DC Pandey Electricity And Magnetism

Previous Years’ Questions (2018-13) 9 Y=1000 – X X According to question τ 2 = − τ1 ⇒|τ 2| = | τ1| G ∴ pE1 sin θ = p 3 E1 cosθ tanθ = 3 ⇒ tan θ = tan 60° ∴ θ = 60° (l – 10) (110 – l) 14. (c) Let there are n capacitors in a row with m such 1000 − X = X rows in parallel. l − 10 100 − (l − 10) n or 1000 − X = X …(ii) n l − 10 110 − l m rows From Eqs. (i) and (ii), we get n 100 − l = l − 10 l 110 − l V=1.0kV (100 − l) (110 − l) = (l − 10)l As voltage not to exceed 300 V 11000 − 100 l − 110 l + l2 = l2 − 10 l ⇒ 11000 = 200 l ∴ n × 300 > 1000 ∴ l = 55 cm Substituting the value of l in Eq. (i), we get [a voltage greater than 1 kV to be withstand] ⇒ n > 10 ⇒ n = 4 (or 3.33) X = 1000 − 55 55 100 − 55 3 ⇒ 20 X = 11000 ∴ X = 550 Ω Also, CEq = mC = 2 µF n ⇒ m = 2 ⇒ m = 8 [Q C = 1 µ F] n So, total number of capacitors required 10. (c) A potential drop across each resistor is zero, so = m × n = 8 × 4 = 32 the current through each of resistor is zero. 15. (b) In steady state no current flows through the 11. (d) For a voltmeter, Ig Rs capacitor. G So, the current in circuit I = E Ig (G + R s) = V V ⇒ R = V −G r + r2 Ig Q Potential drop across capacitor = Potential drop ⇒ R = 1985 = 1.985 kΩ or R = 1.985 × 103 Ω across r2 = Ir2 = r Er2 + r2 12. (a) In a balanced Wheatstone bridge, there is no effect ∴Stored charge of capacitor, Q = CV = CEr2 on position of null point, if we exchange the battery r + r2 and galvanometer. So, option (a) is incorrect. 13. (b) Torque applied on a Y 16. (b) Induced constant, I = e dipole τ = pE sin θ R where θ = angle p Here, e = induced emf = dφ dt between axis of dipole 90 – q and electric field. q X I = 1 =  dφ  ⋅ 1 R dt R For electric field E1 = E$i it means field is directed ⇒ dφ = IRdt along positive X direction, so angle between dipole ⇒ φ = ∫ IRdt ∴ Here, R is constant ⇒ φ = R ∫ Idt and field will remain θ, therefore torque in this direction ∫ I ⋅ dt = Area under I− t graph E1 = pE1 sin θ = 1 × 10 × 0.5 = 2.5 2 In electric field E2 = 3 E$j, it means field is directed along positive Y-axis, so angle between dipole and ∴ φ = R × 2.5 = 100 × 2.5 = 250 Wb. field will be 90 − θ 17. (a) Torque in this direction T2 = pE sin (90 − θ). = p 3 E1 cos θ

10 Electricity & Magnetism S 20. (d) B at centre of a circle = µ 0I I − Ig 2R I B at centre of a square = 4× µI l [sin 45° + sin 45° ] = 4 2 µ 0I 4π ⋅ 2 πl Ig G 2 In parallel, current distributes in inverse ratio of Now, R = L and l = L (as L = 2 πR = 4l) 2π 4 resistance. Hence, where, L = length of wire. I − Ig = G S = GIg Ig S ⇒ I − Ig ∴ BA = µ 0I = πµ 0I = π  µ0 I  2⋅ L L  L  As Ig is very small, hence 2π S = GIg ⇒ b = (100) (1 × 10−3 ) = 0.01 Ω I 10 BB = 4 2 µ 0I = 8 2µ 0I = 82 µ 0I πL π  L  2 π  L  18. (a) As E is constant, 4 Hence, Ea = Eb ∴ BA = π 2 BB 8 2 Q 21. (d) We need high retentivity and high coercivity for r electromagnets and small area of hysteresis loop for transformers. 22. (d) V 2 = VR2 + VL2 ⇒ 2202 = 802 + VL2 dr Solving, we get As per Guass theorem, only Qin contributes in electric VL ≈ 205 V field. XL = VL = 205 = 20.5 Ω = ωL k Q + A I 10 ∫kQ b4 πr2dr ⋅ r  ∴ = ∴ L = 20.5 = 0.065 H Here, a2 a 2 π × 50 b2 k= 1 23. (a) Theoretical question. Therefore, no solution is 4 πε0 required. b Q b2  r 2   b2 − a2  V V a2  a   2  R ⇒ =Q+ 4 πA  2 = Q + 4 πA ⋅ 24. (c) i = neAvd or = neAvd ⇒ = neAvd  b2   ρl   a2  A ⇒ Q = Q + 2 πA(b2 − a2 ) ∴ ρ = V = resistivity of wire nelvd  b 2− a 2  Q  a2  2 πa2 ⇒ Q = 2 πA(b2 − a2 ) ⇒ A= Substituting the given values we have 19. (c) 3 µF and 9µF = 12µF ρ = (8 × 1028)(1.6 × 5 × 10−4) 10−19 )(0.1)(2.5 4µ F and 12 µF = 4 × 12 = 3µF 4 + 12 ≈ 1.6 × 10−5 Ω-m Q = CV = 3 × 8 = 24µC (on 4µF and 3µF ) 25. (b) Now, this 24µC distributes in direct ratio of capacity 6V 2Ω between P 3µF and 9µF. Therefore, i1+ i2 1 1Ω 2 i1 i2 9V Q 9 µF = 18µC ∴ Q 4µF + Q 9 µF = 24 + 18 = 42 µC = Q (say) 3Ω Q 3Ω E = kQ = 9 × 109 × 42 × 10−6 = 420 N/C Applying Kirchhoff’s loop law in loops 1 and 2 in the R 2 302 directions shown in figure we have

Previous Years’ Questions (2018-13) 11 6 − 3(i1 + i 2 ) − i 2 = 0 ...(i) This is also the potential difference across2 µF. 9 − 2 i1 + i 2 − 3i1 = 0 ...(ii) Solving Eqs. (i) and (ii) we get, ∴ Q2 = (2 µF)(V2µF )   2CE  2 i 2 = 0.13 A or Q2 =  C + 3 =   E Hence, the current in 1 Ω resister is 0.13 A from Q to P.  1 + 3  C 26. (b, c) V0 = potential on the surface = Kq From this expression of Q2, we can see that Q2 will R increase with increase in the value of c (but not linearly). where, K = 1 and q is total charge on sphere. Therefore, only options (a) and (b) may be correct. 4 πε0 Further, d (Q2 ) 2 E(C(C+ 3) − C  6E Potential at centre = 3 Kq = 3 V0 dc = + 3)2  = (C + 3)2 2R 2  Hence, R1 = 0 = Slope of Q2 versus C graph. From centre to surface potential varies between 3 V0 i.e. slope of Q2 versus C graph decreases with 2 increase in the value of C. Hence, the correct graph and V0 F5Vro0mwisllubrfeacpeotteonitniafilnaittya, itpvoainritebsebtweteweenecneVn0tre is (a). and 0, 4 29. (d) and surface. At any point, at a distance r(r ≤ R ) from P centre potential is given by V = Kq  3 R 2 − 1 r 2  Q R3 2 2 = V0  3 R 2 − 1 r 2  (as V0 = Kq ) R R2 2 2 R Putting V = 5 V0 and r = R2 in this equation, we get S 4 If we calculate the force on inner solenoid. Force on Q R2 = R due to P is outwards (attraction between currents in 2 same direction. Similarly, force on R due to S is also outwards. Hence, net force F1 is zero) 3V0 and V0 are the potentials lying between V0 and Force on P due to Q and force on S due to R is 4 4 inwards. Hence, net force F2 is also zero. Alternate Thought Field of one solenoid is uniform zero hence these potentials lie outside the sphere. At and other solenoid may be assumed a combination of circular closed loops. In uniform magnetic field, net a distance r(≥ R ) from centre potential is given by force on a closed current carrying loop is zero. V = Kq = V0R rr Putting V = 3 V0 and r = R3 in this equation we get, 4 4R 30. (a) R3 = 3 θ L θ r = Lsinθ T V0 Further putting V = 4 and r = R4 in above equation, F = Magnetic force F r (repulsion) per unit length we get R4 = 4R = µ 0 I2 = µ 0 I2 Thus, R1 = 0, R2 = R , R 3 = 4R and R4 = 4R with 2 π 2r 4π Lsinθ λg 2 3 λg = weight per unit length these values, option (b) and (c) are correct. 27. (d) Electric field lines originate from position charge Each wire is in equilibrium under three concurrent forces and termination negative charge. They cannot form as shown in figure. Therefore, applying Lami’s theorem. closed loops and they are smooth curves. Hence the most appropriate answer is (d). F = λg or µ 0 I2 sin(180 − θ) sin(90 + θ) 4π Lsinθ = λg 28. (a) Resultant of 1 µF and 2 µF is 3 µF. Now in series, sinθ cosθ potential difference distributes in inverse ratio of ∴ I = 2 sinθ πλgL µ 0 cosθ capacity. ∴ V3µF = C or V3 µF = C  E 31. (b) Direction of magnetic dipole moment M is given by Vc 3    C + 3  screw law and this is perpendicular to plane of loop.

12 Electricity & Magnetism In stable equilibrium position, angle between M and B 37. (c) After connecting C to B hanging the switch, the is 0° and in unstable equilibrium this angle is 180°. circuit will act like an L-R discharging circuit. 32. (c) Steady state current i 0 was already flowing in the R L -R circuit when K1 was closed for a long time. Here, i0 = V = 15 V = 0.1 A L R 150 Ω Now, K1 is opened and K 2 is closed. Therefore, this i 0 Applying Kirchhoff’s loop equation, will decrease exponentially in the L-R circuit. Current i VR + VL = 0 ⇒ VR = − VL ⇒ VR = − 1 VL −t at time t will be given by i = i 0 e τ L where, τL = L ⇒∴ − Rt 38. (b) When force exerted on a current carrying R i = i 0e L conductor Fext = BIL Average power = Work done Substituting the values, we have Time taken −( 0.15 × 10 3 )(10 −3 ) i = (0.1) e ( 0. 03) = (0.1)(e −5 ) ∫ ∫P = 1 2 Fext. . dx = 1 2 B( x)IL dx t 0 t 0 = 0.1 = 6.67 × 10−4 A 150 1 2 5 × 10−3 = 0.67 mA ∫= 3 × 10−4e −0. 2x × 10 × 3 dx 33. (c) Total power (P) consumed 0 = (15 × 40) + (5 × 100) + (5 × 80) + (1 × 1000) = 9 [1 − e −0. 4 ] = 9 1 − 1  = 2.967 ≈ 2.97 W = 2500 W e 0. 4  As we know, Power i.e. P = VI ⇒ I = 2500 A = 125 = 11.3 A 39. (d) Statement I is false and Statement II is true. 220 11 40. (d) As, P = V 2 Minimum capacity should be 12 A. R 34. (c) As we know, potential difference VA − VO is where, P = power dissipates in the circuit, V = applied voltage, dV = − Edx R = net resistance of the circuit R = 120 × 120 = 240 Ω [resistance of bulb] VA 2 ⇒ VA − VO = − 30 ×  x3 2  3 0 60 ⇒ ∫ dV = − ∫ 30 x2dx Vo 0 60 W 6Ω 240 Ω 6Ω = − 10 × [2 3 − (0)3 ] = − 10 × 8 = − 80 J 240 Ω 35. (a) When free space between parallel plates of ⇒ 60 Ω capacitor, E = σ 120 V 120 V ε0 54 Ω 48 Ω 6Ω When dielectric is introduced between parallel plates ⇒ of capacitor, E′ = σ 120 V 120 V Kε0 Req = 240 + 6 = 246 Ω ⇒ i1 = V = 120 Electric field inside dielectric, σ = 3 × 104 Req 246 Kε [before connecting heater] 0 R = V 2 = 120 × 120 where, K = dielectric constant of medium = 2.2 R 240 ε0 = permitivity of free space = 8.85 × 10−12 ⇒ σ = 2.2 × 8.85 × 10−12 × 3 × 104 = 6.6 × 8.85 × 10−8 = 5.841 × 10−7= 6 × 10−7 C/m 2 36. For solenoid, the magnetic field needed to be magnetised the magnet. B = µ 0nI ⇒ R = 60 Ω [resistance of heater] where, n = 100, l = 10cm = 10 m = 0.1 m So, from figure, 100 V1 = 240 × 120 = 117.073 V [Q V = IR] 246 ⇒ 3 × 103 = 100 × I ⇒ I = 3 A 0.1

Previous Years’ Questions (2018-13) 13 ⇒ i2 = V = 120 ⇒ V2 = 48 × 120 = 106.66 V 44. (b) Bnet = B1 + B2 + BH R2 54 54 N V1 − V2 = 10.04 V BH 41. (a) Fnet = 2 F cosθ B1 S S B2 N y NO –q/2 F θθ F S qa a qx Bnet = µ0 (M1 + M2) + BH 4π r3 2 kq (q /2 ) y Fnet = ( y2 + a2 )2 ⋅ y2 + a2 = 10−7 (1.2 + 1) + 3.6 × 10−5 = 2.56 × 10−4 Wb / m2 (0.1)3 Fnet = 2 kq (q /2 ) y F sin θ F sin θ 45. (a) Magnetic field at the centre of smaller loop ( y2 + a2 )3/ 2 2F cos θ µ iR 2 B= 0 2 kq 2 y ⇒ a3 ∝ y 2( R 2 + x2 )3/ 2 2 42. (d) L L Area of smaller loop S = πR12 ∴Flux through smaller loop φ = BS O x A dx B Substituting the values, we get, φ ≈ 9.1 × 10−11 Wb 2L kdQ = 2L k  Q  dx =Q 2L  1  ∫46. (d) e = 3l (ωx) Bdx = Bω [(3l)2 − (2 l)2 ] = 5Bωl 2 Lx L 4 πε0L L x ∫ ∫ ∫V = dx 2l 2 2 Lx ω 2l l = Q [loge x]2LL = Q [loge 2L − loge L] x dx 4 πε0L 4 πε0L 47. (c) For charging of capacitor q = CV (1 − et / τ ) = Q ln(2 ) 4 πε0L At t = 2 τ ; q = CV (1 − e −2 ) 43. (b, c) Polarity should be mentioned in the question. 48. (c) Amplitude decreases exponentially. In 5 s, it Potential on each of them can be zero if, qnet = 0 remains 0.9 times. Therefore, in total 15 s it will remains (0.9) (0.9) (0.9) = 0.729 times its original or q1 ± q 2 = 0 value. or 120C1 ± 200C 2 = 0 or 3C1 ± 5C 2 = 0 JEE Advanced 1. In the figure below, the switches S1 and S2 RL R 2L are closed simultaneously at t = 0 and a V IV current starts to flow in the circuit. Both the batteries have the same magnitude of S1 S2 the electromotive force (emf) and the polarities are as indicated in the figure. (a) Imax = V (b) Imax = V Ignore mutual inductance between the 2R 4R inductors. The current I in the middle wire reaches its maximum magnitude Imax (c) τ = L ln 2 (d) τ = 2L ln 2 at time t = τ. Which of the following R R statements is (are) true? (More than One Correct Option, 2018)

14 Electricity & Magnetism 2. Two infinitely long straight wires lie in B2k$ . A positively charged particle is projected from the origin along the the xy-plane along the lines x = ±R. The wire located at x = +R carries a constant positive Y-axis with speed v0 = π ms−1 at current I1and the wire located at x = −R t = 0, as shown in figure. Neglect gravity carries a constant current I2. A circular loop of radius R is suspended with its in this problem. Let t = T be the time centre at ( 0, 0, 3R ) and in a plane when the particle crosses the X-axis from parallel to the xy-plane. This loop carries below for the first time. If B2 = 4B1, the a constant current I in the clockwise average speed of the particle, in ms−1, direction as seen from above the loop. The current in the wire is taken to be positive, along the X-axis in the time interval T if it is in the +$j-direction. Which of the following statements regarding the is............ . (Numerical Value, 2018) magnetic field B is (are) true? Y (More than One Correct Option, 2018) B1 V0=π ms–1 (a) If I1 = I2 , then B cannot be equal to zero at the X origin (0,0,0) B2 (b) If I1 > 0 and I2 < 0, then B can be equal to zero 5. An infinitely long thin λZ at the origin (0,0,0) non-conducting wire is P parallel to the Z-axis (c) If I1 < 0 and I2 > 0, then B can be equal to zero and carries a uniform R at the origin (0,0,0) line charge density λ. It O (d) If I1 = I2 , then the z-component of the 120° magnetic field at the centre of the loop is pierces a thin  − µoI  2R non-conducting spherical Q shell of radius R in such 3. Three identical V0 a way that the arc PQ S1 capacitors C1 ,C2 subtends an angle 120° at the centre O of and C3 have a capacitance of S2 the spherical shell, as shown in the C1 1.0µF each and C2 C3 figure. The permittivity of free space is ε0. they are Which of the following statements is (are) true? (More than One Correct Option, 2018) uncharged (a) The electric flux through the shell is 3 Rλ / ε0 . (b) The z-component of the electric field is zero. initially. They are at all the points on the surface of the shell. connected in a circuit as shown in the (c) The electric flux through the shell is 2 Rλ / ε0 . figure and C1 is then filled completely (d) The electric field is normal to the surface of with a dielectric material of relative the shell at all points. permittivity εr. The cell electromotive force (emf) V0 = 8V. First the switch S1 is 6. A particle of mass 10−3kg and charge 1.0 C closed while the switch S2 is kept open. When the capacitor C3 is fully charged, S1 is initially at rest. At time t = 0, the is opened and S2 is closed simultaneously. particle comes under the influence of an When all the capacitors reach equilibrium, electric field E (t) = E0 sin ωt $i, where E0 = 1.0 NC−1 and ω = 103 rad s−1. Consider the charge on C3 is found to be 5µC. The the effect of only the electrical force on value of εr = ............ (Numerical Value, 2018) the particle. Then, the maximum speed in 4. In the xy-plane, the region y > 0 has a m s−1, attained by the particle at uniform magnetic field B1k$ and the region y < 0 has another uniform magnetic field subsequent times is .......... . (Numerical Value, 2018)

Previous Years’ Questions (2018-13) 15 7. A moving coil galvanometer has 50 turns (a) P → 5; Q → 3,4; R → 1; S → 2 and each turn has an area 2 × 10−4 m2. The (b) P → 5; Q → 3; R → 1, 4; S → 2 magnetic field produced by the magnet (c) P → 5; Q → 3; R → 1, 2; S → 4 inside the galvanometer is 0.02 T. The (d) P → 4; Q → 2, 3; R → 1; S → 5 torsional constant of the suspension wire is 10 −4N - m rad−1. When a current flows Passage (Q. Nos. 9-10) through the galvanometer, a full scale deflection occurs, if the coil rotates by Consider a simple RC circuit as shown in 0.2 rad. The resistance of the coil of the Figure 1. galvanometer is 50 Ω. This galvanometer Process 1 In the circuit the switch S is is to be converted into an ammeter closed at t = 0 and the capacitor is fully capable of measuring current in the range charged to voltage V0 (i.e. charging 0 − 1.0 A. For this purpose, a shunt continues for time T >> RC). In the process resistance is to be added in parallel to the some dissipation (ED ) occurs across the galvanometer. The value of this shunt resistance R. The amount of energy finally resistance in ohms, is ............. . stored in the fully charged capacitor is Ec. (Numerical Value, 2018) Process 2 In a different process the voltage is first set to V0 and maintained for 8. The electric field E is measured at a point P (0, 0, d) generated due to various charge 3 distributions and the dependence of E on a charging time T >> RC. Then, the voltage d is found to be different for different is raised to 2V0 without discharging the charge distributions. List-I contains different relations between E and d. 3 List-II describes different electric charge capacitor and again maintained for a time distributions, along with their locations. T >> RC. The process is repeated one more Match the functions in List-I with the time by raising the voltage to V0 and the related charge distributions in List-II. capacitor is charged to the same final voltage V0 as in Process 1. These two (Matching Type, 2018) processes are depicted in Figure 2. List-I List-II S (Passage Type, 2017) P. E is 1. A point charge Q at the origin V + R V – C V0 Process 1 independent Figure 1 2V0/3 Process 2 of d V0/3 T >> RC Q. E ∝ 1 2. A small dipole with point t d charges Q at (0, 0, l) and − Q at T 2T (0, 0, − 1). (Take, 2 l << d) Figure 2 3. An infinite line charge 9. In Process 1, the energy stored in the coincident with the X-axis, with R. E ∝ 1 uniform linear charge density λ. capacitor EC and heat dissipated across d2 resistance ED are related by (a) EC = ED ln2 (b) EC = ED (c) EC = 2ED 1 S. E ∝ 1 4. Two infinite wires carrying a (d) EC = 2 ED d3 uniform linear charge density 10. In Process 2, total energy dissipated parallel to the X- axis. The one across the resistance ED is along ( y = 0, z = l) has a charge density + λ and the (a) ED = 1  1 CV02  (b) ED = 3  1 CV02  one along ( y = 0, z = − l) has a 3 2 2 charge density – λ. (Take, 2 l << d). 1 2 5. Infinite plane charge coincident (c) ED = 3CV02 (d) ED = CV02 with the xy-plane with uniform surface charge density.

16 Electricity & Magnetism Directions (Q.Nos. 11 to 13) Matching the diametrically opposite vertices of the star information given in the three columns of the is 4a. The magnitude of the magnetic field following table. at the center of the loop is A charged particle (electron or proton) is introduced at the origin (x = 0, y = 0, z = 0) (Single Correct Option, 2017) with a given initial velocity v. A uniform electric field E and a uniform magnetic field (a) µ 0I 6[ 3 − 1] (b) µ 0I 6[ 3 + 1] B exist everywhere. The velocity v, electric 4 πa 4 πa field E and magnetic field B are given in columns 1, 2 and 3, respectively. The (c) µ 0I 3[ 3 − 1] (d) µ 0I 3[2 − 3] quantities E0 , B0 are positive in magnitude. 4 πa 4 πa (Matching Type, 2017) 15. A uniform magnetic field B exists in the region between x = 0 and x = 3R (region 2 Column 1 Column 2 Column 3 2 (I) Electron with (i) E = E0 $z (P) B = − B0 $x in the figure) pointing normally into the plane of the paper. A particle with charge v = 2 E0 $x (ii) E = − E0 $y (Q) B = B0 $x +Q and momentum p directed along B0 X-axis enters region 2 from region 1 at (iii) E = − E0 $x (R) B = B0 $y point P1 ( y = − R). (II) Election with (iv) E = E0 $x (S) B = B0 $z v = E0 $y Which of the following option(s) is/are B0 correct? (More than One Correct Option, 2017) (III) Proton with y v=0 Region 1 Region 2 Region 3 (IV) Proton with B v = 2 E0 $x B0 11. In which case would the particle move in a O x straight line along the negative direction P2 of Y -axis? +Q P1 (y=–R) (a) (IV) (ii) (S) (b) (II) (iii) (Q) (c) (III), (ii) (R) (d) (III) (ii) (P) (3R/2) 12. In which case will the particle move in a (a) When the particle re-enters region 1 through the longest possible path in region 2, the straight line with constant velocity? magnitude of the change in its linear momentum between point P1 and the farthest (a) (II) (iii) (S) (b) (III) (iii) (P) point fromY -axis is p 2 (c) (IV) (i) (S) (d) (III) (ii) (R) (b) For B = 8 p , the particle will enter region 3 13. In which case will the particle describe a 13 QR helical path with axis along the positive through the point P2 on X-axis (c) For B > 2 p , the particle will re-enter region 1 z-direction? 3 QR (a) (II) (ii) (R) (b) (III) (iii) (P) (d) For a fixed B, particles of same charge Q and (c) (IV) (i) (S) (d) (IV) (ii) (R) same velocity v, the distance between the point P1 and the point of re-entry into region 1 14. A symmetric star I is inversely proportional to the mass of the shaped conducting particle wire loop is carrying 4a a steady state current I as shown in the figure. The distance between the

Previous Years’ Questions (2018-13) 17 16. In the circuit shown, L = 1 µH, C = 1 µF shown. There is S and R = 1 kΩ . They are connected in series no mutual with an AC source V = V0 sin ωt as shown. inductance + V R Which of the following options is/are between the two – L1 L2 correct? (More than One Correct Option, 2017) inductors. The L=1µH C=1µF R=1kΩ switch S is initially open. At t = 0, the switch is closed and current begins to V0 sin ωt flow. Which of the following options is/are correct? (More than One Correct Option, 2017) (a) At ω ∼ 0 the current flowing through the circuit (a) After a long time, the current through L1 will becomes nearly zero   be V  L2  (b) The frequency at which the current will be in R  L1 + L2  phase with the voltage is independent of R (b) After a long time, the current through L2 will (c) The current will be in phase with the voltage if   ω = 104 rads−1 be V  L1  R  L1 + L2  (d) At ω >> 106 rads−1, the circuit behaves like a capacitor (c) The ratio of the currents through L1 and L2 is fixed at all times (t > 0) 17. A circular insulated copper B (d) At t = 0, the current through the resistance R wire loop is twisted to form is V R two loops of area A and 2A area A as shown in the figure. At 19. The instantaneous voltages at three the point of crossing the terminals marked X , Y and Z are given wires remain electrically area 2A by VX = V0 sin ωt,VY = V0 sinωt + 2π  and insulated from each other. 3  The entire loop lies in the = V0 sinωt + 4π  3  plane (of the paper). A w VZ . uniform magnetic field B An ideal voltmeter is configured to read points into the plane of the paper. At rms value of the potential difference between its terminals. It is connected t = 0, the loop starts rotating about the between points X and Y and then between Y and Z. The reading(s) of the voltmeter common diameter as axis with a will be (More than One Correct Option, 2017) constant angular velocity ω in the magnetic field. Which of the following options is/are correct? (More than One Correct Option, 2017) (a) V rms = V0 1 YZ 2 (a) the emf induced in the loop is proportional to 3 the sum of the areas of the two loops (b) V rms = V0 2 XY (b) The rate of change of the flux is maximum when the plane of the loops is perpendicular (c) independent of the choice of the two terminals to plane of the paper (d) V rms = V0 (c) The net emf induced due to both the loops is XY proportional to cos ωt 20. An infinite line charge of uniform electric (d) The amplitude of the maximum net emf induced due to both the loops is equal to the charge density λ lies along the axis of an amplitude of maximum emf induced in the electrically conducting infinite cylindrical smaller loop alone shell of radius R. At time t = 0, the space 18. A source of constant voltage V is inside the cylinder is filled with a connected to a resistance R and two ideal inductors L1 and L2 through a switch S as material of permittivity ε and electrical conductivity σ. The electrical conduction in the material follows Ohm's law. Which one of the following graphs best describes

18 Electricity & Magnetism the subsequent variation of the magnitude their conducting surface, the balls will get charge, will become equipotential with the of current density j (t) at any point in the plate and are repelled by it. material? (Single Correct Option, 2016) j(t) j(t) The balls will A eventually collide with the top plate, – HV (a) (b) where the + (0, 0) t coefficient of j(t) (0, 0) j(t) t restitution can be taken to be zero due to te soft nature of the material of the balls. The electric field in (a) (d) the chamber can be considered to be that of a parallel plate capacitor. Assume that there t t are no collisions between the balls and the (0, 0) (0, 0) interaction between them is negligible. (Ignore gravity) (Passage Type, 2016) 21. In the circuit shown below, the key is 22. Which one of the following statements is pressed at time t = 0. Which of the following correct? statement(s) is (are) true? (a) The balls will execute simple harmonic (More than One Correct Option, 2016) motion between the two plates 25 kΩ40 µF (b) The balls will bounce back to the bottom plate 50 kΩ − carrying the same charge they went up with V + (c) The balls will stick to the top plate and remain there A 20 µF (d) The balls will bounce back to the bottom plate carrying the opposite charge they went up with +– 23. The average current in the steady state Key 5 V registered by the ammeter in the circuit (a) The voltmeter display −5 V as soon as the key will be is pressed and displays + 5 V after a long time (b) The voltmeter will display 0 V at time t = ln 2 (a) proportional toV02 (b) proportional to the potentialV0 seconds (c) zero (d) proportions toV01/2 (c) The current in the ammeter becomes 1/e of the initial value after 1 second 24. An incandescent bulb has a thin filament (d) The current in the ammeter becomes zero after of tungsten that is heated to high a long time temperature by passing an electric current. Passage (Q. Nos. 22-23) The hot filament emits black-body radiation. Consider an evacuated cylindrical chamber The filament is observed to break up at of height h having rigid conducting plates at the ends and an insulating curved surface random locations after a sufficiently long as shown in the figure. A number of spherical balls made of a light weight and time of operation due to non-uniform soft material and coated with a conducting material are placed on the bottom plate. The evaporation of tungsten from the filament. balls have a radius r << h. Now, a high voltage source (HV) connected across the If the bulb is powered at constant voltage, conducting plates such that the bottom plate is at +V0 and the top plate at −V0. Due to which of the following statement(s) is (are) true? (More than One Correct Option, 2016) (a) The temperature distribution over the filament is uniform (b) The resistance over small sections of the filament decreases with time

Previous Years’ Questions (2018-13) 19 (c) The filament emits more light at higher band 27. Two inductors L1 (inductance 1mH, of requencies before it breaks up internal resistance 3 Ω) and L2 (inductance 2 mH, internal resistance (d) The filament consumes less electrical power towards the end of the life of the bulb 4 Ω), and a resistor R (resistance 12 Ω) 25. A conducting loop in the shape of a right are all connected in parallel across a 5V angled isosceles triangle of height 10 cm is kept such that the 90° vertex is very battery. The circuit is switched on at time close to an infinitely long conducting wire (see the figure). The wire is electrically t = 0. The ratio of the maximum to the insulated from the loop. The hypotenuse of the triangle is parallel to the wire. The minimum current (Imax/ Imin ) drawn from current in the triangular loop is in counter- clockwise direction and increased the battery is (Single Integer Type, 2016) at a constant rate of 10 As−1. Which of the 28. A rigid wire loop of square shape having following statement(s) is (are) true? side of length L and resistance R is moving along the x-axis with a constant (More than One Correct Option, 2016) velocity v0 in the plane of the paper . 10 cm 90° At t = 0, the right edge of the loop enters a region of length 3L where there is a (a) There is a repulsive force between the wire uniform magnetic field B0 into the plane of and the loop the paper, as shown in the figure. For sufficiently large v0, the loop eventually (b) If the loop is rotated at a constant angular crosses the region. Let x be the location of the right edge of the loop. Let v(x ), I(x) speed about the wire, an additional emf of and F(x) represent the velocity of the loop, current in the loop, and force on the loop, (µ 0 / π) volt is induced in the wire respectively, as a function of x. Counter- clockwise current is taken as positive. (c) The magnitude of induced emf in the wire is (More than One Correct Option, 2016)  µ0  volt π (d) The induced current in the wire is in opposite R direction to the current along the hypotenuse L 26. Consider two identical galvanometers and v0 two identical resistors with resistance R. If the internal resistance of the x galvanometers Rc < R/ 2, which of the 0 L 2L 3L 4L following statement(s) about anyone of the galvanometers is (are) true? Which of the following schematic plot(s) is (are) correct? (Ignore gravity) (More than One Correct Option, 2016) F(x) I(x) (a) The maximum voltage range is obtained when all the components are connected in series (a) 0 L 2L 3L 4L x (b) 3L 4L x (b) The maximum voltage range is obtained when 0 L 2L the two resistors and one galvanometer are connected in series, and the second v(x) I(x) galvanometer is connected in parallel to the v0 (d) first galvanometer (c) (c) The maximum current range is obtained when all the components are connected in parallel x x 0 L 2L 3L 4L 0 L 2L 3L 4L (d) The maximum current range is obtained when the two galvanometers are connected in series, and the combination is connected in parallel with both the resistors

20 Electricity & Magnetism 29. An infinitely long uniform line charge shown in the figure, d/2 distribution of charge per unit length λ lies parallel to the y-axis in the y-z plane the capacitance at z = 3 a (see figure). If the magnitude 2 becomes C2. The ε2 S/ 2 S/ 2 of the flux of the electric field through the ratio C2 is rectangular surface ABCD lying in the x-y plane with its centre at the origin is λL C1 nε0 (a) 6 (b) 5 ε1 (ε0 = permittivity of free space), then the 5 3 d value of n is = 6) (Single Integer Type, 2015) (c) 7 (d) 7 z 5 3 (Single Correct Option, 2015) D L C √3 a 32. In an aluminium (Al) bar of square cross 2 section, a square hole is drilled and is filled with iron (Fe) as shown in the ay figure. The electrical resistivities of Al and Fe are 2.7 × 10−8 Ωm and A B 1.0 × 10−7 Ωm, respectively. The electrical x resistance between the two faces P and Q of the composite bar is (Single Correct Option, 2015) 30. Consider a uniform R2 Al P spherical charge 50 mm a R1 distribution of radius O R1 centred at the origin O. In this distribution, a Fe spherical cavity of 2 mm radius R2, centred at 7 mm (b) 1875 µ Ω P with distance OP = a = R1 − R2 (see (a) 2475 µ Ω 64 figure) is made. If the electric field inside 64 (d) 2475 µ Ω the cavity at position r is E (r ), then the (c) 1875 µ Ω 132 correct statement(s) is/are 49 (Single Correct Option, 2015) (a) E is uniform, its magnitude is independent of 33. In the following circuit, the current R2 but its direction depends on r through the resistor R(= 2 Ω ) is I (b) E is uniform, its magnitude depends on R2 amperes. The value(SoifnIgliesInteger Type, 2015) and its direction depends on r R(=2Ω) 1Ω (c) E is uniform, its magnitude is independent of ‘a’ but its direction depends on a 2Ω 8Ω (d) Eis uniform and both its magnitude and 6Ω 2Ω direction depend on a 4Ω 31. A parallel plate capacitor having plates of 6.5V 10Ω area S and plate separation d, has 12Ω 4Ω capacitance C1 in air. When two dielectrics of different relative permittivities (ε1 = 2 and ε2 = 4) are introduced between the two plates as

Previous Years’ Questions (2018-13) 21 34. The figures below depict two situations in Passage (Q. Nos. 36-37) which two infinitely long static line charges of constant positive line charge In a thin rectangular metallic strip a density λ are kept parallel to each other. constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are λ λλ λ l, w and d, respectively. A uniform magnetic field B is applied on the strip along the x x positive y-direction. Due to this, the charge +q –q carriers experience a net deflection along the z-direction. This results in accumulation of charge carriers on the surface PQRS and In their resulting electric field, point appearance of equal and opposite charges on charges q and −q are kept in equilibrium between them. The point charges are the face opposite to PQRS. A potential confined to move in the x direction only. If they are given a small displacement about difference along the z-direction is thus their equilibrium positions, then the correct statements is/are developed. Charge accumulation continues (Single Correct Option, 2015) until the magnetic force is balanced by the (a) both charges execute simple harmonic electric force. The current is assumed to be motion uniformly distributed on the cross section of (b) both charges will continue moving in the direction of their displacement the strip and carried by electrons. (c) charge + q executes simple harmonic motion (Passage Type, 2015) while charge − q continues moving in the direction of its displacement ly (d) charge − q executes simple harmonic motion Iw K I x while charge + q continues moving in the M R direction of its displacement S d z P Q 35. A conductor (shown in the figure) 36. Consider two different metallic strips (1 carrying constant current I is kept in the and 2) of the same material. Their lengths x-y plane in a uniform magnetic field B. If are the same, widths are w1 and w2 and F is the magnitude of the total magnetic thicknesses are d1 and d2, respectively. force acting on the conductor, then the Two points K and M are symmetrically correct statements is/are located on the opposite faces parallel to the x-y plane (see figure). V1 and V2 are (More than One Correct Option, 2015) the potential differences between K and M in strips 1 and 2, respectively. Then, I π/6 RR y x for a given current I flowing through L π/4 L them in a given magnetic field strength B, the correct statements is/are RR (a) If w1 = w 2 and d1 = 2d2, thenV2 = 2V1 (a) if B is along z$, F ∝ (L + R) (b) If w1 = w 2 and d1 = 2d2, thenV2 = V1 (b) if B is along x$ , F = 0 (c) If w1 = 2w 2 and d1 = d2, thenV2 = 2V1 (c) if B is along y$ , F ∝ (L + R) (d) If w1 = 2w 2 and d1 = d2, thenV2 = V1 (d) if B is along z$, F = 0 37. Consider two different metallic strips (1 and 2) of same dimensions (length l, width w and thickness d) with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed in magnetic field B2, both along

22 Electricity & Magnetism positive y-directions. Then V1 and V2 are 40. Charges Q, 2Q and 4Q are uniformly the potential differences developed between K and M in strips 1 and 2, distributed in three dielectric solid respectively. Assuming that the current I is the same for both the strips, the correct spheres 1, 2 and 3 of radii R / 2, R and 2R options is/are respectively, as shown in figure. If (a) If B1 = B2 and n1 = 2n2, thenV2 = 2V1 (b) If B1 = B2 and n1 = 2n2, thenV2 = V1 magnitudes of the electric fields at point P (c) If B1 = 2B2 and n1 = n2, thenV2 = 0.5V1 (d) If B1 = 2B2 and n1 = n2, thenV2 = V1 at a distance R from the centre of spheres 38. A parallel plate capacitor has a dielectric 1, 2 and 3 are E1, E2 and E3 respectively, slab of dielectric constant K between its plates that covers 1/3 of the area of its then (Single Correct Option, 2014) plates, as shown in the figure. The total capacitance of the capacitor is C while P P P that of the portion with dielectric in R R R between is C1. When the capacitor is charged, the plate area covered by the Q 2Q 4Q dielectric gets charge Q1 and the rest of R/2 the area gets charge Q2. The electric field 2R in the dielectric is E1 and that in the other portion is E2. Choose the correct Sphere-1 Sphere-2 Sphere-3 option/options, ignoring edge effects. (a) E1 > E 2 > E 3 (b) E 3 > E1 > E 2 (More than One Correct Option, 2014) (c) E 2 > E1 > E 3 (d) E 3 > E 2 > E1 Q1 E1 41. Four charges Q1, Q2, Q3 and Q4 of same magnitude are fixed along the x-axis at E2 x = − 2a, − a, + a and +2a respectively. A Q2 positive charge q is placed on the positive y-axis at a distance b > 0. Four options of the signs of these charges are given in Column I. The direction of the forces on the charge q is given in Column II. Match Column I with Column II and select the correct answer using the code given below the lists. (Matching Type, 2014) +q (0,b) (a) E1 = 1 (b) E1 = 1 Q1 Q2 Q3 Q4 E2 E2 K (–2a,0) (–a,0) (+a,0) (+2a,0) (c) Q1 = 3 (d) C = 2 + K Q2 K C1 K 39. Let E1(r ), E2(r ) and E3(r ) be the respective Column I Column II electric fields at a distance r from a point P. Q1, Q2, Q3, Q4 all positive 1. +x charge Q, an infinitely long wire with −x +y constant linear charge density λ, and an Q. Q1, Q2 positive; Q3, Q4 negative 2. −y infinite plane with uniform surface R. Q1, Q4 positive; Q2, Q3 negative 3. charge density σ. If E1(r0) = E2(r0) = E3(r0) S. Q1, Q3 positive; Q2, Q4 negative 4. at a given distance r0, then Codes (More than One Correct Option, 2014) PQ (a) Q = 4σπr02 (b) r0 = λ (a) 3, 1, RS 2 πσ (b) 4, 2, 4, 2 (c) 3, 1, 3, 1 (c) E1  r0  = 2E2  r0  (d) E 2  r0  = 4E 3  r0  (d) 4, 2, 2, 4 2 2 2 2 1, 3

Previous Years’ Questions (2018-13) 23 42. Two ideal batteries of emf V1 and V2 and 45. A galvanometer gives full scale deflection three resistances R1, R2 and R3 are connected as shown in the figure. The with 0.006 A current. By connecting it to current in resistance R2 would be zero if a 4990 Ω resistance, it can be converted (More than One Correct Option, 2014) into a voltmeter of range 0-30 V. If V1 R1 R2 connected to a 2n Ω resistance, it 249 V2 R3 becomes an ammeter of range 0-1.5 A. The (a) V1 = V2 and R1 = R2 = R3 value of n is (Single Integer Type, 2014) (b) V1 = V2 and R1 = 2R2 = R3 (c) V1 = 2V2 and 2R1 = 2R2 = R3 Passage (Q. Nos. 46-47) (d) 2V1 = V2 and 2R1 = R2 = R3 The figure shows a circular loop of radius a 43. During an experiment with a metre bridge, the galvanometer shows a null with two long parallel wires (numbered 1 point when the jockey is pressed at 40.0 cm using a standard resistance of 90 Ω, as and 2) all in the plane of the paper. The shown in the figure. The least count of the scale used in the metre bridge is 1 mm. distance of each wire from the centre of the The unknown resistance is loop is d. The loop and the wires are (Single Correct Option, 2014) carrying the same current I. The current in R 90 Ω the loop is in the counter-clockwise direction if seen from above. (Passage Type, 2014) 40.0cm 46. When d ≈ a but wires are not touching the loop, it is found that the net magnetic (a) 60 ± 0.15 Ω field on the axis of the loop is zero at a (c) 60 ± 0.25 Ω height h above the loop. In that case (a) current in wire 1 and wire 2 is the direction PQ and RS, respectively and h ≈ a (b) current in wire 1 and wire 2 is the direction PQ and SR, respectively and h ≈ a (c) current in wire 1 and wire 2 is the direction PQ and SR, respectively and h ≈ 1.2a (d) current in wire 1 and wire 2 is the direction PQ and RS, respectively and h ≈ 1.2a (b) 135 ± 0.56 Ω 47. Consider d >> a, and the loop is rotated (d) 135 ± 0.23 Ω about its diameter parallel to the wires by 44. Two parallel wires in the plane of the 30° from the position shown in the figure. paper are distance X0 apart. A point If the currents in the wires are in the charge is moving with speed u between opposite directions, the torque on the loop the wires in the same plane at a distance X1 from one of the wires. When the wires at its new position will be (assume that carry current of magnitude I in the same the net field due to the wires is constant direction, the radius of curvature of the over the loop) path of the point charge is R1. In contrast, if the currents I in the two wires have (a) µ 0I 2a2 (b) µ 0I 2a2 directions opposite to each other, the d 2d radius of curvature of the path is R2. If X0 = 3, and value of R1 is (c) 3 µ 0I2a2 (d) 3 µ 0I2a2 X1 R2 d 2d (Single Integer Type, 2014) 48. At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating current I(t) = I0 cos (ωt), with I0 = 1 A and

24 Electricity & Magnetism ω = 500 rad s−1 starts flowing in it with S1 S2 S3 V0 2V0 C1 C2 the initial direction shown in the figure. At t = 7π , the key is switched from B to D. (a) the charge on the upper plate of C1 is 2 CV0 (b) the charge on the upper plate of C1 is CV0 6ω (c) the charge on the upper plate of C2 is 0 (d) the charge on the upper plate of C2 is −CV0 Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If C = 20 µF, R = 10 Ω and the battery is ideal with emf of 50 V, identify the correct statement(s). (More than One Correct Option, 2014) BD 51. Two non-conducting A 50V spheres of radii R1 ρ –ρ C=20µF and R2 and carrying R1 R2 uniform volume R = 10 Ω (a) Magnitude of the maximum charge on the charge densities + ρ capacitor before t = 7 π is 1 × 10−3 C and − ρ, respectively, are placed such that 6ω they partially overlap, as shown in the (b) The current in the left part of the circuit just before t = 7 π is clockwise figure. At all points in the overlapping 6ω region (More than One Correct Option, 2013) (c) Immediately after A is connected to D, the current in R is 10 A (a) the electrostatic field is zero (b) the electrostatic potential is constant (d) Q = 2 × 10−3 C (c) the electrostatic field is constant in magnitude (d) the electrostatic field has same direction 49. Two non-conducting solid spheres of radii 52. A particle of mass M and positive charge Q, moving with a constant velocity R and 2 R, having uniform volume charge u1 = 4ims−1, enters a region of uniform densities ρ1 and ρ2 respectively, touch static magnetic field normal to the x-y each other. The net electric field at a plane. The region of the magnetic field extends from x = 0 to x = L for all values distance 2 R from the centre of the of y. After passing through this region, the particle emerges on the other side smaller sphere, along the line joining the after 10 milliseconds with a velocity centre of the spheres, is zero. The ratio ρ1 u2 = 2( 3i + j) ms−1. The correct ρ2 statement(s) is (are) can be (More than One Correct Option, 2013) (More than One Correct Option, 2013) (a) − 4 (b) − 32 (c) 32 25 25 (d) 4 50. In the circuit shown in the figure, there (a) the direction of the magnetic field is −z direction. are two parallel plate capacitors each of (b) the direction of the magnetic field is + z capacitance C. The switch S1 is pressed direction first to fully charge the capacitor C1 and then released. The switch S2 is then (c) the magnitude of the magnetic field is 50πM pressed to charge the capacitor C2. After 3Q some time, S2 is released and then S3 is pressed. After some time units. (More than One Correct Option, 2013) (d) the magnitude of the magnetic field is 100πM units. 3Q

Previous Years’ Questions (2018-13) 25 53. A steady current I flows along an 55. The magnitude of the induced electric infinitely long hollow cylindrical conductor of radius R. This cylinder is field in the orbit at any instant of time placed coaxially inside an infinite solenoid of radius 2R. The solenoid has n during the time interval of the magnetic turns per unit length and carries a steady current I. Consider a point P at a distance field change is r from the common axis. The correct statement(s) is (are) (a) BR (b) − BR (c) BR (d) 2BR 4 2 (More than One Correct Option, 2013) Passage (Q. Nos. 56-57) (a) In the region 0 < r < R, the magnetic field is non-zero A thermal power plant produces electric power of 600 kW at 4000 V, which is to be (b) In the region R < r < 2R, the magnetic field is transported to a place 20 km away from the along the common axis power plant for consumers’ usage. It can be transported either directly with a cable of (c) In the region R < r < 2R, the magnetic field is large current carrying capacity or by using a tangential to the circle of radius r, centered combination of step-up and step-down on the axis transformers at the two ends. The drawback of the direct transmission is the large (d) In the region r > 2R, the magnetic field is energy dissipation. In the method using non-zero transformers, the dissipation is much smaller. In this method, a step-up Passage (Q. Nos. 54-55) transformer is used at the plant side so that the current is reduced to a smaller value. At A point charge Q is moving in a circular orbit of the consumers’ end a step-down transformer radius R in the x-y plane with an angular is used to supply power to the consumers at velocity ω. This can be considered as equivalent the specified lower voltage. It is reasonable to a loop carrying a steady current Qω ⋅ A to assume that the power cable is purely resistive and the transformers are ideal 2π with a power factor unity. All the current uniform magnetic field along the positive z-axis and voltages mentioned are rms values. is now switched on, which increases at a constant rate from 0 to B in one second. Assume (Passage Type, 2013) that the radius of the orbit remains constant. The applications of the magnetic field induces 56. If the direct transmission method with a an emf in the orbit. cable of resistance 0.4 Ω km −1 is used, the The induced emf is defined as the work done by power dissipation (in %) during an induced electric field in moving a unit positive charge around a closed loop. It is known transmission is that, for an orbiting charge, the magnetic dipole moment is proportional to the angular (a) 20 (b) 30 (c) 40 (d) 50 momentum with a proportionality constant γ. 57. In the method using the transformers, (Passage Type, 2013) assume that the ratio of the number of turns in the primary to that in the 54. The change in the magnetic dipole secondary in the step-up transformer is 1 : 10. If the power to the consumers has to moment associated with the orbit, at the be supplied at 200V, the ratio of the number of turns in the primary to that in end of the time interval of the magnetic the secondary in the step-down transformer is field change, is (a) γBQR 2 (b) −γ BQR 2 2 (a) 200 : 1 (b) 150 : 1 (c) γ BQR 2 2 (d) γBQR 2 (c) 100 : 1 (d) 50 : 1

Answer with Explanations 1. (b, d) I1 = V  e −tR  (d) Z B1 R 1− L   Bx B2 RL I1 I R I2 2L V I1 I2 V X I2 I1 I2 = V  e − tR  At centre of ring, B due to wires is along X-axis. R 1− 2L  Hence Z-component is only because of ring which  B = µ 0i (−k$ ). From principle of superposition, 2R I = I1 − I2 3. (1.50) V − tR  − tR  +8µC ⇒ I = R e 2L  1 − e 2L  …(i) 1µF   V0=8V Before –8µC I is maximum when dI = 0, which gives e − tR = 1 or 2L dt 2 t = 2L ln2 3µC + C R – +5µC Substituting this time in Eq. (i), we get After 1µF Imax = V 3µC –5µC 4R –3µC 1µF 2. (a,b,d) Applying loop rule, Z 5− 3 −3=0 ⇒ 3 =2 ⇒ εr = 1.50 1 εr 1 εr I (0, 0, √3 R) C 4. (2 m/s) If average speed is considered along X-axis, Y R1 = mv 0 , R 2 = mv 0 = mv 0 ⇒ R1 > R2 qB1 qB2 4qB1 X=–R X O X=R R1 C2 C1 R2 –Y (a) At origin, B = 0 due to two wires if I1 = I2, hence Distance travelled along x-axis, ∆x = 2 (R1 + R2)= 5mv 0 (Bnet ) at origin is equal to B due to ring. which is 2 qB1 non-zero. Total time = T1 + T2 = πm + πm (b) If I1 > 0 and I2< 0, B at origin due to wires will be 2 2 qB1 qB2 along + k$ . Direction of B due to ring is along − k$ = πm + πm = 5πm direction and hence B can be zero at origin. qB1 4qB1 4qB1 (c) If I1 < 0 −akn$ daIn2d> 0, B at origin due to wires is 5mv 0 along also along − k$ due to ring, hence Magnitude of average speed = 2 qB1 = 2 m/s 5 πm B cannot be zero. 4qB1

Previous Years’ Questions (2018-13) 27 5. (a, b) PQ = (2 ) R sin 60° (4) E = 2 λ − l) − 2 λ + l) = λ(2 l) πε0(d πε0(d 2 πε0d 2 lZ P ⇒ E ∝ 1 d2 R 120° O (5) E = σ ⇒ E is independent of d 2 ε0 R Q 9. (b) When switch is closed for a very long time capacitor will get fully charged and charge on capacitor will be q = CV = (2 R ) 3 = ( 3R ) Energy stored in capacitor, 2 EC = 1 CV 2 K(i) q enclosed = λ ( 3R ) 2 We have, φ = q enclosed ⇒ φ =  3 λR  R   ε0  ε0  S Also, electric field is perpendicular to wire, so C Z-component will be zero. 6. (2) a = F = qE = 103 sin(103t ) mm Work done by a battery, vt W = Vq = VCV = CV 2 dv = 103 sin(103t ) dt ∫ ∫dv = 103 sin(103t ) ⇒ dt 0 0 Energy dissipated across resistance ∴ v = 103 [1 − cos (103t )] ED = (work done by eq. battery) − (energy stored) 103 1 CV 2 1 CV 2 ED = CV 2 − 2 = 2 K(ii) Velocity will be maximum when cos(103t ) = − 1 vmax = 2 m /s From Eqs. (i) and (ii), we get 7. (5.55) Given, N = 50, ED = EC A = 2 × 10−4 m2, C = 10−4, R = 50 Ω, 10. (a) For process (1) B = 0.02 T, θ = 0.2 rad Charge on capacitor = CV0 3 ∴ Ni g AB = Cθ Energy stored in capacitor = 1 C V02 = CV02 ⇒ ig = Cθ = 10−4 × 0.2 0.02 = 0.1 A 2 9 18 N AB 50 × 2 × 10−4 × Work done by battery = CV0 × V = CV02 ∴ Vab = i g × G = (i − i g ) S 33 9 0.1 × 50 = (1 − 0.1) × S Heat loss = CV02 − CV02 = CV02 S 9 18 18 i – ig For process (2) Charge on capacitor = 2CV0 3 i a ig G b Extra charge flow through battery = CV0 3 5 = 0.9 × S Work done by battery = CV0 ⋅ 2 V0 = 2CV02 ∴ S = 50 Ω = 5.55 Ω 33 9 9 1  2 V0  2 4CV02 2 3 18 Final energy stored in capacitor = C = 8. (b) List-II Energy stored in process 2 = 4CV02 − CV02 = 3CV02 18 18 18 (1) E = 1 Q ⇒ E ∝ 1 4 πε0 d2 ⇒ d2 Heat loss in process (2) = work done by battery in (2) Eaxis = 1 2Q(2 l) E ∝ 1 4 πε0 d3 d3 process (2) − energy store in capacitor process (2) = 2CV02 − 3CV02 = CV02 (3) E = λ ⇒ E∝1 9 18 18 2 πε0d d

28 Electricity & Magnetism For process (3) α = 60° and β = − 30° Charge on capacitor = CV0 = µ 0I  3 − 1  Extra charge flown through battery 4 πd  2 2    = CV0 − 2CV0 = CV0 B12 = µ 0I  3 − 1 3 3 4 πd    2  Work done by battery in this process d=a =  CV0  (V0 ) = CV02 B0 = 12 B12 3 3 µ 0I  3− 1 µ 0I 6 [ Final energy stored in capacitor = 1 CV02 = 12 × 4 πd  2  = 4 πa 3 − 1] 2   Energy stored in this process 15. (b,c) (a)|∆P|= 2 p = 1 CV02 − 4CV02 = 5CV02 2 18 18 p Heat loss in process (3) p p d = CV02 − 5CV02 = CV02 3 18 18 Now, total heat loss (ED) = CV02 + CV02 + CV02 = CV02 (b) r(1 − cosθ) = R ⇒ r sinθ = 3R 18 18 18 6 2 Final energy stored in capacitor = 1 CV02 2 2 sin θ cos θ So we can say that ED = 1  1 CV02  sinθ = 3 2 2 = 3 3 2 1 − cos θ 2 2 2 sin2 θ 11. (c) For particle to move in negative y-direction, either 2 its velocity must be in negative y-direction (if initial θr velocity ≠ 0) and force should be parallel to velocity or r it must experience a net force in negative y-direction only (if initial velocity = 0) cot θ = 3 ⇒ tan θ = 2 22 23 12. (a) Fnet = Fe + FB = qE + qv × B 2  2  4 12 For particle to move in straight line with constant 3 5 velocity, Fnet = 0 ⇒ tanθ = = 3 = 1− 4 5 ∴ qE + qv × B = 0 99 13. (c) For path to be helix with axis along positive z-direction, particle should experience a centripetal acceleration in xy-plane. For the given set of options only option (c) satisfy the condition. Path is helical with increasing pitch. 14. (a) 2 13 12 a1 θ aα β Od 5 sinθ = 12 r  12  = 3R ; r= 13R = P ;B= 8P 13 13 2 8 QB 13QR (c) P < 3R , B > 2 P QB 2 3QR B12 = µ 0I [sinα + sinβ ] (d) r = mv , d = 2 r = 2 mv ⇒ d ∝ m 4 πd QB QB

Previous Years’ Questions (2018-13) 29 16. (a,b) At ω ≈ 0, XC = 1 = ∞. Therefore, current is 19. (b,c) VXY = V0sin ωt + 2π  − V0 sin ωt ωC 3 nearly zero. = V0sinωt + 2π  + V0 sin(ωt + π) 3 Further at resonance frequency, current and voltage ⇒ φ = π − 2π = π are in phase. This resonance frequency is given by, 33 ωr = 1= 1 = 106 rad /s ⇒ V0 ′ = 2 V0 cos  π  = 3 V0 LC 10−6 × 10−6 6 We can see that this frequency is independent of R. ⇒ VXY = 3V0 sin(ωt + φ) Further, XL = ωL, XC = 1 ⇒ (VXY )rms = (VYZ )rms = 3 V0 ωC 2 At, ω = ωr = 106 rad /s, X L = XC. 20. (d) Suppose charger per unit length at λ For ω > ωr, X L > XC. So, circuit is inductive. any instant 17. (b,d) The net magnetic flux through the loops at time t is λ. is Initial value of λ is suppose λ 0. Electric field s at a distance r at any φ = B(2 A − A)cos ωt = BA cos ωt instant is so, dφ = BωAsinωt dt  E= λ 2 πεr ∴dφ is maximum when φ = ωt = π /2 J = σE = σ λ dt  2 πεr The emf induced in the smaller loop, i = dq = J( A) = − Jσ2 πrl dt εsmaller = − d (BA cos ωt ) = BωA sinωt dt d λl = − λ × σ2 πrl dt 2 πεr (q = λl ) ∴ Amplitude of maximum net emf induced in both the −σt loops ∫ ∫λ d λ = − σ t dt ⇒ λ = λ 0e ε = Amplitude of maximum emf induced in the smaller λ0 λ ε0 loop alone. 18. (a,b,c) S R J= σ λ = σλ 0 − σt = −σt 2 πεr 2 πεr eε J0e ε V + L1 L2 Here, J0 = σλ 0 – 2 πεr ∴ J(t )decreases exponentially as shown in figure below. Since inductors are connected in parallel j(t) VL1 = VL2 ; L1 dI1 = L2 dI2 dt dt L1I1 = L2I2 ; I1 = L2 I2 L1 t Current through resistor at any time t is given by (0, 0) I= V (1 − e − RT ), where L = L1L2 21. (a,b,c,d) Just after pressing key, L R L1 + L2 5 − 25000i1 = 0 After long time I = V 5 − 50000i 2 = 0 (As charge in both capacitors = 0) R ⇒ i1 = 0.2 mA ⇒ i 2 = 0.1 mA I1 + I2 = I K(i) L1I1 = L2I2 K(ii) and VB + 25000i1 = VA ⇒ VB − VA = − 5 V From Eqs. (i) and (ii), we get After a long time, i1 and i 2 = 0 (steady state) I1 = V L2 ⇒ I2 = V L1 ⇒ 5 − q1 = 0 ⇒ q1 = 200µC R + L2 R L1 + L2 40 L1 5− q2 = 0 (d) Value of current is zero at t = 0 and 20 ⇒ q 2 = 100µC Value of current is V / R at t = ∞ VB − q2 = VA ⇒ VB − VA = + 5 V 20 Hence option (d) is incorrect.

30 Electricity & Magnetism ⇒ (a) is correct. 24. (c,d) Because of non-uniform evaporation at different For capacitor 1, q1 = 200[1 − e −t /1] µC section, area of cross-section would be different at different sections. i1 = 1 e −t /1 mA 5 Region of highest evaporation rate would have rapidly reduced area and would become break up For capacitor 2, q 2 = 100[1 − e −t /1] µC cross-section. i2 = 1 e −t /1 mA Resistance of the wire as whole increases with time. 10 Overall resistance increases hence power decreases. ⇒ VB − q2 + i1 × 25 = VA 20  V2 1 constant  .  P = R or P ∝ R as V is At break up ⇒ VB − VA = 5[1 − e −t ] − 5e −t  = − 5[1 − 2e −t ] junction temperature would be highest, thus light of highest band frequency would be emitted at those At t = ln2, VB − VA = 5[1 − 1] = 0 cross-section. ⇒ (b) is correct. 25. (a,c) By reciprocity theorem of mutual induction, it can At t = 1, i = i1 + i2 = 1 e −1 + 1 e −1 = 3 ⋅1 be assumed that current in infinite wire is varying at 5 10 10 e 10A/s and EMF is induced in triangular loop. At t = 0, i = i1 + i2 = 1 + 1 = 3 5 10 10 i ⇒ (c) is correct. y After a long time, i1 = i 2 = 0 ⇒ (d) is correct. dy 2y 22. (d) Balls will gain positive charge and hence move Flux of magnetic field through triangle loop, if current in towards negative plate. On reaching negative plate, balls will attain negative infinite wire is φ, can be calculated as follows: charge and come back to positive plate. dφ = µ 0i ⋅ 2 ydy ⇒ dφ = µ 0i dy and so on, balls will keep oscillating. But oscillation is not S.H.M., 2 πy π As force on balls is not ∝ x. ⇒ option (d) is correct. ⇒ φ = µ 0i  l  π 2 23. (a) As the balls keep on carrying charge form one ⇒ EMF = dφ = µ0  l  ⋅ di dt π 2 dt plate to another, current will keep on flowing even in steady state. When at bottom plate, if all balls attain = µ0 (10 cm) 10 A  = µ0 volt π s π charge q, kq = V0  k = 1 ⇒ q = V0 r If we assume the current in the wire towards right then r  4 πε0  k as the flux in the loop increases we know that the induced current in the wire is counter clockwise. Inside cylinder, electric field Hence, the current in the wire is towards right. E = [V0 − (− V0)]h = 2 V0h. ⇒ Acceleration of each ball, Field due to triangular loop at the location of infinite wire is into the paper. Hence, force on infinite wire is a = qE = 2 hr ⋅ V02 away from the loop. m k m ⇒ Time taken by balls to reach other plate, By cylindrical symmetry about infinite wire, rotation of triangular loop will not cause any additional EMF. 2h = 2 h. k m 1 km t= a 2 hrV02 = V0 r 26. (a,c) For maximum range of voltage resistance should If there are n balls, then be maximum. So, all four should be connected in series. For maximum range of current, net resistance Average current, should be least. Therefore, all four should be connected in parallel. iav = nq = n× V0r × V0 r ⇒ iav ∝ V02 t k km

Previous Years’ Questions (2018-13) 31 27. (8) L1=1mH r1=4Ω dφ = 0 e = 0, i = 0 L2=2mH r2=4Ω dt R=12 Ω F = 0 ⇒ x > 4L ⇒ e = Blv v ε=5V Imax = ε = 5A (Initially at t = 0) R 12 (finally in steady state) Imin = ε = ε  1 + 1 + 1 X R eq  r1 r2 R  Force also will be in left direction. = 5  1 + 1+ 1  = 10 A i = BLv (clockwise) 3 4 12 3 R Imax = 8 a = − B2L2v = v dv Imin mR dx = B2L2v x B2L2 dx = vf − dv 28. (b,c) v ∫ ∫F R L mR vi ⇒ − B2L2 ( x − L) = vf − vi mR R B2L2 ( x mR vf = vi − − L) (straight line of negative slope) x I = BLv → (Clockwise) (straight line of negative slope) R When loop was entering (x < L) 29. (6) ANBP is cross-section of a cylinder of length L. The φ = BLx line charge passes through the centre O and e = − d φ = − BL dx perpendicular to paper. dt dt P |e| = BLv i = e = BLv (anticlockwise) O RR F = ilB (Left direction) = B2L2v (in left direction) 30° 30° R AMB ⇒ a = F = − B2L2v ⇒ a = v dv N m mR dx a v dv = − B2L2v v − B2L2 x AM = a, MO = 3a dx mR 22 dv dx ∫ ∫⇒ = v0 mR 0 ⇒v = v0 − B2L2v x ∴ ∠AOM = tan−1  AM  = tan−1 1  = 30° mR OM 3 (straight line of negative slope for x < L) Electric flux passing from the whole cylinder I = BL v ⇒ (I vs x will also be straight line of negative φ1 = q in = λL R ε0 ε0 slope for x < L) L ≤ x ≤ 3L ∴ Electric flux passing through ABCD plane surface (shown only AB) = Electric flux passing through cylindrical surface ANB =  60°  (φ1 ) = λL 360° 6 ε0 ∴ n=6

32 Electricity & Magnetism 30. (d) The sphere with cavity can be assumed as a 33. (1) complete sphere with positive charge of radius R1 + 6.5V 2Ω 1Ω another complete sphere with negative charge and radius R2. 6Ω 2Ω 8Ω 2Ω 4Ω E + → E due to total positive charge E − → E due to total negative charge. 10Ω 12Ω E =E + + E− 4Ω If we calculate it at P, then E − comes out to be zero. ∴ E=E+ ⇒⇒ 1 q and E+ = 4 πε0 R13 (OP), in the direction of OP. 2Ω 1Ω Here, q is total positive charge on whole sphere. 6Ω 2Ω 2Ω 4Ω It is in the direction of OP or a . 6.5V 12Ω 10Ω 6.5V Now, inside the cavity electric field comes out to be 4Ω uniform at any point. This is a standard result. 31. (d) C d/2 2Ω 10Ω 2Ω C′ 6Ω 4Ω + E1 E2 S/ 2 12Ω – E1 S/ 2 C′′ 2Ω d 6Ω C C′ 6.5V 2Ω 12Ω 4Ω +– 2Ω 4.5Ω C′′ 1 ⇒ ε0s, C 2 ε0 s 2 ε0s 6.5V d 2 d C1 = = = d /2 4ε0 s 4ε0s 2 C′= = d /2 d 2 ε0 s ε0s 34. (c) At the shown position, net force on both charges is 2 and C ′′ = = zero. Hence they are in equilibrium. But equilibrium of dd +q is stable equilibrium. So, it will start oscillations when displaced from this position. These small C2 = CC ′ + C ′′ = 4 ε0s + ε0s oscillations are simple harmonic in nature. While C + C′ 3 d d equilibrium of −q is unstable. So, it continues to move in the direction of its displacement. = 7 ε0s C 2 = 7 3 d C1 3 35. (a,b,c) 32. (b) 1 = 1 + 1 =  AAl + AFe  1 y R Al R Fe  ρAl ρFe  l Qx R PI 2(L+R) (7 2 − 2 2) 2 2  10−6 =  +  × 1 Force on the complete wire = force on straight wire PQ  2.7 10  10−8 50 × 10−3 carrying a current I. Solving we get, R = 1875 × 10− 6 Ω = 1875 µΩ F = I(PQ × B) 64 64 = I[{2(L + R )$i} × B]

Previous Years’ Questions (2018-13) 33 This force is zero if B is along i$ direction or x-direction. ⇒ (d) is incorrect If magnetic field is along $j direction or k$ direction, 40. (c) E1 = kQ , where k = 1 R2 4 πε0 |F| = F = (I)(2 )(L + R )Bsin 90° or F = 2 I(L + R )B E2 = k(2Q ) ⇒ E2 = 2 kQ or F ∝ (L + R ) R2 R2 ∴ Options (a), (b) and (c) are correct. E3 = k(4Q) R ⇒ E3 = kQ E3 < E1 < E2 (2 R )3 2R2 36. (a,d) FB = Bev = Be I = BI 41. (c) (P) Component of forces along x-axis will vanish. nAe nA Net force along positive y-axis Fe = eE ⇒ Fe = FB eE = BI ⇒ E= B F4 F3 F2 F1 nA nAe +q V = Ed = BI ⋅ w = BIw = BI nAe n(wd )e ned V1 = d 2 +Q +Q +Q +Q V2 d1 (Q) Component of forces along y-axis will vanish. Net ⇒ if w1 = 2 w2 and d1 = d 2 force along positive x-axis V1 = V2 F2 F1 ∴Correct answers are (a) and (d). +q 37. (a,c) V = BI ⇒ V1 = B1 × n2 F4 ned V2 B2 n1 F3 If B1 = B2 and n1 = 2 n2, then V2 = 2 V1 If B1 = 2 B2 and n1 = n2, then V2 = 0.5V1 ∴Correct answers are (a) and (c). 38. (a,d) C = C1 + C2 C1 = Kε0 A / 3 , C = ε02 A / 3 d d 2 +Q +Q –Q –Q ⇒ C = (K + 2 )ε0A ⇒ C = K + 2 (R) Component of forces along x-axis will vanish. Net 3d C1 K force along negative y-axis Also, E1 = E2 = V , where V is potential difference F4 F1 d +q between the plates. 39. (c, d) Q = λ = σ F2 F3 4 πε0r02 2 πε0r0 2 ε0 Q = 2 πσr02 (a) is incorrect, r0 = λ +Q –Q –Q +Q πσ (b) is incorrect, E1  r0  = 4E1( r0 ) (S) Component of forces along y-axis will vanish. Net 2 force along negative x-axis As E1 ∝ 1 F3 F1 r2 +q F4 E2  r0  = 2 E2(r0) as E2 ∝ 1 2 r F2 ⇒ (c) is correct E3  r0  = E3(r0) = E 2( r0 ) 2 +Q –Q +Q –Q as E3 ∝ r0

34 Electricity & Magnetism 42. (a,b,d) Let us take VP = 0. Then potentials across Substituting x1 = x0 (as x0 = 3) 3 x1 R1, R2 and R3 are as shown in figure (ii) In the same figure B1 = 3µ 0 I − 3µ 0I = 3µ 0I 2 πx0 4 πx0 4 πx0 V1 i1 R1 R1 = mv and B2 = 9µ 0I qB1 4 πx0 V1 R1 P R2 O O R2 = mv ⇒ R1 = B2 = 9 = 3 i2 R3 qB2 R2 B1 3 V2 R2 i3 R3 o –V2 45. (5) ig 4990 Ω V (i) (ii) G i1 + i 2 = i 3 i g (G + 4990) = V ⇒ 6 (G + 4990) = 30 ∴ V1 − V0 + 0 − V0 = V0 − (−V2) R1 R2 R3 1000 Solving this equation we get ⇒ G + 4990 = 30,000 = 5000 6 V1 + 0 − V2 R1 R3 ⇒ G = 10 Ω V0 = 1+ 1 +1 R1 R2 R3 Vab = Vcd ⇒ ιγΓ = (1.5 − ιγ ) Σ c S Current through R2 will be zero if d V0 = 0 ⇒ V1 = R1 (1.5–ig) V2 R3 1.5 A a ig In options (a), (b) and (d) this relation is satisfied. G 43. (c) For balanced meter bridge b X= l (where, R = 90 Ω ) ⇒ 6 × 10 = 1.5 − 10600  S R (100 − l) 1000 ∴ X = 40 90 100 − 40 ⇒ S = 60 = 2 n ⇒ n = 249 × 30 = 2490 = 5 1494 249 1494 498 ∴ X = 60 Ω 46. (c) BR = B due to ring l B1 = B due to wire-1 BR X=R B2 = B due to wire-2 (100 − l) ∆X = ∆l + ∆l = 0.1 + 0.1 P X l 100 − l 40 60 ∆X = 0.25 θθ B2 B1 So, X = (60 ± 0.25) Ω rh 44. (3) B2 = µ 0I + µ0I 2 πx1 2 π ( x0 − x1 ) (when currents are in opposite directions) B1 = µ0I − 2π µ0I x1 ) 2 πx1 ( x0 − 1a 2 II µ 0I 2 πr In magnitudes B1 = B2 = Resultant of B1 and B2 x1 x0 x1 = 2B1 cos θ = 2  µ 0I   h  = µ 0Ih (when currents are in same direction) 2 πr r πr 2

Previous Years’ Questions (2018-13) 35 BR = µ 0IR 2 = 2µ 0Iπa2 Substituting the values of Q1, C and R we get 2(R 2 + x2)3/2 4 πr3 I = 10 A As, R = a, x = h and a2 + h2 = r2 In steady state Q2 = CV = 1 mC ∴ Net charge flown from battery = 2 mC For zero magnetic field at P, 49. (b,d) At point P µ 0Ih = 2µ 0Iπa2 ⇒ πa2 = 2 rh ⇒ η ≈ 1.2α πr 2 4 πr3 47. (b) Magnetic field at mid-point of two wires Q C1 P C2 = 2 (magnetic field due to one wire) = 2 µ0 I 2R 2R 2 π d  = µ 0I ⊗ πd Magnetic moment of loop M = IA = I πa2 If resultant electric field is zero then Torque on loop = MB sin 150° = µ 0I2a2 KQ1 = KQ2 R ⇒ ρ1 = 4 2d 4R 2 8R 3 ρ2 48. (c,d) dQ = I ⇒ Q = ∫ I dt = ∫ (I0 cos ωt ) dt At point Q dt If resultant electric field is zero then ∴ Qmax = I0 = 1= 2 × 10−3 C KQ1 + KQ2 = 0 ω 500 4R 2 25R 2 Just after switching ρ1 =− 32 (ρ1 must be negative) ρ2 25 –Q1 50 V 50. (b, d) After pressing S1 charge on upper plate of C1 is Q1=1mC+ ++ + + 2 CV0. Q1 After pressing S 2 this charge equally distributes in two +– capacitors. Therefore charge an upper plates of both capacitors will be + CV0 . R=10Ω When S 2 is released and S 3 is pressed, charge on upper plate of C1 remains unchanged (= + CV0) but In steady state charge on upper plate of C 2 is according to new battery (= − CV0). Q2 + ++ +–Q2 51. (c,d) For electrostatic field, 50 V Field at P EP = E1 + E2 = ρ C1P + (− ρ) C 2 P 3ε0 3ε0 R=10Ω = ρ (C1P + PC 2) 3ε0 At t = 7 π or ωt = 7 π 6ω 6 EP = ρ C1C 2 3ε0 Current comes out to be negative from the given expression. So, current is anti-clockwise. For electrostatic. Since, electric field is non-zero so it is not equipotential. Charge supplied by source from t = 0 to 7π ⇒ Q = 6 ω cos (500t ) dt ∫t = 7 π 52. (a,c) u = 4i; v = 2( 3i + j) 6ω 0 = sin 500t 7π = sin 7 π = − 1 mC vj  500 6 6ω  0 500 Apply Kirchhoff’s loop law just after changing the switch to position D ui L 50 + Q1 − IR = 0 C

36 Electricity & Magnetism According to the figure, magnetic field should be in ⊗ where γ = Q =  Q  (Iω) 2m 2m direction, or along − z direction. Further, tanθ = vy = 2 = 1 =  Q  (mR 2ω) = QωR 2 2m 2 vx 2 3 3 Induced electric field is opposite. Therefore, ∴ θ = 30° or π ω′ = ω − αt 6 = angle of v with x-axis = angle rotated by the particle (Q )  BR  R 2 = Wt =  BQ  t α = τ = (QE )R = = QB M I mR 2 2m mR 2 ∴ B = πM = 50 πM units (as t = 10−3 second) 6Qt 3Q ∴ ω′ = ω − QB ⋅ 1 = ω − QB 2m 2m 53. (a,d) Q Qω′ R 2 Q ω QB  R2 2 2m 2 P Mf = = − P → Hollow cylindrical ∴ ∆M = Mf − Mi = − Q 2BR 2 conductor 4m R Q → Solenoid M = − γ QBR 2  as γ Q  2R 2 2m = In the region, 0 < r < R 55. (b) The induced electric field is given by, BP = 0, ∫ E ⋅ dl = − dφ or El = − s  dB  BQ ≠ 0, along the axis dt dt ∴ Bnet ≠ 0 In the region, R < r < 2 R ∴ E(2 πR ) = − ( πR 2)(B) BP ≠ 0, tangential to the circle of radius r, centred on or E = − BR the axis. 2 BQ ≠ 0, along the axis. 56. (b) P = Vi ∴ Bnet ≠ 0 neither in the directions mentioned in options (b) or (c). ∴ i = P = 600 × 103 = 150 A V 4000 In region, r > 2 R Total resistance of cables, BP ≠ 0 R = 0.4 × 20 = 8Ω BQ ≠ 0 ∴ Bnet ≠ 0 ∴Power loss in cables = i 2R = (150)2 (8) 54. (b) M = Q y = 180000 W = 180 kW L 2m Q This loss is 30% of 600 kW. E 57. (a) During step-up, x Np = Vp Ns Vs or 1 = 4000 10 Vs ∴ M =  Q  L ⇒ M ∝ L, or Vs = 40,000 V 2m In step, down transformer, Np = Vp = 40000 = 200 N s Vs 200 1