potential difference (V ) = 2  di 2 And L dt = 3 1 3 di dt = + 2  3 L = 2  3 3 10–3 = 2000 A/s 21.   C1 = 0 A / 2 d , C2 = 0 A / 2 d/ 2  d = 4 0 A 5d C = C1 + C2 13 = 10 0 A d k 2 22. Current in the element = J(2r . dr) Current enclosed by Amperian loop of radius a 2 a / 2 J r 2J  a 3 J a2 I =  a . 2r . dr = 0   = 0 0 3a Applying Ampere's law  2  12 B . 2 . a 2 = µ0 . J0a2 12  B = 0 J0a 12 On putting values B = 10 µT