DC Pandey Electricity And Magnetism

690 — Electricity and Magnetism 17. (a) From energy conservation, 23. In first quadrant magnetic field due to I1 is U θ = Kθ = U 0° + K0° outwards and due to I2 is inwards. So, net or (− MB cosθ) + 0 = (− MB cos 0° ) + K0° magnetic field may be zero. Substituting the given values, we can calculate. Similarly, in third quadrant magnetic field due to (b) To other side also it rotates upto the same I1 is inwards and due to I2 magnetic field is angle. outwards. Hence, only in first and third quadrants magnetic field may be zero. 18. (a) T = 2πr Let magnetic field is zero at point P(xy), then v (b) I = qf = (e)  2vπr BI1 = BI2 ∴ µ0 I1 = µ0 I2 =  2eπvr (πr2) = evr 2 2π y 2π x (c) M = IA ∴ y = I1 x I2 19. Assume equal and opposite currents in wires cf 24. Two straight wires produces outward magnetic field and eh. by arc of circle produces inward magnetic field. 20. Assume equal and opposite currents in wires PQ Due to straight wires, and RS, then find M. 2 µ0 i sin 90° ) z 4π R Q B1 = (sin θ + P y = µ0i (outwards) S 2πR R Due to circular arc, B2 = θ  µ2R0i (inwards) 2π x For net field to be zero, Now, B (2$j) ⇒ τ = M × B B1 = B2 or θ = 2 rad 21. Bnet = B2 + B2 = 2B 25. (a) If currents are in the same direction, then above where, B = µ0 i (sin 0° + sin 90° ) 4π r and below the wires magnetic fields are in the 22. Magnetic fields at O due to currents in wires ab same direction. Hence, they can’t produce zero and cd are zero. magnetic field. a In between the wires, let B = 0 at a distance (r) cm from the wire carrying 75 A current. Then, b α µ0  7r5 =  µ0   25 π  O 2π 2π 40 − cβ Solving, we get r = 30 cm. d r1 (b) If currents are in opposite direction, then in r2 between the wire magnetic field are in the same direction. So, they cannot produce zero Magnetic field due to current in wire da (say B2) is magnetic field. The points should be above or inwards due to current in wire bc (say B1) is below the wires, nearer to wire having smaller outwards. current. Let it is at a distance r from the wire having 25 A current. Then, µ0 i B1 = 4π r1 (sin α + sin β) µ0  2r5 =  µ0   75  2π 2π  40 + r µ0 i B2 = 4π r2 (sin α + sin β) Solving this equation, we get r = 20 cm B1 > B2 as r1 < r2 26. Apply B = µ 0NiR2 ∴ Bnet = B1 − B2 2(R2 + x2)3/ 2 (outwards)

Chapter 26 Magnetics — 691 27. I2 produces inwards magnetic field at centre. (b) The above calculated magnetic field can be written as Hence, I1 should produce outward magnetic field. Or current should be towards right. Further,  sin π /n 2 B = µ0i  π /n  µ 0I2 =  µ0   I1  2R 2π D 2r cos π /n ∴ I1 =  πD  I2 As, n→ ∝, π → 0 R n 28. (a) B1 = µ 0Ni Hence, lim  sin π/ n → 1 2R  π /n π / n→ 0 µ 0NiR2 (b) B2 = 2(R2 + x2)3/ 2, given B2 = B1 and lim (cot π /n)→ 1 2 π / n→ 0 µ 0i 29. 10 A and 8 A current produce inward magnetic or B → 2r field. While 20 A current produces outward 32. Current per unit area, magnetic field. Hence, current in fourth wire should be (20 − 10 − 8) A or 2 A and it should σ = πa2 − I − (a/ 2)2 produce inward magnetic field. So, it should be π (a/2)2 downwards toward the bottom. = 2I 30. (a) B at origin BKLM = BKNM πa2 = µ0I (−i$ + $j) Total area is (πa2). Therefore, the total current is 4R I1 = (σ) (πa2) = 2I Now, we can apply F = q (v × B) for finding force on it. Cavity area is π (a/2)2. Therefore, cavity current is (b) In uniform magnetic field, I2 = (σ) (πa2 / 4 ) = I 2 FKLM = FKNM = FKM = I (l × B) Now, the given current system can be assumed as shown below. = I [{2R (− k$ )} × {B0$j}] = (2B0 IR) $i 2 I/2 ∴ Net force is two times of the above value. 1+ ππ 3 I/2 nn 31. (a) 2I x (a) At P1 , B1 = µ0 2I = µ0I (towards left) and 2π 5 πr (towards right) a (towards right) µ0 I /2 1  2nπr πr B2 = 2π (r − a/2) 2 n a= = π  πnr  π  B2 = µ0 (r I /2 n n 2π + a/2) x = a cot = cot  µ0 i  sin π π   ∴ Bnet = B1 − B2 − B2 (towards left)  4π x n n  B = n + sin = µ0I 1 − 1− 1 π  r 4r − 2a 4r + 2a = n  µ0 i π/n (2 sin π /n) = µ0I 16r3 − 4a2 − 4r2 − 2ar − 4r2 + 2ar   4π (πr / n) cot π  r(16 r2 − 4a2)     πn  πn  µ 0in2 sin 2π 2r tan = µ0I  2r2 − a2  πr =  4 r2 − a2  (towards left)  

692 — Electricity and Magnetism (b) B1 a √r 2+ a2/4 r v 2 θ θ B2 P2 θ θθ B3 r 36. θ +v Bnet = B1 − 2B2 cos θ] (towards the top) L = µ0 2I − 2  µ 0 I /2  r Deviation, θ = sin−1 Lr  for L < r   where, r = mv 2π R  2π r2 + a2 /4  r2 + a2 /4 Bq = µ0I  2r2 + a2  (towards the top) πr    4r2 + a2  33. B vv CC vv B in the above situation is given by L=r L>r B = µ0λ and θ = π if L ≤ r 2 37. aE = qE0 (along negative z-direction) 1 m P B2 B1 2 Electric field will make z-component of velocity zero. At that time speed of the particle will be Q B1 minimum and that minimum speed is the other B2 component, i.e. v0. At point P, B1 and B2 are in opposite directions. This is minimum when, Hence, BP = 0 vz = uz + az t At point, Q , B1 and B2 are in same direction. or 0 = v0 − qE0 t m Hence, BQ = 2  µ 0λ  = µ0λ 2 or t = mv0 34. F = µ 0 I1I2 qE0 L 2π a ∴ F = µ 0 I1I2 L (Repulsion or upwards) 38. Path is helix and after one rotation only 2π a x-coordinate will change by a distance equal to pitch. M of the loop is inwards and magnetic field to I1 ∴ x = p = (v0 cosθ)  2Bπqm on the plane of loop is outwards. Hence, τ = 0, as  τ = M × B and angle between two is 180°. 35. x- axis 39. M = i(CO × OA) θ = i(CO × CB) O = 4[(− 0.1 i$) × (0.2 cos 30° $j + 0.2sin 30°k$ ) = (0.04 $j − 0.07 k$ ) A-m2 Electrons touch the x-axis again after every pitch. y Therefore, the asked distance is A  2Bπqm B  40. d = p = v11T = (v cosθ) τ For paraxial electron θ ≈ 0° and q = e, O 30° x ∴ d = 2πmv C Be z

Chapter 26 Magnetics — 693 M = Ni (OA × AB) LEVEL 2 = Ni (OA × OC) = (100) (1.2) [(0.4$j) × (0.03 cos 30° $i Single Correct Option + 0.3 sin 30°k$ )] 1. τmg about the left end (from where string is = (7.2i$ − 12.47k$ ) A-m2 connected) τ=M×B = |M × B| = MB sin 90° = [(7.2 i$ − 12.47 k$ ) × (0.8 $i )] or (mgR) = (NiA)B0 = i (πR2)B0 = (− 9.98$j) N-m or i = mg ∴ | τ | = 9.98 N-m Torque vector and expected direction of rotation is πRB0 shown in figure. 2. In uniform field, 41. BA = BB = BC = BD y = B = µ0 i = (2 × 10−7 ) (5) 2π r (0.2/ 2) PQ = (5.0 2) × 10−6 T –2 O x (m) AD +2 Magnetic force on POQ = magnetic force on straight wire PQ having the same current. Hence, F = i (l × B) = i (PQ × B) = 2 [(4i$) × (− 0.02 k$ )] = (0.16 $j) BA+BC BB +BD ∴ a = F = (0.16 $j) = (1.6 $j) m/s2 m 0.1 Net 3. Linear impulse = mv BC Net magnetic field or F∆t = m 2gh or (ilB) ∆t = m 2gh = (BA + BC )2 + (BB + BD )2 But, i ∆t = ∆q ∴ (∆q) (lB) = m 2gh = 2 2B Hence, m 2gh ∆q = = 2.0 × 10−5 T (towards bottom as shown) 4. M = q Bl L 2m r r (2πr dr) (br) = 2π br3 (2πr dr) j = ∫ ∫42. i = 3 00 ∴ M =  2qm L =  2qm (Iω) (a) For r1 < R =  2qm  2 mR2ω = 1 qR2ω B = µ 0 iin 5 5 2π r1 = µ0  2πbr13 / 3 5.  2π  r1  v = µ 0br12 r θ 3 θ (b) For r2 > R d B = µ 0 iin 2π r2 +v x=0 x=d =  µ0   2πbR3 / b = µ 0bR3 sin θ = d = d ⇒ q = v sin θ 2π  3r2  r2  r (mv/ Bq) m Bd

694 — Electricity and Magnetism 13. FMNPQ = FMQ and this force should be upwards to 6. balance the weight. Force NP MQ Variation of magnetic force on wire ACB is as ∴ ilB = mg, where shown in figure. Point of application of net force l = MQ = a lies some where between A and C. 2 7. At a distance X from current I2, ∴ i (a/2) B = mg or i = 2Mg aB B = µ0 I2 2π X Force is upwards if current is clockwise or current Magnetic force of small element dX of wire AB in MQ is towards right. dF = I2 (dX )B sin 90° 14. r = radius of circular path x = 2a 2 = mv = mv ∫∴ F = dF Bq (µ0ni) (q) x=a ∴ v = µ0qr ni 8. Apply screw law for finding magnetic field around 2m a straight current carrying wire. 9. B1 = µ 0IR3 (R2 + x2)3/ 2 2 15. (a) B ⊥ v, so it may along y - axis B2 = µ 0I (2R )2 (b) F ⊥ v, 2 [(2R)2 + (2x)2 ]3/ 2 ∴ a⊥v=0 B1 = 2 B2 or a ⋅ v = 0 10. (b − a) = radius of circular part or a1 b1 = a2 b2 (c) See the logic of option (a). = mv Bq (d) Magnetic force cannot change the kinetic ∴ V = Bq (b − a) energy of a particle. m + 16. Magnetic force is always perpendicular to 11. M = q b–a velocity. So, it will always act in radial direction L 2m which will change tension at different points. But, time period and θ will remain unchanged. M =  2qm (Iω) =  2qm  ml 2 (2πf ) = πqfl2 17. Force on the wires parallel to x-axis will be   3  3 obtained by integration (as B ∝ x and x coordinates vary along these wires). But on a loop 12. At x = 0, there are two such wires. Force on them will be y equal and opposite. P Forces on two wires parallel to y-axis can be obtained directly (without integration) as value of N x B is same along these wires. But their values will M be different (as x-coordinate and therefore B is different). Fnet = ∆F (on two wires) y = ± 2m = Ia (∆B) = Ia (B0) (∆x) FMNP = FMP = i [MP × B ] = Ia B0(a) = IB0a2 = 3 [(4$j) × (5k$ )] This is indecent of x = 60 $i ∴ F1 = F2 = IBla2 ≠ 0

Chapter 26 Magnetics — 695 18. Bx = µ0 I in 23. qE = Bqv 2π x ∴ v=E where, I in = I −  I  [ π (x 2 − b2)] B (c2   π − b2 )  r = mv = m (E / B)  Bq B(q) = I (c2 − x2) E (c2 − b2) B2S = (where, S = q/m) 19. Fe = qE = (1.6 × 10− 19 ) (− 102.4 × 103) k$ 24. F = weight per unit length = (− 1.6384 × 10− 14 k$ )N l Fm = q (v × B) ∴ µ0 i1i2 = 0.01 × 10 = (1.6 × 10− 19 )[(1.28 × 106i$) × (8 × 10− 2$j)] 2π r = (1.6384 × 10− 14k$ )Ν or (2 × 10− 7 ) (100 × 50) = 0.01 × 10 r Now, we can see that or r = 0.1 m Fe + Fm = 0 When B wire is displaced downwards from 20. Due to $j component of B, magnetic force is zero. It equilibrium position, magnetic attraction from A wire will decrease (which is upwards). But, weight is only due to $i component. (which is downwards). So, net force is downwards, l is towards − y direction, B is towards i$ direction. in the direction of displacement from the mean position or away from the mean position. Hence, Hence, l × B (the direction of magnetic force is equilibrium is unstable. + k$ ) 1 25. Magnetic field due to current in the wire along F = ∫ dF = ∫0 (i)(dy) (B) z-axis is zero. Magnetic field due to wire along ∫= 1 (2 × 10− 3) (dy) (0.3 y) x-axis is along $j direction and magnetic field due 0 to wire along y - axis is along − $i direction. Both = 3 × 10− 4 N wires produce 21. r = mv (during circular path) B = µ0 i 2π a Bq ∴ v = Bqr 26. | FABC | + = |FAC | = ilB m = (2) (5) (2) = 20 N Now, qE = Bqv 27. E= 1 qx ∴ E = Bv = B2qr 4 πε 0 (R2 + x2)3/ 2 m µ 0 iR 2 ∴ E = (0.1)2 (20 × 10− 6) (5 × 10− 2) B = 2 (R2 + x2)3/ 2 (20 × 10− 9 ) i = qf = q  2VπR = 0.5 V/m where, 22. B1 = µ 0i1 = (4 π × 10− 7 ) (5) c= 1 2R1 (2)  ε0µ0 5 × 10− 2 2 = 2 2π × 10− 5 T More than One Correct Options B2 = µ 0i2 = (4π × 10− 7 ) (5 2) 1. B1 = µ 0N 1i1 2R2 (2) (5 × 10− 2) 2R`1 = 2 2π × 10− 5 T = (4 π × 10− 7 ) (50) (2) 2 (5 × 10− 2) Bnet = B12 + B22 = 4π × 10− 4 T

696 — Electricity and Magnetism B2 = µ 0N 2i2 8. KE = qV or KE ∝ V 2R2 r = 2qVm or r ∝ V = (4π × 10− 7 ) (100) (2) Bq (2) (10 × 10− 2) T = 2πm or T is independent of V . = 4π × 10− 4 T Bq When currents are in the same direction, then 9. (a) Point a lies to the right hand side of ef and fg. Bnet = B1 + B2 Hence, both wires produce inward magnetic When currents are in the opposite directions, then field. Hence, net magnetic field is inwards. Bnet = B1 − B2 Same logic can be applied for other points also. 2. (a) v is parallel on anti-parallel to B. 10. See the hint of Q.No-3 of Assertion & Reason (c) qE + q (v × B) = 0 section for Level 1 or E = − (v × B) = (B × v) Match the Columns 3. r = mV = (1) (10) = 5 m 1. (a) F = q(v × B) Bq (2) (1) q is negative, v is along + i$ and B along + $j. T = 2πm = (2) (π ) (1) = π Therefore, F is along negative z. Bq (2) (1) (b) Same logic is given in (a). = 3.14 s (c) B is parallel to v. So, magnetic force is zero. Charge is negative so, electric force is opposite Plane of circle is perpendicular to B, i.e. to E. xy- plane. (d) Charge is negative. So, electrostate force is in 4. θ = 180° opposite direction of E. τ = MB sin θ = 0 2. For direction of magnetic force apply Fleming’s U = − MB cos θ = + MB = maximum left hand rule. According to that w and x are 5. If current flows in a conductor, then positively charged particles and y and z negatively charged particle. E≠0 (for inside points) Secondly, E=0 (for outside points) r = 2Km B = µ0 i r (for inside points) Bq 2π R2 ∴ r∝ m (K → same) B = 0 at r = 0, i.e. at centre q B = µ0 i for outside points. 3. Force on a current carrying loop is zero for all 2π r angles. τ is maximum when Q, then angle 6. Fab = upwards between M and B is 90° minimum potential energy is at θ = 0°. Positive potential energy is for obtuse Fabc = leftwards angle. Direction of M is obtained by screw law. ∴ Net force on loop is neither purely leftwards or 4. Two currents are lying in the plane of paper. Its rightwards or upwards or downwards. point is lying to the right hand side of the current 7. Fm = q (v × B) carrying wire, magnetic field is inward (− k$ direction). Depending on sign of q, Fm may be along positive z-axis or along negative z-axis. If point lies to left hand side, field is outward (k$ direction). Fe = qE Again, depending on the value of q it may be along 5. Let us take F = µ0 ⋅ i. i positive z-axis or along negative z-axis. 2π r If q is positive, v × B and Fm comes along negative z-axis also. But, Fe comes along positive z-axis. So, Current in same direction means attraction and it may also pass undeflected. current in opposite direction means repulsion. Let

Chapter 26 Magnetics — 697 us find force on wire-2 by other three wires 1,3 = µ0 I and 4. π 2a + 3x F231 = 2 x=a cos 60°) (−$i ) F2 = F ∫∴ (dF F1 = F x=0 ∫=  aµ0 . I . I . (dx ) 1 (− i$ ) 2 0 π 2a +  F  3x 2  F3 = 2 =  µ 0I 2 . 1 [ln (2a + 3x)]0a (− $i )   π 3 Fnet is rightwards µ 0I 2  2 + 3  ln   6. FABC = FADC = FAC = 3π  2  (− $i ) ∴ Floop = 2 FAC µ 0I 2 2π = I (l × B) and F12 = (i$ )  2 = µ 0I 2 1 1  2 + 3   = I (AC × B) ∴ FTotal π  − ln    (i$ ) Ans.  2 3 2  Hence, AC = (l i$ + l $j) 2. Magnetic moment According to Bohr’s hypothesis, angular momentum in nth orbit is L = n  h  . 2π BC Further, M = q or e L 2m 2m ∴ Magnetic moment, AD M =  2em (L) =  2em  n h  =  4nπemh  2π Now, putting value of B we can find net force in Magnetic field induction different cases. mv2 1 (e)(e) Subjective Questions r = 4 πε 0 r2 …(i) …(ii) 1. Force on wire 12 will be  µ0 aI  (Ia) in positive From Bohr’s hypothesis : 2π mvr = nh 2π x-direction. Solving these two equations, we find 2 v = e2 and r = ε 0n2h2 x 2ε0nh πme2 60° Now, magnetic field induction at centre r 60° 3 B = µ0i dF 2r dF Here, i = qf = (e)  2vπr 1  µ2r0  2eπvr µ 0e  v  4π r2 ∴ B = = Force on 2-3 and 3-1 : r = a + x sin 60° = a + 3 x =  µ40πe  e2   π 2m2e4  2    ε 20n4h4   2ε0nh   µ0 I µ0 I Br = 2π r = 2π µ 0πm2e7 x 8ε 30h5n5 a + 3 = Ans. 2

698 — Electricity and Magnetism 3. r = 9 + 7 = 4 cm Further r= P or Bq B = µ0 i (2 sin α ) (as P = constant) 4π r r∝ 1 q = (10−7 )  30   2 × 53  × 10−2  Since, charge has become two times 4  ∴ r′ = r = 0.2 m = 9.0 × 10−5 T 2 θB At t = (π/40) second, particle will be at P in √7 cm xy -plane. r ∴ x = r′ = 0.2 m y = r′ = 0.2 m Ans. × 3 cm z-coordinate Mass of combined body has become (a) 5 times of the colliding particle. Therefore, from αβ r conservation of linear momentum, velocity component in z-direction will become 1 times. Or r = 4 cm 5 vz = 1 × 40 m/s = 8 m/s 5 π π ∴ z = vz t = 8 × π = 0.2 m Ans. π 40 (b) 5. Fe = Fm or eE = eBv Net field = 4 B sin θ ∴ v = E = 120 × 103 = 4 × 9.0 × 10−5 × 3 B 50 × 10−3 4 Ans. = 2.4 × 106 m/s = 2.7 × 10−4 T 4. xy-plane T = 2πr = 2π (0.4) = 8 π Let n be the number of protons striking per second. v 5 50 Since, Then, T = 2πm or T ∝ m Bq q ne = 0.8 × 10−3 y or n = 0.8 × 10−3 1.6 × 10−19 t=0 = 5 × 1015 m/s 5 m/s p Force imparted = Rate of change of momentum rN x = nmv r = 5 × 1015 × 1.67 × 10−27 × 2.4 × 106 = 2.0 × 10−5 N Ans. 6. (a) Speed of particle at origin, v = 2qV = vx m After collision mass has become 5 times and t= x = a =a m 4 vx 2qV 2qV charge two times. m  5 21 5 8π =π y = 1 ayt 2 = 1  qE  a2 × m  = a2E Ans. 4 8 50 10 2 2  m  2qV  4V ∴ T′ = × T = × s Given time t = T ′ , i.e. combined mass will (b) Component parallel to B is 4 vy = ayt =  qmE  m  q complete one-quarter circle. a  = Ea 2mV  2qV 

Chapter 26 Magnetics — 699 Now, pitch = component parallel to B × time y S period = =  q   2Bπqm R vyT  Ea 2mV   x   = πEa 2m Ans. PQ B qV 7. To graze at C Using equation of trajectory of parabola, y = x tan θ − ax2 θ …(i) Angular acceleration, α = |τ | 2v2 cos2 I qE 10−6 × 10−3 where, I = moment of inertia of loop about QS. m 10−10 Her , a = = I QS = I PR = I ZZ (From the theorem of perpendicular axis) = 10 m/s2 But, I QS = I PR Substituting in Eq. (i), we have ∴ 2I QS = I ZZ = 4 ML2 10 × (0.17)2 3 0.05 = 0.17 tan 30° − 2v2 × ( 3 /2)2 2 ML2 I QS = 3 Solving this equation, we have I 0L2 B 2/ 3ML2 v = 2 m /s ∴ α = |τ | = = 3 I0B I 2 M In magnetic field, AC = 2r or 0.1 = 2r ∴ Angle by which the frame rotates in time or r = 0.05 m = mv cos 30° ∆t is Bq θ = 1α (∆t)2 ∴ B = mv cos 30° 2 (0.05)q or θ = 3 I0B . (∆t)2 Ans. 4M (10−10)(2)( 3 /2) = (0.05)(10−6) 9. In equilibrium, = 3.46 × 10−3 T 2T0 = mg or T0 = mg …(i) 2 = 3.46 mT Ans.  ω Q (πR2) 2π 8. Magnetic moment of the loop, M = (iA)$j Magnetic moment, M = iA = = (I0L2)k$ τ = MB sin 90° = ωBQR2 2 Magnetic field, B = (B cos 45°) $i + (B sin 45° )$j = B ($i + $j) Let T1 and T2 be the tensions in the two strings 2 when magnetic field is switched on (T1 > T2). For translational equilibrium of ring in vertical (a) Torque acting on the loop, τ = M × B direction, = (I0L2k$ ) ×  B (i$ + $j) T1 + T2 = mg …(ii)  2 For rotational equilibrium, ∴ τ = I0L2B ($j − $i ) or |τ | = I0L2B Ans. (T1 − T2 ) D = τ = ωBQR2 2 2 2 (b) Axis of rotation coincides with the torque and or T1 − T2 = ωBQR2 …(iii) since torque is along $j − $i direction or parallel 2 to QS. Therefore, the loop will rotate about an Solving Eqs. (ii) and (iii), we have axis passing through Q and S as shown in the figure. T1 = mg + ωBQR2 2 2D

700 — Electricity and Magnetism As T1 > T2 and maximum values of T1 can be 3T0 , ∴ 0.0442 = (kx)(l sin 53° ) = (4.8)(x)(0.2) 45 2 We have 3T0 ωmax BQR2 or x = 0.057 m 2 2D U = 1 kx2 = 1 × (4.8)(0.057)2 = T0 + 22 ∴ ω max = DT0 Ans. BQR2 = 7.8 × 10–3J Ans. 10. dB = µ0 (i/ω). dx 13. Let the direction of current in wire PQ is from P to 2π x Q and its magnitude be I. ∫∴ B = µ0i d + ω dx 2πω d x z B = µ 0i ln  d + ω (upwards) Ans. 2πω d dB P Qy x xS 1 dx R 11. θ = tan −1  R  = tan −1  R = tan −1  BmqvR The magnetic moment of the given loop is r  mv/ Bq M = − Iabk$ Deviation = 2θ = 2tan−1 BmqvR Torque on the loop due to magnetic force is ×× × τ1 = M × B ×× = (− Iabk$ ) × (3i$ + 4k$ )B0i$ × r = − 3IabB0$j × Cθ αO Torque on weight of the loop about axis PQ is × r 90° R × τ2 = r × F=  a i$ × (− mgk$ ) 2 × × 12. (a) Yes, magnetic force for calculation of torque = mga $j 2 can be assumed at centre. Since, variation of torque about P from one end of the rod to the We see that when the current in the wire PQ is other end comes out to be linear. from P to Q , τ1 and τ2 are in opposite directions, so they can cancel each other and the loop may (IlB) 2l  Il2B remain in equilibrium. So, the direction of current 2 ∴ τ = = I in wire PQ is from P to Q. Further for equilibrium of the loop = (6.5)(0.2)2(0.34) |τ1 | = |τ2 | 2 or 3IabB0 = mga 2 = 0.0442 N-m Ans. (b) Magnetic torque on rod will come out to be I = mg Ans. clockwise. Therefore, torque of spring force 6bB0 should be anti-clockwise or spring should be stretched. Magnetic force on wire RS is Ans. F = I (l × B) (c) In equilibrium, = I[(− b$j) × {(3$i + 4k$ )B0}] F = IbB0(3k$ − 4i$) Clockwise torque of magnetic force = anti-clockwise torque of spring force

27. Electromagnetic Induction INTRODUCTORY EXERCISE 27.1 8. S = [(5 × 10− 4) k$ ] m2 1. ⊗ magnetic field passing through loop is φ = |B ⋅S | = 9× 10− 7 Wb increasing. Hence, induced current will produce INTRODUCTORY EXERCISE 27.2 magnetic field. So, induced current should be anti-clockwise. 1. Q e = Bvl = 1.1 × 5 × 0.8 2. It is true that magnetic flux passing through the = 4.4 V Apply right hand rule for polarity of this emf. loop is calculated by integration. But, it remains constant. 2. Q e = Bvl 3. dφB = [Potential or EMF ] i = e = Bvl RR dt = [ML2A− 1 T− 3 ] F = ilB = B2l2v R 4. u is increasing. Hence, ⊗ is produced by the = (0.15)2 (0.5)2 (2) induced current. So, it is clockwise. 3 5. By increasing the current in loop-1, magnetic field = 0.00375 N in ring-2 in downward direction will increase. 3. VA − VC = Bωl2 Hence, induced current in ring-2 should produce 2 upward magnetic field. Or current in ring should be in the same direction. VD − VC = Bω (2l)2 2 12 From these two equations, we find B VA − VD = − 3Bωl2 /2 6. 2πR = 4L 4. Circuit is not closed. So, current is zero or magnetic force is zero. ∴ L = πR = (π ) (10) INTRODUCTORY EXERCISE 27.3 22 = (5π ) cm 1. |e | = L ∆i or L di ∆S = Si − S f = (πR2) − L2 ∆t dt = π (0.1)2 − (5π )2 × 10− 4 Here, L = 1H = 0.0067 m2 and di = 3 [sin t + t cos t ] dt = ∆φ =  ∆S  e ∆t B ∆t ∴ |e| = 3 (t cos t + sin t) = 100 × 0.0067 2. VL = + L di = (2) d (10e−4t ) 0.1 dt dt = 6.7 V = − 80 e−4t 7. ∆φ = 2 (NBS ) Further, Va − iR − L di = Vb dt ∆q = ∆φ = 2NBS RR ∴ Va − Vb = iR + L di dt = 2 × 500 × 0.2 × 4 × 10− 4 50 or Vab = (10e−4t ) (4) − 80e−4t = 1.6 × 10− 3C = − 40e−4t

702 — Electricity and Magnetism 3. (a) dI /dt = 16A/s INTRODUCTORY EXERCISE 27.5 ∴ L = e = 10 × 10− 3 1. Q E = 1 Li2 = i2Rt dI /dt 16 2 = 0.625 × 10− 3 H = 0.625 mH ∴ L has the units of time. R (b) At t = 1s, I = 21 A U = 1 LI 2 = 1 × (0.625 × 10− 3)(21)2 2. (a) τL = L = 2 = 0.2 s 22 R 10 = 0.137 J (b) i0 = E = 100 = 10 A P = Ei = (10 × 10− 3) (21) R 10 = 0.21 J/s (c) i = i0 (1 − e− t/ τ L ) 4. (a) = 10 (1 − e− 1/ 0.2) l = 9.93 A L = µ0N 2S 3. E = VR + VL l INTRODUCTORY EXERCISE 27.6 N=l d 1. U = 1 Li2 = 1 q2 ∴ L = µ 0lS = (4 π × 10− 7 ) (0.4) (0.9 × 10− 4) 2 2C d2 (0.1 × 10− 2)2 ∴ LC = q = it = t = 4.5 × 10− 5 H ii (b) e = L ∆i = (4.5 × 10− 5) (10) = 4.5 × 10− 3 V 3. (a) VL = VC ∆t 0.1 L di = q INTRODUCTORY EXERCISE 27.4 dt C q = (LC ) di 1. Q M = e2 = (50 × 10− 3) dt di1/dt (8/0.5) = (0.75 × 18 × 10− 6) (3.4) = 45.9 × 10− 6 C = 3.125 × 10− 3 H = 3.125 mH e1 = M  ∆i2  = (3.125 × 10− 3) (6) = 45.9 µC ∆t 0.02 (b) VL = VC = q C = 0.9375 V 4.2 × 10− 4 2. (a) M = N 2φ 2 = (1000) (6 × 10− 23) = 2 H = 18 × 10− 6 i1 3 (b) |e2| = M  ∆i1  = (2) (3) = 30 V = 23.3 V ∆t 0.2 4. 1 Lim2 ax = 1 CVm2ax 2 2 (c) L1 = N1φ1 = (600) (5 × 10− 3) = 1H i1 3  L ∴ Vmax =   imax 3. (a) |e2| = M  ddit1  C = (3.24 × 10− 4) (830)  20 × 10− 3  (0.1) =  0.5 × 10− 6  = 0.27 V (b) Result will remain same. = 20 V

Chapter 27 Electromagnetic Induction — 703 INTRODUCTORY EXERCISE 27.7 F = qE = eR2 (6t2 − 8t) 2r2 1. In the theory we have already derived mutual Substituting the values, we have inductance between solenoid and coil, F = (1.6 × 10− 19 ) (2.5 × 10− 2) [6 (2)2 − 8 (2)] 2 × 5 × 10− 2 M = µ 0N 1N 2 (πR12) l1 = 8.0 × 10− 21N = µ 0N 1N 2S1` ⊗ magnetic field at the given instant is l1 increasing. Hence, induced current in an |e2| = M di = µ 0N 1N 2S1 di1 imaginary circular loop passing through P2 dt l1 dt should produced u magnetic field. Or, current = (4 π × 10− 7) (25) (10) (5 × 10− 4) (0.2) 10− 2 in this should be anti-clockwise. Hence, = 3.14 × 10− 6 V electrons should move in clockwise direction. (b) El = dφ = e Or electron at P2 should experience force in ∴ dt downward direction (perpendicular to r2). E=e= e (b) At P1 l 2πR2 El = dφ = 3.14 × 10− 6 dt (2π ) (0.25) ∴ E (2πr1) = s dB = (πr12)  ddBt  = 2 × 10− 6 V/m dt ∴ E = r1  ddBt  2 2. (a) At P2 El = dφ = r1 [6t2 − 8t ] dt 2 ∴ = 0.02 [6(3)3 − 8(3)] 2 E (2πr2) = πR2 ⋅ dB = 0.3 V/m dt As discussed in the above part, direction of E = R2  ddBt  electric field is in the direction of induced 2r2 current (anti-clockwise) in an imaginary circular conducting loop passing through P1. Exercises LEVEL 1 2. At time t0, magnetic field is negative or and Assertion and Reason increasing. 1. Due to non-uniform magnetic field (a function of Hence, induced current will produce ⊗ magnetic field. Or induced current should be clockwise. x) magnetic flux passing through the loop obtained If dφ = constant. Then, e = constant by integration. But that remains constant with dt time. Hence, dφ = 0 ∴ i or rate of flow of charge is constant. dt 3. It can exert a force on charged particle. or e = 0 4. di = 2 A/s Magnetic field is along − k$ direction or in ⊗ dt magnetic is increasing. Hence, induced current Va − Vb = L di dt should produce u magnetic field. Or induced = (2) (2) = 4 V current should be anti-clockwise.

704 — Electricity and Magnetism 5. Comparing with spring-block system  dI  5. dt is acceleration. amax = ω2A = ω (ωA) If magnetic field in the shown cylindrical region is changing, then induced electric field exists even = ω (vmax ) outside the cylindrical regions also where magnetic field does not exist. ∴  dI  =ω I max dt 6. i = dq = (8t)A max dt 6. Applying RHR, we can find that di = 8 A/s dt Va > Vb At t = 1 s, q = 4 C, i = 8 A 8. Ferromagnetic substance will attack more number and di = 8 A/s of magnetic lines through it. So, flux passing dt through it will increase. Hence, coefficient of self-inductance will increase. L depends on number of turns in the coil’s radius Charge on capacitor is increasing. So, charge on of coil etc. It does not depend on the current passing through it. positive plate is also increasing. Hence, direction 9. Current developed in the inductor wire will of current is towards left. decrease exponentially through wire ab. a 2Ω 4V 2H 2F b 10. VL1 = VL2 –+ i ∴ L di1 = L2 di2 Now, Va + 2 × 8 − 4 + 2 × 8 + 4 = Vb dt dt 2 or L1di1 = L2di2 ∴ Va − Vb = − 30 V or L1i1 = L2i2 7. dI = I0ω cos ωt or i ∝ 1 dt L e=M dI = MI0ω cos ωt dt Objective Questions ∴ emax = MI0ω = 0.005 × 10 × 100π 1. U = 1 Li2 = (5π ) V 2 8. 1 Li2 = 1CV 2 U   ML2T− 2   i2   A2  22 ∴ [L ] = =   = [ML2T− 2A− 2 ] ∴ V= L ⇒ i= 4 2 (2) C × 10− 6 2. M ∝ N 1 N 2 = 2 × 103 V 3. SN 9. e = Bωl2 = constant NS 2 When brought closer induced effects should 10. ∆q = ∆φ produced repulsion. So, currents should increase, so that pole strength increases. Hence, repulsion R increases. or i∆t = ∆φ 4. Magnetic field of ring is also along its axis, or in R ∴ ∆φ = i(∆t)R = (10 × 10− 3) (5) (0.5) the direction of velocity of charged particle. Hence, no magnetic force will act on charged = 25 × 10− 3 Wb particle. But, due to g velocity of charged particle will increase. 11. ⊗ magnetic field is increasing. Therefore, induced electric lines are circular and anti-clockwise. Force on negative charge is opposite to electric field.

Chapter 27 Electromagnetic Induction — 705 12. e = dφ = (aτ − 2at) 21. N S = IP = VS dt N P IS VP i = e = aτ − 2at VP = Vi =  N P  V0 RR  N S  τ = 2 × 20 = 40 V 1 H = ∫ i2Rdt 0 IP =  N S  IS  N P  13. e = L di = L (Slope of i-t graph) dt Initially, slope = 0 ⇒ e = 0 =  12 (4) = 2 A Then in remaining two regions slopes are constants 22. Relative velocity = 0 but of opposite signs. Hence, induced emfs are constants but of opposite signs. ∴ Charge in flux = 0 14. VA − 1 × 5 + 15 + (5 × 10− 3) (103) = VB 23. In case of free fall, ∴ VB − VA = 15 V d = 1 gt2 2 15. I = (10t + 5)A dI = 10 A/s = constant = 1 (10) (1)2 = 5m dt 2 At, t = 0, I = 5 A Here due to repulsion from induced effects Now, VA − 3 × 5 − 1 × 10 + 10 = VB a<g ∴ VA − VB = 15 V ∴ d<5m 16. (VC )max = (VL )max 24. VA − VB = L di dt q0  dI  C = L dt = L (− α ) = − αL max 25. = E = 12 = R 0.3  dI  q0 i0 40 A dt LC ∴ = 1 1 2 2 max U = Li02 = × 50 × 10− 3(40)2 17. VL = L di dt = 40 J 18. φi = BS cos 0° = 2 Wb 26. Value remains 1 th in 20 ms times. Hence, two φ f = BS cos 180° = − 2 Wb 4 |∆φ| = 4Wb half-lives are equal to 20 ms. So, one half-life is |∆q| = |∆φ| 10 ms. R = 4 = 0.4 C t1/ 2 = (ln 2) τC = (ln 2) L 10 R 19. S = (ab) k$ → perpendicular to x y -plane ∴ R = (ln 2) L t1/ 2 φ = B⋅S = (50) (ab) = constant = (ln 2) (2) = (100 ln 4) Ω dφ = 0 10 × 10− 3 dt 27. Q i= e ∴ e=0 R 20. Back emf = Applied voltage potential drop across = N (∆φ / ∆t) = NS (∆B/∆t) RR armature coil = 10 (10 × 10− 4) (104) = 200 − iR 20 = 200 − 1.5 × 20 =5A = 170 V

706 — Electricity and Magnetism 28. In steady state, whole current passes through the ∆t in second case is large. Hence, induced emf is less. inductor. 29. If current is passed through the straight wire, 2. e = N∆φ =  NS∆∆t B cos 30° ∆t magnetic lines are circular and tangential to the loop. So, no flux is linked with the loop. ∴ S =  Ne∆∆Bt  sec 30° 30. In second position, ∆φ = 0 ∴ |Q2| = ∆φ = 0 =  (80 × 10− 3) (0.4)   2  = 1.85 m R  (50) (400   3   × 10− 6 )  31. From Lenz's law, induced effects always oppose ∴ Side of square = 1.36 m the cause due to which they are produced. So, Total length of wire = 50 (4 × 1.36) when the first loop is moved towards the smaller loop, it will face repulsion. = 272 m 32. τL = L = 2 s 3. φ =BS = B0 S e−at R Induced emf = dφ = aB0 Se− at i0 = E = 3A , t = 2 s dt R 4. (a) At a distance x from the wire, magnetic field i = i0 (1 − e− t/ τ L ) over the wire ab is Substituting the given values, we can find i. B = µ0 i 2π x 33. In AB, l is parallel to its v. Hence, PD = 0 34. v is parallel to l. dV = Bvdx =  2µπ0xi  vdx 35. For wire ab, velocity vector is parallel to l. ∫∴ Total emf = x=d+ l dV 36. Current increases with time. So, flux passing x= d through B will increase with time. From Lenz's (b) Magnetic field due to current i over the wire ab law, it should have a tendency to move away from is inwards. Velocity of wire ab is towards the coil to decrease flux. right. Applying right hand rule, we can see that a point is at higher potential. 37. For E ≠ 0, φ must change (c) Net change in flux through the loop abcd is or dφ ≠ 0 zero. Hence, induced emf is zero. So, induced dt current is zero. 38. Even if radius is doubled, flux is not going to 5. At t = 0, inductor offers infinite resistance. Hence, change. M current through inductor wire is zero. Whole B⊥ current passes through two resistors of 4Ω each. 39. i1 = 10 = 1.25 A 4+4 B Velocity B At t = ∞ , inductor offers zero resistance.  Rnet = 4 + 8× 4 N 8+ 4 B | | is parallel to MN (or l) and B⊥ is parallel or = 6.67 Ω antiparallel to velocity. So, main current Subjective Questions i2 = 10 = 1.5A Rnet 1. When switch is opened current suddenly This distributes in 4Ω and 3Ω in inverse ratio of decreasing from steady state value to zero. When resistance. Hence, current through 4Ω is 1A and switch is closed, it takes time to increase from 0 to through 8Ω is 0.5A. steady state value. For equivalent τL of the circuit Rnet across inductor e = L ∆i after short-circuiting, the battery is 10 Ω. ∆t

Chapter 27 Electromagnetic Induction — 707 ∴ τL = 1 = 1 = 0.1 s Rnet 10 iL 0.5 A i e1 e2 9. t −t xa iL = 0.5 (1 − e 0.1 ) enet = e1 − e2 = 0.5 (1 − e−10t ) i = B1va − B2va 1.5 A i = Current through = (B1 − B2) va 1.25 A battery =  µ0 i − µ0 i  va t  2π x 2π x+ a  i = 1.25 + 0.25 (1 − e−t/ 0.1) = µ0 2ia2v = 1.5 − 0.25 e−10t 4π x (x + a) 6. Similar to above problem 10. B = µ0 i ∆q = 2NBS 2π r R dV = Bvdr = µ 0iv ∫ dr 2π r ∴ B = (∆q)R 2NS ∫V r2 = µ 0iv ln  r2  2π  r1  (4.5 × 10− 6) (40) = dV = (2) (60) (3 × 10− 6) r1 11. VL = L di dt = 0.5 T 1 7. e = dφ = s  ddBt  = πR2 (Slope of B - t graph) ∴ di = L (VLdt) dt 1 (a) e = (π ) (0.12)2  02.5 = 0.011 V/m ∴ ∫ di = i = L ∫ VLdt (b) Slope of B-t graph is zero. Hence, or i = 1 (area under VL versus t graph) L e=0 (a) At t = 2 ms (c) Slope is just opposite to the slope of part (a). i = (150 × 10−3)−1 1 × 2 × 10−3 × 5 8. Induced emf (e = BvL) and therefore induced 2 current is developed only during entering and = 3.33 × 10−2A during existing from the magnetic field. (b) At t = 4 ms i = e = BvL RR Area is just double. Hence, current is also double. F = iLB = B2L2v 12. (a) L = e = 0.016 = 0.25 H R di/dt 0.064 Further, magnetic force always opposes the change. Hence, external force is always positive. (b) L = Nφ i During entering into the field, ⊗ magnetic field increases. Hence, induced current should produce ∴ φ = Li = (0.25) (0.72) N 400 u magnetic field. Or it should be anti-clockwise. = 4.5 × 10− 4 Wb During existing from the magnetic field case is just opposite.

708 — Electricity and Magnetism 13. (a) M = N 2φ2 = (400) (0.032) = 1.96 H Now, apply i1 6.52 i = i0 (1 − e− t/ τ L ) (b) M = N 1φ1 where, τL = L i2 R φ2 = Mi2 = (1.96) (2.54) 17. Steady state current developed in the inductor N1 700 = E = i0 (say ) = 7.12 × 10−3 Wb r L (a) Now this current decreases to zero R 14. τL = = 0.1 s exponentially through r and R. The given time t = 0.1 s is one time constant. ∴ i = i0e− t/ τ L The desired ratio is iVL (Q P = Vi) where, τL = R L r iE + = VL Energy stored in inductor, E U0 = 1 Li02 =  1 L  Er  2 E 2 2 After on time constant VL = e as Now, this energy dissipates in r and R in direct VL = Ee− t/ τ L ratio of resistances. Hence, the desired ratio is 1 ≈ 0.37. ∴ Hr =  R r  U 0 = E2L r) e  + r 2r (R + 15. i0 = V = 3.24 = 0.253 A 18. In steady state, main current from the battery is R 12.8 i0 = E = 20 = 4A L 3.56 R 5 τL = R = 12.8 = 0.278 s (a) After one time constant (t = 0.278 s = τC) Now, this current distributes in inverse ratio of inductor. i = 1 − 1e i0 ∴ i5 =  10  (4A) = 8 A  10 + 5 3 ≈ 0.63 i0 19. (a) 1 Li02 = 1 CV02 = 0.16 A 2 2 Power supplied by battery = Ei P = (3.24) (0.16) = 0.518 W ∴ L = CV02 = (4 × 10− 6) (1.5) (b) PR = i2R i02 (50 × 10− 3)2 = (0.16)2 (12.8) = 0.328 W = 3.6 × 10− 3 H (b) f = 1 = 1 (c) PL = P − PR = 0.191W 2π LC 2π (3.6 × 10− 3) (4 × 10− 6) 16. (a) After one half-life, = 0.133 × 104 Hz t = t1/ 2 = (ln 2) τL = 1.33 kHz = 0.693 L R (c) t =T = 1 = 4 1 s = (0.693) (1.25 × 10− 3) 4 4f × 1.33 × 103 50 = 1.73 × 10− 5 s = 0.188 × 10− 3 s (b)  1 Li2 =  1 Li02 /2 = 0.188 ms 2 2 20. (a) ω = 2π , T = 1 ff ∴ i = i0 (b) At t = 0, q = q0 = CV0 = (100 µC) 2 Now, q = q0 cos ωt

Chapter 27 Electromagnetic Induction — 709 (c) ω = 1 ⇒ L = 1 Area, S = l2 = (a + 2v0 t)2 LC ω 2C φ = BS = B (a + 2v0t)2 (d) |i | = dq = q0ω sin ωt dt dφ e = dt = 4 Bv0 (a + 2v0t) Average value of current in first quarter cycle R = λ [4l ] = 4λ (a + 2v0t) ∴ i = e = Bv0 T/4 Rλ = ∫0 dt T /4 21. (a) 1 qi2 + 1 LI 2 = 1 CV02 6. At time t 2 C 2 i 2 ∴ V0 = qi (as Ii = 0) Side of square l = li − dt C S = l2 = (li − αt)2 (b) 1 CV02 = 1 LI 2 At given time 2 2 0 l = li − αt = a ∴ I0 = C V0 φ = BS = B (li − αt)2 L e= dφ = 2Bα (li − αt) 1 2 dt (c) U max = 2 LI 0 1L  I20 2 But, (li − αt) = a 2 ∴ e = 2a α B (d) U L = UC = 1 q2 = U max −UL 7. 2 C LEVEL 2 Single Correct Option ω = v = v = 2v R l/2 l 1. 1 mv02 = 1 Lim2 ax 2 2 e = Bωl2 = B  2lv l2 = Bvl 22 ∴ imax = m v0 L 8. From right hand rule, we can see that 2. VC = Bvl VA > VB ∴ qA is positive and qB is negative. ∴ q = CVC = BvlC = constant q = CV = C (Bvl) ∴ IC = dq = 0 = (20 × 10− 6) (0.5) (0.2) (0.1) dt = 0.2 × 10− 6C = 0.2 µC UC = 1 CV 2 = 1 CB 2L2v 9. i = Bvl 2 2 R 3. From right hand rule, we can see that P and Q l points are at higher potential than O. v 4. At mean position, velocity is maximum. Hence, Let λ = resistance per unit length of conducting motional emf Bvl is also maximum. v oscillates rod, then simple harmonically. Hence, motional emf will also move simple harmonically. Further, polarity i = Bvl = Bv = constant of induced emf will keep on changing. λl λ 5. At t = t side of square, l = (a + 2v0 t)

710 — Electricity and Magnetism 10. At time t, angle rotated by loop is θ = ωt. This is x also the angle between B and S. Then, i c φ = BS cos θ = Bb2 cos ωt 15. e = dφ = b2Bω sin ωt dx dt B = µ0 i ⇒ dS = cdx 11. El = dφ = S dB 2π x dt dt dφ = BdS = µ0 ic dx ∴ E (2πr) = (πr2) dB 2π x dt ∫φ = b dφ = µ 0ic ln  ba or E = r dB a 2π 2 dt M = φ = µ 0c ln  ba F = qE = qr dB i 2π 2 dt 16. Bx = µ0 I W = F d = (2πr) 2π x = πr2q  ddBt  µ0 I 2π x =  272 (1)2 (10− 6) (2 × 10− 3) de = Bx vdx = v dx = 2π × 10− 9 J ∫e = b = µ 0Iv ln  ba 2π de a 12. Initial current = 10 = 1 A i = e = µ 0Iv ln  ba = Induced current R 2πR 10 ∴ φi = L(Ii ) = 500 mWb = 0.5 Wb dF = (i) (dx) Bx Final current = 20 = 4A = µ 0Iv ln  ab   µ0 I  dx 5  2πR   2π x  φ f = L (I f ) = (0.5) × 4 = 2 Wb b ∴ ∆φ = 1.5 Wb F = ∫a dF 13. 1 Li2 = 1 1 Li02  2 2  2  17. El = dφ = S dB ∴ i= i0 = i0 (1 − e− t/ τL ) dt dt 2 ∴ E (2πr) = πr2 dB e− t/ τ L = 1 − 1 = 2 − 1 dt 22 or E = r dB or E ∝ r 2 dt ∴ t = ln  2  18. Magnetic field through Q (by I2) is downwards.   τL  2 − 1 By decreasing I1, downward magnetic field through Q will decrease. Hence, induced current in or t = τL ln  2  Q should produce magnetic field in same    2 − 1 direction. = L ln  2  19. i = 0 (1 − e− t./ τ L ) = E (1 − e− t/ τ L )   R  2 − 1 R 14. B = µ0 i = E − Ee− t/ τ L = i0 −  VL  (as VL = Ee− t/ τ L) R R R 2π a F = Bqv sin 90° = µ0 i (qv) ∴ VL = (i0R) − (R) i 2π a i.e. VL versus i graph is a straight line with positive intercept and negative slope.

Chapter 27 Electromagnetic Induction — 711 20. e = Bvl = 0.5 × 4 × 0.25 = 0.5 V ∴ τ net =L = 2L Rnet R 12 Ω and 4Ω are parallel. Hence, their net resistance R = 3 Ω. 28. At t = 0, i = E/R i = e = 0.5 = 0.1 A Now, this current will decay in closed loop in R+ r 3+ 2 anti-clockwise direction. So, |i2| = i2 = E/R in 21. El = dφ = S dB upward or opposite direction. dt dt Hence, i2 = − E E (2πR) = (πr2) (β) R ∴ E = r2 β ⇒ F = qE 29. 1 Li2 = 1 1 Li02  2R 2 4  4  and τ = FR = qER = 1 qr2β So, i = i0 , half value 2 2 22. ∴ t = t1/ 2 = (ln 2) τL = (ln 2)  RL v 30. Steady state current through inductor in E. 2R R ω = So, at t = 0, current in closed loop (confiding of capacitor) will remain same. PD = Bωl2 = (B) (v/2R) (4R)2 = 4 BvR 31. At t = 0, VL = − E 22 Bω (2R)2  η   η  32. VA − V0 = 2 = 2BωR2 …(i)  + 1  + 1 23. L1 = L, R1 = R η η Bω (2R)2 2 BωR 2 2 1  1  V0 − VC = = …(ii)  η + 1  η + 1 L2 = L, R2 = R CA Lnet = L1L2 L1 + L2 Similarly, Rnet = R1R2 R1 + R2 O O Lnet =L ω ω Rnet R τL = Adding these two equations, we get VA − VC = 4 BωR2 24. i = i0e− t/ τ L v2 β i0 = i0e− T / τ L 33. T ∴ τL = ln (1/β) dx v x P τ= L 25. P = i02R ⇒ i02 = R ⇒ R ∴ L = τR v1 Heat dissipated = 1 Li02 = 1 (τR)  RP = 1 Pτ v = v1 +  v2 − v1 x 2 2 2 l 26. In decay of current through L-R circuit, current can Small potential difference = Bv (dx) not remain constant. l 27. By short-circuiting the battery, net resistance ∴ Total potential difference = ∫0 Bvdx across inductor is R (R and R in parallel). = 1 B (v1 + v2 ) l 2 2

712 — Electricity and Magnetism 34. At time t = 0, resistance offered by a capacitor = 0 ∆φ = |Qi− Qf | = µ 0ia ln  b + a 2π  b − a and resistance offered by an inductor = α. Rnet = R + R = 5R = 5 Ω ∆q = ∆φ = µ 0ia ln b+ a 2 3 6 R 2πR  b − a ∴ Current from the battery, i = E = 5 =1A More than One Correct Options Rnet 5 1. e = Bvl, where l = L 35. τL = L = 0.01 = 10− 3 s 2 R 10 For polarity of this motional emf, we can use right hand rule. τC = CR = (0.1 × 10− 3) (10) = 10− 3s x 20 (i0 )L = 10 = 2 A (i0 )C = 20 = 2 A i 10 2. (a) The given time is the half-life time of both the dx circuits. ∴ iL = iC = 2=1A µ0 i 2 2π x Bx = or total current is 2A. µ0 2π 36. | e | = dφ = S dB = (4b2 − πa2) B0 dφ = (Bx ) dS =  xi  (adx) dt dt i = |e| = (4b2 − πa2) B0 ∫φ = 2a = µ 0ia ln 2 RR dφ a 2π ⊗ magnetic field is increasing. So, M = φ = µ0a ln 2 i 2π u magnetic field is produced. (c) B b + a µ0 i µ0ia ln  b + a Wire produces u magnetic field over the loop. If b 2π x 2π b the loop is brought closer to the wire, u magnetic ∫37. φi = (adx) = field passing through the loop increases. Hence, x induced current produces ⊗ magnetic field so, induced current is clockwise. 3. (a) L = Nφ ⇒ φ = Li iN So, SI unit of flux is Henry-ampere. a (c) L = − e = − − e∆t ∆i/ ∆t ∆i dx Hence, SI unit of L is V - s . ampere 4. τL = L = 2 = 1s R 2 Similarly, φf = µ0ia ln  b − a t1 = (ln 2) τL = (ln 2) s 2π b 2

Chapter 27 Electromagnetic Induction — 713 Hence, the given time is half-life time. Va − Vb = L di =1× 4 = 4 V ∴ i = i0 = 8/2 = 2 A dt 22 Vb − Vc = q = 2 =1V C 2 Rate of energy supplied by battery = Ei = 8 × 2 = 16 J/s Vc − Vd = iR = 4 × 4 = 16 V Va − Vd is summation of above three, i.e. 21 V. PR = i2R = (2)2(2) = 8 J/s 10. Va − Vc = 0 as l is parallel to v. Va − Vb = E − iR = 8 − 2 × 2 = 4 V Va − Vb = Vc − Vb = Bωl2 5. According to Lenz's law, induced effects always 2 oppose the change i1 and i2 both are in same Comprehension Based Questions direction. Hence, magnetic lines from B due to both currents are from right to left. By bringing A 1. dB = (6t2 + 24) T/s closer to B or increasing i1 right to left magnetic field from B will increase. So, i2 should decrease. dt 6. φi = BS cos 0° = (4) (2) = 8 Wb At t = 2 s, dB = 48 T/s dt φ f = BS cos 90° = 0 ∆φ = 8 Wb El = dφ =S  ddBt  |e| = ∆φ = 8 = 80 V dt ∆t 0.1 i = |e| = 20 A or E (2πr) = πr2  ddBt  R ∆q = ∆φ = 2 C ∴ E = r ⋅ dB …(i) R 2 dt This current is not constant. So, we cannot find the F = qE = qr dB heat generated unless current function with time is 2 dt not known. = (1.6 × 10−19 ) (1.25 × 10−2) (48) 7. imax = ωq0 =  1  q0 2 LC = 48 × 10− 21N  ddti max  1  2. From Eq. (i) of above problem, we can see that LC = ω2q0 = q0 E∝r 8. If ⊗ magnetic field increases, then induced electric i.e. E - r graph is a straight line passing through origin. lines are anti-clockwise. If ⊗ magnetic field decreases, then induced electric lines are 3. ⊗ Magnetic field is increasing. Hence, u magnetic clockwise (both inside and outside the cylindrical field is produced by a conducting circular loop region). placed there. For producing, magnetic field induced current should be anti-clockwise. On positive charge, force is in the direction of E. Direction of induced circular electric lines are also On negative charge, force is in the opposite anti-clockwise. direction of E. 4. |e | = dφ = S dB = (πa2) B0 9. q = 2t2 dt dt i = dq = 4t 5. El = dφ E (2πa) = (πa2)B0 dt dt E = 1 aB0 di = 4 A/s ∴ 2 dt or At t = 1 s, q = 2C, i = 4A and di = 4 A /s 6. F = qE = 1 qaB0 2 dt

714 — Electricity and Magnetism τ = F a = 1 qa2 B0 PR1 = e2 2 R1 α=τ =  1 qa2 B0 = qB0 ∴ 0.76 = 0.36 v2 …(ii) I 2 ma2 2m R1 7. ω = αt =  qB0  t PR2 = e2 2m R2 P = τω =  1 qa2 B0  q2Bm0 ∴ 1.2 = 0.36 v2 …(iii) 2 R2 = q2B02a2 R1 and R2 are in parallel. 4m ∴ Rnet = R1R2 …(iv) 8. |e| = dφ = S dB R1 + R2 …(v) dt dt i= e = (0.2 × 0.4) (2) = 0.16 V Rnet i = |e| = (1) (40 + 0.16 × 10− 2 Solving these five equations, we can get the R 40 + 20) results. = 0.16 A Match the Columns u Magnetic field passing through the loop is 1. (a) B = F increasing. So, induced current should produce il ⊗ magnetic field. Hence, induced current is clockwise.  MLT− 2  [ MT − 2A−  AL  [ B ] =   = 1 ] 9. At t = 2 s, rod will move 10 cm. Hence, 40 cm side (b) U = 1 Li2 2 will become 30 cm. |e| = e1 (say ) = S  ddBt  ∴ [L] = U  =  ML2T− 2  = [ML2T− 2A− 2 ]  i2    (0.2 × 0.3) (2) = 0.12 V  A2  At t = 2 s , B = 4 T (c) ω = 1 ⇒ [ LC ] = L  = [ T2 ] LC ω2  ∴ e2 = Bvl = (4) (5 × 10− 2) (0.2) (d) [φ ] = [ BS ] = [MT− 2A− 1L2 ] = 0.04 V 2. VL = Ee− t/ τ L = 10e− t/ τ L ∴ enet = e1 − e2 = 0.08 V τL = L = R 10. i = enet = 0.08 1s R 30 + 20) (1) (30 + × 10− 2 ∴ VL = 10e− t = 0.1 A VR = E − VL = 10 (1 − e− t ) F = ilB Now, we can put t = 0 and t = 1second. = (0.1) (0.2) (4) = 0.08 N 3. ω = 1 = 2 rad /s 11 to 13 LC At terminal velocity, (a) imax = ωq0 = 8 A iLB = mg ∴ i = mb = 0.2 × 98 (b) di = ω2q0 = 16 A/s dt LB 1 × 0.6 max i = 3.27 A …(i) (c) VL = VC = q = 2 = 8V C 1/ 4 e = BvL (v = terminal velocity) = (0.6) (v) (1) (d) VC = VL = L di = (1)  126 = 8V dt e = 0.6v

Chapter 27 Electromagnetic Induction — 715 4. Steady state current through inductor, and τC = CR, τC = CR2 = C . L = 1 τL L L C i0 = 9 = 3 A 3 ∴ τL = τC Now, this current decays exponentially across inductor and two resistors. ∴ For the given condition τL = τC = τ (say) L 9 Now, in L-R circuit R + τL = = 6 3 =1s I1 = V (1 − e− t / τ ) R t1/ 2 = (ln 2) τL = (ln 2) s In CR circuit, I2 = V e− t / τ Given time is half-life time. Hence, current will R remain 1.5 A. ∴ I = I1 + I2 = V = constant Ans. i = i0e− t/ τ L = 3e− t R  − ddit = 3e− t 2. Motional emf, V = Bvl In the beginning  −ddt i = 3 A/s Net resistance of the circuit = R + R1R2 After one half-life time  −ddt i = 1.5 A/s R1 + R2 (a) VL = L  −ddt i = 9 × 1.5 = 13.5 V ∴ Current through the connector, i = Bvl Ans. R1R2 R + R1 + R2 3. θ = ωt (b) V3Ω = iR = 1.5 × 3 = 4.5 V i (c) V6Ω = iR = 1.5 × 6 = 9 V (d) Vbc = VL − V3Ω = 9 V dx A d – x sin θ = d – x sin ωt 5. φ = 2t x x sin θ θ (a) e = dφ = 2 V dt O (b) i = e = 1 A = constant B R de = B (ωx) dx (c) ∆q = i∆t = 1 × 2= 2 C (d) H = i2R∆t = (1)2(2) (2) = 4J Here, B = µ0 i 2π d – x sinωt ∴ de = µ0iω x ⋅ dx 6. (a) If current is increased, ⊗ magnetic field 2π d – x sinωt passing through loop will increase. So, induced a µ 0iω ax dx current will produce u magnetic field. Hence, de ∫ ∫VOA = V0 – VA = = 0 d – x sinωt induced current is anti-clockwise. 0 2π Now, i and I currents in PQ are in opposite = – µ 0iω d ln  d – adsin ωt  + a directions. Hence, they will repel each other. 2π sinωt  sinωt Same logic can be applied for (b) part. Similarly, (c) situation is similar to (b) situation and µ 0iω a x dx (d) situation is similar to (a) situation. 2π 0 d + x sinωt ∫VOB = = VO – VB P = µ 0iω a – d ln  d + adsin ωt   2π sinωt sin ωt  iI Q ∴ VAB = VOB – VOA Subjective Questions = µ 0iω  + d ln d – aassiinnωωtt  2π sinωt 2a sin ωt  d +    L 1. τL = R Ans.

716 — Electricity and Magnetism Current through inductor will increase Note This function is discontinuous at ωt = nπ. exponentially from 0 to 5 A. 4. At t = 0, equivalent resistance of an inductor in ∴ i = 5  − − 2000t  infinite and at t = ∞, equivalent resistance is zero. 1 3  e 10 Ω 5Ω 6. These are two independent parallel circuits across the battery. (a) Vab = E = 120 volt (at all instants) (b) a is at higher potential. ∴ Initial current through inductor = 0 and (c) Vcd will decrease exponentially from 120 V to zero. Final current through inductor = 36 = 3.6 A 10 ∴ Vcd = 120 volt, just after the switch is closed. (d) c will be at higher potential. To find equivalent time constant, we will have to short circuit the battery and find net resistance (e) When switch is opened, current through R1 across inductor. will immediately become zero. While through Rnet = 10 × 5 = 10 Ω R2, will decrease to zero from the value 10 + 5 3 E = 2.4 A = i0 (say), exponentially. Path of L = 3 ms R2 τL = Rnet 10 this decay of current will be cdbac. Current through inductor will increase ∴ Just after the switch is opened, exponentially from 0 to 3.6 A. Vab = − i0R1 = − 2.4 × 30 = − 72 volt (f) Point b is at higher potential. ∴ i = 3.6(1 − e− t / τ L ), where τL = 3 ms = 300 µs 10 (g) Vcd = − i0(R1 + R2) = − 2.4(80) = − 192 volt Current through 10 Ω will vary with time. Ans. (h) This time point d will be at higher potential. 5. At t = 0, Current through inductor will be zero. 7. q1 = 8 CV0 : q2 = CV0 At t = ∞, net emf = 2/2 + 4/1 = 10 V 1/2 + 1/1 3 q1 + q2 = 9CV0 Net resistance = 2 × 1 = 2 Ω In the absence of inductor, this 9C0V will 2+1 3 distribute as 6 CV0 in 2C and 3CV0 in C. Thus, mean position of q1 is 6 CV0 and mean position of ∴ i = 10/3 = 5 A q2 is 3CV0. 2/3 –+ +– 2Ω q1 q2 L 1Ω At t = 0, q1 is 2CV0 more than its mean position and q2 is 2CV0 less. Thus, q0 = 2CV0 To find equivalent time constant short circuit, both C net = 2C 3 the batteries and find net resistance across inductor. ∴ ω= 1 = 3 Rnet = 2 ×1 = 2 Ω LC net 2LC 2+1 3 (a) Imax = q0ω = 1 × 10−3 ∴ τL = L 2/3 = 3s (b) V1 = 6CV0 = 3V0 Rnet 2000 2C

Chapter 27 Electromagnetic Induction — 717 and V2 = 3CV0 (d) F v = 0.8 × 7.5 = 6 W Ans. C (c) i = q0 sinωt  BRvl 2R B2l2 .v2 = 3V0 i2R = = R VL = (0.8)2(0.5)2 × (7.5)2 = 6 W Ans. 1.5 So, we can see that both rates are equal. 8. 11. (a) Magnitude of induced electric field due to change in magnetic flux is given by ∫ E ⋅ dl = dφ = S. dB dt dt 20 V (a) VL = L di = (1 × 10−3) d (20t) = 0.02 V or El = πR2(2B0t)  dB = 2 B0t dt dt dt = 20 mV Here, E = induced electric field due to change in magnetic flux t t 10t2 i dt = (20t) ∫ ∫(b) q = dt = or E(2πR) = 2πR2B0t 00 V =q = 10t2 = (10+7 t2) V or E = B0Rt C 10−6 Hence, F = QE = B0QRt (c) q2 > 1 Li2 This force is tangential to ring. Ring starts 2C 2 rotating when torque of this force is greater or (10t 2 )2 > 1 × 10−3 × (20t)2 than the torque due to maximum friction 2 × 10−6 2 ( fmax = µmg) or when or t > 63.2 × 10−6 s τF ≥ τ fmax Taking the limiting case or t > 63.2 µs Ans. τF = τ fmax or FR = (µmg) R or F = µmg 9. In steady state when switch was closed, or B0QRt = µmg i0 = E/R = (1/ 5)A = 0.2 A It is given that ring starts rotating after 2. So, After switch is opened, it becomes L-C circuit in putting t = 2, we get which peak value current is 0.2 A. µ = 2B0RQ mg ∴ 1 Li02 = 1 CV02 Ans. 2 2 or L = V02 .C (b) After 2 i02 τF > τ fmax = (150)2 × 0.5 × 10−6 Therefore, net torque is (0.2)2 τ = τF − tan fmax = B0QR2t − µmgR = 0.28 H Ans. Substituting µ = 2B0QR, we get mg 10. (a) e = Bvl = 0.8 × 7.5 × 0.5 = 3 V τ = B0QR2(t − 2) or I  ddωt  = B0QR2(t − 2) (b) Current will flow in anti-clockwise direction, as magnetic field in ⊗ direction passing or mR2  ddωt  = B0QR2(t − 2) through the closed loop is increasing. Therefore, induced current will produce magnetic field in u direction. (c) F = = = e lB =  BRvl = B2l2 v ω = B0Q 4 R R dω (t − 2) dt Fm ilB Bl ∫ ∫or 0 m2 = (0.8)2(0.5)2 × 7.5 = 0.8 N or ω = 2B0Q …(i) 1.5 m

718 — Electricity and Magnetism Now, magnetic field is switched off, i.e only 13. With key K1 closed, C1 and C2 are in series with retarding torque is present due to friction. So, the battery in steady state. angular retardation will be ∴ Cnet = 1µF or q0 = Cnet V = 20 µC (a) With K1 opened and K2 closed, charge on α = τ fmax = µmgR = µg I mR2 R C2 will remain as it is, while charge on C1 will oscillate in L-C1 circuit. Therefore, applying ω= 1 ω2 = ω 2 − 2αθ LC 1 0 =1  2 B0Q  2  µRg 0.2 × 10−3 × 2 × 10−6 m or 0 = − 2 θ or θ = 2B02Q 2R = 5 × 104 rad/s Ans. µm2g Substituting µ = 2B0RQ (b) Since, at t = 0, charge is maximum (= q0). mg Therefore, current will be zero. We get θ = B0Q Ans. 1 Li2 = 1  1 q2  m   2 3 2C 12. Let v be the velocity of connector at some instant or i = q = qω 3LC 3 of time. Then, b From the expression, i i = ω q02 − q2 Fv Fm – q We have, qω = ω q02 − q2 R + 3 i1 a i2 or q= 3 q0 2 Vab = Bvl, i1 = Bvl , q = C (Bvl) Since at t = 0, charge is maximum or q0, so we R can write ∴ i2 = dq = CBl dv q = q0 cos ωt or 3q0 = q0 cos ωt dt dt 2 Now, i = i1 + i2 = Bvl + CBl dv or ωt = π or t = π = 6× π R dt 6 6ω 5 × 104 Magnetic force, Fm = ilB = B2l2 ⋅ v + B2l2C ⋅ dv = 1.05 × 10−5 s R dt 3 3 × 20 = 10 Further, Fnet = F − Fm (c) q = 2 q0 = 2 3 µC Ans. or m dv = F − B2l2 − B2l2C dv 14. In the capacitor, v dt R dt qi = CVi = (10 × 10−3)(5) = 0.05 C v dv t dt − B2l2 0 m + B2l2C ∫ ∫∴0 = qf = CV f = (10 × 10−3)(10) = 0.1 C F R v Integrating we get, ∴ Charge in capacitor will increase from 0.05 C to 0.1 C exponentially. FR  −  B 2l2 2l 2C  t  Time constant for this increase would be B2l2 1 + RB  v = − e mR  τC = CR = 1 s. ∴ Charge at time t will be   q = 0.05 + (0.1 − 0.05)(1 − e−t/ τ C ) Terminal velocity in this case is : vT = FR Ans. B2l2 = 0.1 − 0.05 e−t/ τ C

Chapter 27 Electromagnetic Induction — 719 (a) At t = 1 s, q = 0.1 − 0.05 e−1 = 0.0816 C 17. Let at time t velocity of ring be v (downwards) V = q = 0.0816 = 8.16 V Ans. e = Bv (2r) = 2B v r C 10 × 10−3 (Two batteries of emf 2Bvr are connected in (b) This charge 0.0816 C is also the maximum parallel) R charge q0 of L-C oscillations. From energy conservation equation, ii T i/2 1 q02 = 1 Li02 we have, 2 C 2 Fm α v, a i0 = q0 = 0.0816 mg LC 25 × 10−3 × 10 × 10−3 = 5.16 A Ans. i/2 Further, ω = 1 or f = ω = 1 ∴ i = e = 2Bvr LC 2π 2π LC RR =1 Now, a = mg – Fm – T 2π 25 × 10−3 × 10 × 10−3 m = 10 Hz Ans. Here, Fm = 2  2i  (2r) B = 2irB = 4 B2r2v R 15. (a) Let at time t velocity of rod be v (towards ∴ a = g – 4 B2r2v – T …(i) mR m right) and current in the circuit is i (from a to b). The magnetic force on it is ilB (towards right). α = Tr = T …(ii) Writing the equation of motion of the rod, mr2 mr m⋅ dv = il B =  E0 – Blv lB a = rα = T …(iii) dt R m ∫ ∫v dv = t From Eqs. (i), (ii) and (iii), we get – B2l2 a = g – 2B2r2v 0 E0Bl dt 2 mR 0 mR mR v – B2l2 t v dv t 0 2 B 2r2v ∴ v = E0 (1 – e mR ) Ans. ∫ ∫or g = dt 0 Bl – 2 mR (b) i = E0 – Blv R or v = mgR (1 – – 2B 2r2 t Ans. 4 B2r2 16. Let v be the velocity at some instant. Then, e mR ) motional emf, V = Bvl 2 Bvr mg  − 2B 2r2 t  R 2Br 1  Charge stored in capacitor q = CV = (CBl)v i= = − e mR  Current in the wire = dq = (CBl) dv  dt dt and vT = mgR Ans. dv 4 B2r2 Magnetic force, Fm = ilB = (CB 2l 2 ) dt (upwards) ∴ Net force, Fnet = mg − Fm 18. (a) Suppose v be the velocity of rod ef when it has or m dv = mg − (CB2l2) dv fallen a distance, x. Then, dt dt V fe = Vcb or Bvl = L(di/dt) ∴ dv = acceleration, mg or B(dx/dt)l = L(di/dt) or Bl(dx) = L(di) dt + CB2l2 a = m Integrating, we get Li = Blx or i =  BLl x Since, a = constant …(i) ∴ x = 1 at2 = mgt2 Now, magnetic force opposite to displacement 2 2(m + CB2l2) x will be

720 — Electricity and Magnetism  B 2l 2  (b) From right hand rule, we can see that points a   and b will be at higher potential and c and d at F = Fm = ilB =  L x lower potentials. A constant downward force is mg. Fm = ilB = B2l2a0t (R1 + R2 ) R1R2 x=0 +ve Fm Let F be the external force applied, then, Mean F − Fm = ma0 mg position B2l2a0t ∴ F = Fm + ma0 = R1R2 (R1 + R2) + ma0 So, this is similar situation like spring-block 20. (a) At the given instant, system in vertical position. In which a force A F = kx acts upwards and a constant force mg acts downwards. BC a 3 a θ a/2 Hence, the wire will execute SHM, where O k = B2l2 L Amplitude will be at Fm = mg B or  B2l2  A = mg ⇒ A= mgL   B2l2  L AC = a , OC = a At t = 0, rod is in its extreme position. 22 Therefore, if we write the equation from mean and cosθ = a/ 2 = 1 , θ = π position we will write, a2 3 X = − A cos ωt ∴ Velocity of rod But, x = X + A = A − A cos ωt = A(1 − cos ωt)  + v0  where, ω = k = B2l2 = 2 along the direction of current. m mL Ans. Emf induced across the ends M and N (b) From Eq. (i), a 3+ a 3 imax =  BLl xmax 2 ∫Erod = 3−a 3 vrod B dx a 2 Here, xmax = 2A = 2mgL ∫= vrod 3a 3 µ 0i0 dx B2l2 2  Bl   2Bm2gl2L 2mg a 3 2πx L Bl 2 ∴ imax = = Ans.  v0  µ 0i0  13 = 2 2π ln (c) Maximum velocity,  B2l2   mBg2lL2 mL g mL with end M at higher potential. v0 = ωA =  mL  B2l2 Bl = g = Since, the effective length of both the arcs MAN and MBN is MN. Ans. Vrod Ans. 19. (a) At time t, v = a0t i0 M N x Motional emf, V = Bvl = Ba0lt dx Total resistance = R1R2 a3 R1 + R2 ∴ i = (Ba0lt)(R1 + R2) R1R2

Chapter 27 Electromagnetic Induction — 721 ∴ = = µ 0i0 3a 3 2π EMAN EMBN vloop ln 3 ∫Frod = 2 i dxB a3 µ 0i0 = v0 2π ln 3 2 = iµ 0i0 ln 3 = 9 µ 02i02v0 (ln 3)2 2π 32aRπ 3 with point M at higher potential. Resistance of arc MAN 21. Since, PQ and DC both cut the lines of field. ⇒ R1 = (R)2(aθ) = 2aR π ∴ Motional emf will be induced across both of 3 them. ⇒ Resistance of arc MBN π Integrating, potential difference across 3 ⇒ R2 = (R)a(2π − 2θ) = 4 aR ∫ ∫dx ⇒ 2a  µ 0i0  de = 2πx dx v Equivalent circuit at the given instant is shown in the figure. a eDC = vµ 0i0 ln 2 with D at higher potential 2π –EMAN R1 2vµ 0i0 ePQ = 2π ln 2 with P at higher potential i1 Erod The relative velocity of the rod PQ w.r.t. U frame i N vrel = 2v − v = v M Now, time taken by it to loose the contact t = l i2 v –EMBN R2 AB Current through the rod MN, EPQ l P Q i = (i1 + i2) =  EMAN − Erod  +  EMBN − Rrod   R1   R2  i = (EMAN − 1 + 1 EDC C Erod )  R1 R2  D = v0µ 0 i0(ln 3)  1 + 41 3 From equivalent electrical network 4π 2 aRπ  Net emf in the closed loop QPDC. = 9v0 i0µ 0 ln (3) Ans. e= ePQ − eDC = vµ 0i0 ln 2 16aRπ 2 2π (b) Force on the rod Growth of current in the L-R circuit is given by i = i0(1 − e−tR/ L ) =  Re  (1 − e−tR/ L ) x At time t = l dx v ⇒ i =  Re (1 − − Rl e vL )

28. Alternating Current INTRODUCTORY EXERCISE 28.1 = 1 10−6 (360)2 × 1. (a) X L = 2πfL = 7.7 H (b) L = X L Z =R (When X L = XC ) 2πf ∴ I = V = V = 120 Z R 20 (c) XC = 1 =6A 2πfC INTRODUCTORY EXERCISE 28.2 (d) C = 1 2πfXC 1. f = 1 = resonance frequency 2. VL = V 2 − VR2 2π 2C =1 = (150)2 − (100)2 2π 0.03 × 2 × 10−6 = 111.8 V = 650 Hz VL = IX L = I (2πfL) At resonance, X L = XC and Z = R ∴ L = VL = 111.8 ∴ cos φ = R = 1 or φ = 0° 2πfI 2π × 50 × 10 Z = 0.036 H 2. R = V = 40 = 4Ω 3. φ = 0, if XL = XC I 10 2πf L = 1 Z = V = 200 = 20 Ω or ∴ 2π f C I 10 Power factor, cos φ = R = 4 L = 1 C (2πf )2 Z 20 = 0.2 Exercises LEVEL 1 3. ω = 1 Assertion and Reason LC 1. Z = R2 + (XC − X L )2 By inserting a slab, C will increase. So, ω will decrease. From this expression, we can see that XC may be greater than Z also. 4. Average value = total area under i-t graph 2. At resonance frequency fr, total time interval = 8+ 2+2+ 4+ 2 XC = XL Now, X L = 2πfL or X L ∝ f 6 For f > fr , X L > XC = 18 = 3 A At resonance, X L = XC ⇒ Z = R ∴ cos φ = R = 1 6 Z 5. Z = R2 + (XC ~ X L )2 or φ = 0° X L will increase. So, Z may increase or decrease, depending on the value of XC.Therefore, current may decrease or increase.

Chapter 28 Alternating Current — 723 6. VL = VC ⇒ X L = XC 8. Q P = Vrms Irms cos θ So this resonance condition. =  V0   I 0  cos θ 2 8. P = I 2 R 2 rms =  2  2 = V0I0 cos θ 2 2 (10) = 20 W 9. V0 = 240 V 9. I DC = VDC (r = internal resistance of inductor) ∴ Vrms = V0 = 240 = 170 V r 2 2 I AC = VAC = VAC ω = 120 rad /s Z f = ω = 120 = 19 Hz r2 + X 2 L 2π 2 × 3.14 If VDC = VAC , then IDC > IAC 10. ω = 1 = 1 V 10. I = V = LC 0.5 × 8 × 10−6 R2 2 Z + X L = 500 rad /s =V 11. P = Vrms Irms cos φ R2 + (2πfL)2  1020  100 × 10−3  π  =  ⋅ cos 3 with increase in frequency, I will decrease.  2 tan φ = X L = 2πfL RR = 2.5 W with increase in frequency tan φ and therefore φ 12. XC = 1 will increase. ωC 11. At resonance, X L = XC ω = 0 for DC ∴ XC = ∞ ⇒ Z=R or it becomes a perfect insulator. Hence, I =V =V 13. Substituting t = 1 s in the given equation, we ZR 600 So, current at resonance depends on R. have Objective Questions V = 10 cos (100 π )  6010 2. Average value in AC comes out to be zero. = 10 cos π Z 6 3. =5 3V f =1 1 1 ωC 2πf C f 15. XC = or XC ∝ Impedance first decreases, then increases. At i.e. XC versus f graph is a rectangular hyperbola. resonance frequency Z is minimum. 4. In case of only capacitor and inductor phase 16. sin φ = X = 1 difference between current and voltage should Z3 Z be 90°. ∴ φ = sin−1  13 φ X R 5. I rms = I0 ≈ 0.707 I0 2 17. i = ωt π π 6. φ = 90° between V and I functions. 2 sin + 4 + 2  ∴ P = Vrms Irms cos φ = 0 Hence, phase difference between V and i is π . So, 2 7. Z = R2 + (X L − XC )2 power consumed = 0.

724 — Electricity and Magnetism 18. I DC = VDC 26. P = Vrms Irms cos φ R  V0   I 0  I = 100 ⇒ R = 100 Ω = 2 cos 60° R 2 IAC = VAC =  2220  42  21 R2 + X 2 = 220 W L 0.5 = 100 27. IC is 90° ahead of the applied voltage and IL lags (100)2 + X 2 behind the applied voltage by 90°. So, there is a L phase difference of 180° between IL and IC. ∴ I = IC − IL = 0.2 A or X L = 100 3 Ω = (2π f L) ∴ L = 100 3 = 100 3 28. VL function is cos function, which is 90° ahead of 2πf 2π (50)  3  the current function. Hence, current function = π   H should be sin function.  29. HDC = I 2Rt =1 1 19. XC ωC = × 10−6 = 104 Ω 2  I2 2 Rt = I 2 Rt rms 2 100 HAC = I Rt = I rms = Vrms = (200 2)/ 2 ∴ HDC = 2 XC 104 HAC 1 = 0.02 A = 20 mA 30. VC = V 2 − VR2 = (20)2 − (12)2 20. V = VR2 + VL2 = 16 V = (20)2 + (15)2 = 25 V Subjective Questions But this is the rms value. 1. X L = ωL = 100 Ω ∴ Peak value = 2 Vrms = 25 2 V =1 ωC 22. Resistance does not depend on the frequency of XC = 312.5 Ω AC. Z = R2 + (XC − X L )2 23. An ideal choke coil should have almost zero = (300)2 + (312.5 − 100)2 internal resistance. Otherwise, it will consume = 368 Ω some power. 24. 45° phase angle means, = V0 120 Z 368 XL = R (a) I0 = = 0.326 A ∴ (2πfL) = R ∴ L= R (b) Since, XC > X L, voltage lags the current by an 2πf angle given by = 100 φ = cos−1  ZR = cos−1  336080 = 35.3° (2π ) (103) (c) (V0)R = I0R = (0.326) 300 = 97.8 V = 0.0159 H (V0)L = I0X L = (0.326) (100) = 32.6 V ≈ 16 mH (V0)C = I0XC = (0.326) (312.5) = 102 V 25. T = 1 = 1 s 2. (a) Voltage lags f 50 ∴ XC > XL Power factor, cos φ = R t=T = 1 s 4 200 Z = 5 × 10−3 s =R R2 + (XC − X L )2 = 5 ms

Chapter 28 Alternating Current — 725 To increase the power factor denominator (b) X L = ωL = 950 × 0.8 = 760 Ω should decrease. Hence, X L should increase. Therefore, an inductor is required to be (c) (V0)L = I0X L connected. = (8.33 × 10−3) (760) (b) cos φ = R = 0.72 = 6.33V Z Now, VL function leads the current (or VR ) function by 90°. ∴ R = 0.72 Z = 0.72 × 60 ∴ VL = 6.33 cos (950 t + 90° ) = 43.2 Ω = − 6.33 sin (950 t) (XC − X L ) = (60)2 − (43.2)2 = 41.64 Ω 6. X L = 2π f L = 301 Ω New inductor of inductance 41.64 Ω should be XC = 1 = 55 Ω added in the circuit. 2πfC L = XL Z = R2 + (X L − XC )2 2πf = 41.64 = 0.133 H = (240)2 + (301 − 55)2 2π (50) = 343 Ω 3. (b) f = ω = 6280 = 1000 Hz (a) φ = cos−1  RZ  2π 2 × 3.14 = cos−1  234430 (c) φ = π − π = π or 30° 236 = 45.8° cos φ = R Power factor = cos φ = cos 30° Z =3 = 240 2 343 From the given functions of V and i, we can = 0.697 Since, X L > XC, voltage leads the current. see that current function leads the voltage (b) Impedance = Z = 343Ω (c) Vrms = Irms Z function. = 0.45 × 343 (d) Z = 170 = 20 Ω ...(i) = 155 V 8.5 (d) P = Ir2msR = (0.45)2 (240) cos φ = 3 = R = R 2 Z 20 = 48.6 W (e) P = PR = 48.6 W ∴ R = 10 3 Ω = 17.32 Ω (f) PC = 0 XC = Z2 − R2 (g) PL = 0 = 10 Ω C= 1 = 1 F ωXC 6280 × 10 = 15.92 × 10−6 F 4. I0 = V0 = V0 XL ωL 5. (a) I0 = (V0 )R = 2.5 LEVEL 2 R 300 = 8.33 × 10−3 A Single Correct Option = 8.33 mA 1. I2 = V =V ( here V = rms value) XC 3 Current function and VR function are in phase. Hence, I1 = V = V R 4 I = (8.33 mA) cos [(950 rad /s) t ]

726 — Electricity and Magnetism I2 is 90° ahead of applied voltage function and I1 is ∴ XC = R or ωC = R in phase with it. or C = R = R = 0.01 R I ω 100 In option (b), this condition is satisfied. 7. V = VR2 + (VC − VL )2 = 10 V φ VC > VL, hence current leads the voltage. I1 = V VC – VL 4 V tan φ = V /3 = 4 6 V 10 V V /4 3 ∴ φ = 53° φ VR 8V 2. IR and IL are in same phase and phase difference Power factor = cos φ = 8 = 0.8 between them and applied voltage lies between 0° 10 and 90°. 8. See the hint of miscellaneous example numbers 6 3. X L = ωL = (5 × 10−3) (2000) = 10 Ω and 7 of solved examples. =1 1 XC ωC = (2000) (50 × 10−6) = 10 Ω 9. VS = VR2 + VL2 Since, X L = XC circuit is in resonance. = (70)2 + (20)2 = 72.8 V Z = R = (6 + 4) = 10 Ω tan φ = X L = VL = 20 = 2 I rms = Vrms = (20/ 2) = 1.414 A R VR 70 7 Z 10 10. In first case, XC = V = 220 = 880 Ω This is also the reading of ammeter. I 0.25 V = 4Irms In the second case, R = V = 220 = 880 Ω I 0.25 ≈ 5.6 volt 4. IR = Vrms = 200 = 0.2 A In the combination of P and Q, R 100 tan φ = XC = 1 R XC = 1 = 1 2πfC ∴ φ = 45° (2π ) (5 × 103 )  1 × 10−6 π Since the circuit is capacitive, current leads the = 100 Ω voltage. Further, ∴ IC = Vrms = 200 = 2 A Z= R2 + X 2 = 880 2Ω XC 100 C IC is 90° ahead of the applied voltage and IR is in I = V = 220 = 1 A phase with the applied voltage. Hence, there is a Z 880 2 4 2 phase difference of 90° between IR and IC too. 11. See the hints of miscellaneous example numbers 6 ∴ I= 2 + IC2 and 7 of solved examples. R I 12. i = V , i.e. circuit is in resonance. Hence, = (2)2 + (2)2 R = 283 A VC = VL = 200 V  VZrms 2 5. Average value of 5 sin 100 ωt is zero. But average 13. P = I 2 R = R rms value of 5A (= constant current) is 5 A. Hence,  (V0 / 2 )2  average value of total given function is 5 A.  R2 + ω 2L2    6. V function is sin function. I function is ahead of V = R function. Hence, the circuit should be capacitive in = V02R nature. (R2 + ω2L2) Further, φ = 45° 2

Chapter 28 Alternating Current — 727 14. X L = ωL Solving these three equations, we get VR = 50 V, VL = 86.6 V and If ω is very low, then X L ≈ 0 VC = 206.6 V ∴ VL ≈ 0 Power factor = cos φ = R = VR = 50 = 5 Z V 130 13 or V = VC = V0 Since VC > VL, circuit is capacitive in nature. 15. I max = V (at resonance) R 2. i = 5 sin (ωt + 53°) 6 = 24 45 R ∴ R=4Ω I DC = V r = 12 = 1.5 A R+ 4+4 53° sin ωt 16. VR = V 2 − VC2 = (10)2 − (8)2 = 6 V 3 tan φ = XC = VC = 8 = 4 i0 = 5A X R VR 6 3 ∴ irms = i0 = 5A 2 2 17. Current will lead the voltage function by 90° Mean value of current in positive half cycle is voltage function is cos function. Therefore, current function will be − sin function. 2 i0 =  π2 (5) =  1π0 A π i t0 t In V = Vm sin ωt, current i = 5 sin (ωt + 53° ) leads the voltage function. Hence, circuit is capacitive in nature. Same is the case with part (d). t0 =T = (2π /ω) = π 3. PR = VR i 4 4 2ω ∴ i = PR = 60 = 1 A VR 60 = π =1s Now, VL = V 2 − VR2 2 (π /2) = (100)2 − (60)2 R= 1 18. ωC = XC = 80 V = i X L = i (2πfL) L = 80 ∴ Z = R2 + XC2 = 2 R (as XC = R) 2πfi I0 = V0 = V0 ...(i) Z 2R = 80 = 4 H (2π ) (50) (1) 5π When ω becomes 1 3 If we connect another resistance R in series, then it 3 times, XC will become should consume 40 V, so that remaining 60 V is used by the tube light. times or 3 R. R = V = 40 = 40 Ω Z′ = (R2) + ( 3R)2 = 2R i1 I 0′ = V0 = V0 = I0 4. Power factor, cos φ = R Z′ 2R 2 Z More than One Correct Options ...(i) 1. VR2 + VL2 = 100 ...(ii) When circuit contains only resistance, then ...(iii) Z = R ⇒ cos φ = 1 VL ~ VC = 120 VR2 + (VL ~ VC )2 = 130 When circuit contains only inductance, then R=0 ∴ cos φ = 0

728 — Electricity and Magnetism 5. (a) X L > XC, hence voltage function will lead the Comprehension Based Questions current function. 1 to 3. VDC = IDC R R = VDC = 12 = 3 Ω (b) Z = R2 + (X L − XC )2 ∴ IDC 4 = (10)2 + (20 − 10)2 = 10 2 Ω VAC VAC Z (c) cos φ = R = 1 I AC = = R2 2 Z2 L + X Hence, φ = 45° 2.4 = 12 2 (d) Power factor = cos φ = R = 1 (3)2 L + X Z2 Solving this equation, we get 6. (b) At resonance frequency (ωr ) XL = 4 Ω XL > XC XC =1 = 1 ωC 50 × 2500 × 10−6 In the given values, X L > XC. Hence, ω > ωr =8Ω As, X L = ωL ⇒ X L ∝ ω Z = R2 + (XC − X L )2 = 5Ω and XC = 1 ⇒ XC ∝ 1 ∴ I = VDC = 12 = 2.4 A = I rms ωC ω Z 5 (c) If frequency is increased from the given value, P = I 2 R = (2.4 )2 (3) rms X L will further increase. So, X L − XC will increase. Hence, net impedance will increase. = 17.28 W (d) If frequency is decreased from the given value, At given frequency, XC > X L. If ω is further decreased, XC will increase  as XC ∝ ω1 and X L then XC will increase and X L will decrease. So, X L − XC may be less than the previous will increase (as X L ∝ ω). value or XC − X L may be greater than the previous. So, Z may either increase or Therefore, XC − X L and hence Z will increase. So, current will decrease. decrease. Hence, current may decrease or 4. ω = 1 increase. LC 7. (a) VR = IR = 80 V (b) XC = VC = 100 = 50 Ω =1 I 2 4.9 × 10−3 × 10−6 (c) VL = IX L = 40 V = 105 rad /s 7 (d) V = Vrms = VR2 + (VC − VL )2 = (80)2 + (100 − 40)2 5. XC =1 = 1 = 70 Ω ωC = 60 V  105  (10−6 )    7 ∴ V0 = 2 Vrms = 60 2 V 8. I = V ZP = RP2 + X 2 C Z = (32)2 + (70)2 =V 2 ≈ 77 Ω ωL 1  R2 + ~ ωC 6. At maximum current means at resonance, By increasing R, current will definitely decrease XL = XC, Z = R by change in L or C, current may increase or ∴ Power factor = cos φ = R = 1 decrease. Z

Chapter 28 Alternating Current — 729 Match the Columns ∴ L= 1 − R tan φ (2πf )2C 2πf 2. (a) φ = 0° between voltage function and current = (2π × 1 20 × 10−6 − 3.125 tan 83.8° function. 50)2 × 2π × 50 (b) I = I0 sin (ωt − 90° ) = 0.416 H (b) tan φ = X L − XC i.e. φ = 90° and voltage function leads the current function. R (c) Current function leads the voltage function. So, or 2πfL − 1 = R tan φ Ans. (2πfC ) XC > XL (d) Voltage function leads the current function. 1 R tan φ (2πf )2C (2πf ) So, L= + XL > XC 1 3.125 tan 83.8° 50)2 (2π × 50) 3. See the hint of Q.No. 6 and 8 of section more than = (2 × π × × 20 × 10−6 + one correct options. Then, P = I 2 R = 0.597 H Ans. rms By increasing R, current irms will decrease but the 2. Average current will be zero as positive and 2 power, P = I rms R may increase or decrease. negative half cycles are symmetrical RMS current can also be obtained from 0 to τ / 2. 4. (a) R = VR = 40 = 20 Ω I =  τI/02 t =  2I 0  t I2  τ (b) VC = IXC = 2 × 30 = 60 V (c) VL = I X L = 2 × 15 = 30 V ⇒ I2 =  4 I 2  t2  0    (d) V = VR2 + (VC − VL )2 = 50 V τ2 5. (a) Resistance does not depend on the value of ω. τ / 2 4 I 2 0 ∫= t2dt =1 1 0 τ2 (b) XC ωC or XC ∝ ω ⇒ I2 0− τ/2 τ/2 (c) X L = ωL or X L ∝ ω or I 2 = I 2 (d) Z is minimum at ω = ωr and Zmin = R 0 Below or above ωr 0− τ/2 3 Z = R2 + (X L ~ XC )2 ⇒ I rms = I 2 = I0 0 33 or Z > R 3. (a) 0.5 = R1 Subjective Questions Z1 1. P = Vrms irms cos φ Further, P = Vrms irms cos φ or 200 = 230 × 8 × cos φ or 100 = 230 × 230 × 0.5 Z1 ∴ cos φ = 0.108 ∴ Z1 = 264.5 Ω or φ = 83.8° and R1 = 132.25 Ω Further, P = ir2msR Further, X L = Z12 − R12 = 3 Z1 2 ∴ R = P = 200 = 3.125 Ω ir2ms (8)2 = 229 Ω (a) tan φ = XC − X L In second case, 0.6 = R2 R Z2 ∴ 1 − (2πfL) = R tan φ and 60 = 230 × 230 × 0.6 2π f C Z2

730 — Electricity and Magnetism ∴ Z2 = 529 Ω 6. (a) VR = IR = 80 V,VC = 100 V Ans. and Further, R2 = 317.4 Ω and VL = IX L = 160 V XC = Z22 − R22 ∴ V = VR2 + (VL − VC )2 = 100 V = 423.2 Ω Note Value of XL have been taken from part (b). When connected in series, (b) Since the current is lagging behind, there R = R1 + R2 = 449.65 Ω should be an inductor in the box. XC − X L = 194.2 XC = VC = 100 Ω ∴ Z = (449.65)2 + (194.2)2 I = 489.79 Ω Now, 0.8 = R = 80 Power factor, cos φ = R = 0.92 (leading) Z (80)2 + (X L − 100)2 Z Solving, we get P = Vrms irms cos φ = (230)  48293.079 (0.92) X L = 160 Ω or ωL = 160 = 99 W Ans. ∴ (2πfL) = 160 ∴ L = 160 = 160 (b) Since, XC − X L = 194.2 Ω 2πf (2π ) × 50 Therefore, if 194.2 Ω inductive reactance is to = 1.6 H be added in series, then it will become only R π circuit and power factor will become unity. 7. (a) 4. (a) irms = V1 = 40 = 10 A R 4 i0 = 2irms = 10 2 A Ans. 200√2 V V12 + (V2 − V1)2 = 50 V 45° (b) E0 = 2 θ1 400V ∴ E0 = 50 2 V sin ωt (c) XL = (ωL) = V2 = 40 = 4 Ω Reference circle for voltage irms 10 ∴ L= 4 = 4 H= 1 H Ans. ω 100π 25π XC =1 = V1 = 10 =1Ω ωC irms 10 ∴ C=1= 1 F Ans. 10 A ω 100π 60° 5. cos φ1 = 0.5 θ2 20 A ∴ φ1 = 60° cos φ2 = 3 sin ωt 2 Reference circle for current ∴ φ2 = 30° ω = 2πf = (100π ) rad/s Let R be the effective resistance of the box. Then, From the above two figures, we can write φ1 = XC 3 = XC tan R or R …(i) = 400 sin (ωt + θ1) = 400 sin 100πt + π 4  V Ans. tan φ2 = XC or 1 = XC …(ii) 100πt π R + 10 3 R + 10 6  i = 20 sin (ωt + θ2) = 20 sin + Ans. From these two equations, we get R = 5 Ω

Chapter 28 Alternating Current — 731 (b) Phase difference between V and i X L2 = ωL2 = (2π × 50)(0.08) φ = (π/4 − π/6) = π or 15° = 25.13 Ω 12 ∴ Z2 = R22 + X 2 P = Vrms irms cos φ =  4020  202 cos 15° L2 = 25.15 Ω = 3864 W Ans. ∴ P2 = (irms)2Vrms cos φ2 =  2150.105 (100)  251.15 8. ω = 2 = 2 LC 5 × 10−3 × 20 × 10−6 = 15.8 W = 6324.5 rad/s ∴ PTotal = P1 + P2 = 797 W X L = ωL = (6324.5)(5 × 10−3) = 31.62 Ω 10. Z1 = 115 = 38.33 Ω 3 =1 = 1 = 7.9 Ω XC ωC 6324.5 × 20 × 10−6 cos φ1 = R1 Z1 ∴ Z = X L − XC = 23.72 Ω (a) Maximum voltage across capacitor ⇒ R1 = Z1 cos φ1 = (38.33)(0.6) = 23 Ω = i0 XC = (0.211)(7.9) = 1.67 mV X L = Z12 − R12 = 30.67 Ω ∴ Maximum charge q0 = (20 × 10−6)(1.67 × 10−3) = 33.4 nC Z2 = 115 = 23 Ω 5 (b) i0 = V0 = 5 mA = 0.211 mA Z 23.72 R2 = Z2 cos φ2 = (23) (0.707) = 16.26 (c) Since X L > XC , current in the circuit will lag behind the applied voltage by π/2. XC = Z22 − R22 Further voltage across the inductor will lead = (23)2 − (16.26)2 this current by π/2. Therefore, applied voltage and voltage across = 16.26 Ω inductor are in phase. When connected in series, R = R1 + R2 = 39.26 Ω Voltage across the capacitor will lag the circuit current by π/2. X L − XC = 14.41 Ω ∴ Z = R2 + (X L − XC )2 Therefore, phase difference between VL and VC will be 180°. = 41.82 Ω (a) i = V = 230 = 5.5 A 9. X L1 = ωL1 = (2π × 50)(0.02) = 6.28 Ω Z 41.82 ∴ Z1 = R12 + X 2 (b) P = i2R = (5.5)2(39.26) L1 = (5)2 + (6.28)2 = 8.0 Ω = 1187.6 W ≈ 1.188 kW (c) Power factor = cos φ = R = 39.26 = 0.939 P1 = (Irms)1Vrms cos φ1 =  1080 (100)  58 Z 41.92 = 781.25 W Since X L > XC , this power factor is lagging.

JEE Main and Advanced Previous Years’ Questions (2018-13) JEE Main 1. Three concentric metal shells A, B and C 5. An electron, a proton and an alpha of respective radii a, b and c (a < b < c) particle having the same kinetic energy have surface charge densities + σ , − σ and +σ, respectively. The potential of shell B are moving in circular orbits of radii re, rp, is (2018) rα respectively, in a uniform magnetic field B. The relation between re, rp, rα is (2018) (a) σ a2 − b2 + c (b) σ a2 − b2 + c (a) re > rp = rα (b) re < rp = rα ε0  a ε0  b (c) re < rp < rα (d) re < rα < rp (c) σ b2 − c2 + a (d) σ b2 − c2 + a 6. The dipole moment of a circular loop ε0  b ε0  c carrying a current I, is m and the 2. A parallel plate capacitor of capacitance magnetic field at the centre of the loop is 90 pF is connected to a battery of emf B1. When the dipole moment is doubled by keeping the current constant, the 20 V. If a dielectric material of dielectric magnetic field at the centre of the loop is constant K = 5 is inserted between the B2. The ratio B1 is 3 B2 (2018) plates, the magnitude of the induced (d) 1 charge will be (2018) (a) 2 (b) 3 (c) 2 2 (a) 1.2 nC (b) 0.3 nC (c) 2.4 nC (d) 0.9 nC 3. In an AC circuit, the instantaneous emf 7. For an R-L-C circuit driven with voltage of and current are given by amplitude vm and frequency ω 0 = 1, LC 30 t π e = 100 sin 30 t, i = 20 sin − 4 the current exhibits resonance. The quality factor, Q is given by (2018) In one cycle of AC, the average power (a) ω0L (b) ω0R R L consumed by the circuit and the wattless (c) R (d) CR current are, respectively ω0C ω0 (a) 50 , 10 (b) 1000 , 10 2 8. In a potentiometer experiment, it is found (c) 50 , 0 2 (d) 50 , 0 that no current passes through the galvanometer when the terminals of the 4. Two batteries with emf 12 V and 13 V are cell are connected across 52 cm of the connected in parallel across a load resistor potentiometer wire. If the cell is shunted of 10 Ω. The internal resistances of the by a resistance of 5 Ω, a balance is found two batteries are 1 Ω and 2 Ω, respectively. The voltage across the load lies betwe(2e0n18) when the cell is connected across 40 cm of the wire. Find the internal resistance of (a) 11.6 V and 11.7 V (b) 11.5 V and 11.6 V the cell. (2018) (c) 11.4 V and 11.5 V (d) 11.7 V and 11.8 V (a) 1 Ω (b) 1.5 Ω (c) 2 Ω (d) 2.5 Ω

2 Electricity & Magnetism 9. On interchanging the resistances, the 14. A capacitance of 2 µF is required in an balance point of a meter bridge shifts to electrical circuit across a potential difference of 1kV. A large number of 1 µF the left by 10 cm. The resistance of their capacitors are available which can withstand a potential difference of not series combination is 1 kΩ. How much more than 300 V. The minimum number of capacitors required to achieve this is was the resistance on the left slot before (2017 ) interchanging the resistances? (2018) (a) 990 Ω (b) 505 Ω (c) 550 Ω (d) 910 Ω 10. In the below circuit, the current in each (a) 16 (b) 24 (c) 32 (d) 2 resistance is (2017 ) 2V 2V 2V 15. In the given circuit diagram, when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be (2017 ) 1Ω 1Ω 1Ω Er 2V 2V 2V r1 C (a) 0.25 A (b) 0.5 A (c) 0 A (d) 1 A 11. When a current of 5 mA is passed through r2 a galvanometer having a coil of resistance (b) CE r2 (r + r2 ) 15 Ω, it shows full scale deflection. The (a) CE r1 value of the resistance to be put in series (r2 + r ) (d) CE with the galvanometer to convert it into a (c) CE r1 (r1 + r ) voltmeter of range 0-10 V is (2017 ) (a) 2.045 × 103 Ω (b) 2.535 × 103 Ω 16. In a coil of resistance 100 Ω, a current is (c) 4.005 × 103 Ω (d) 1.985 × 103 Ω induced by changing the magnetic flux 12. Which of the following statements is through it as shown in the figure. The false? (2017 ) magnitude of change in flux through the (a) In a balanced Wheatstone bridge, if the cell coil is (2017 ) and the galvanometer are exchanged, the null point is disturbed 10 (b) A rheostat can be used as a potential divider Current (A) (c) Kirchhoff’s second law represents energy conservation Time 0.5 (s) (d) Wheatstone bridge is the most sensitive when all the four resistances are of the same (a) 225 Wb (b) 250 Wb (c) 275 Wb (d) 200 Wb order of magnitude 17. A galvanometer having a coil resistance of 13. An electric dipole has a fixed dipole 100 Ω gives a full scale deflection when a current of 1 mA is passed through it. The moment p, which makes angle θ with value of the resistance which can convert this galvanometer into ammeter giving a respect to X-axis. When subjected to an full scale deflection for a current of 10 A, electric field E1 = Ei$, it experiences a is (2016 ) torque T1 = τk$ . When subjected to another (a) 0.01 Ω (b) 2 Ω (c) 0.1 Ω (d) 3 Ω electric field E2= 3E1$j, it experiences a torque T2 = − T1. The angle θ is (2017 ) (a) 45° (b) 60° (c) 90° (d) 30°

Previous Years’ Questions (2018-13) 3 18. The region 21. Hysteresis loops for two magnetic materials A and B are as given below: between two (2016 ) concentric spheres Qa BB of radii a and b, respectively (see HH the figure), has volume charge b density ρ = A, r where, A is a constant and r is the distance from the centre. At the centre of (A) (B) the spheres is a point charge Q. The value These materials are used to make magnets for electric generators, of A, such that the electric field in the transformer core and electromagnet core. Then, it is proper to use region between the spheres will be (a) A for electric generators and transformers constant, is (2016 ) (b) A for electromagnets and B for electric (a) Q (b) Q generators 2 πa2 2 π(b2 − a2 ) (c) A for transformers and B for electric (c) 2Q (d) 2Q generators π(a2 − b2 ) πa2 (d) B for electromagnets and transformers 19. A combination of capacitors is set-up as shown in the figure. The magnitude of the electric field, due to a point charge Q 22. An arc lamp requires a direct current of (having a charge equal to the sum of the 10 A at 80 V to function. If it is connected charges on the 4 µF and 9µF capacitors), to a 220 V (rms), 50 Hz AC supply, the at a point distant 30 m from it, would series inductor needed for it to work is equal to (2016 ) close to (2016 ) 3µF (a) 80 H (b) 0.08 H 4µF (c) 0.044 H (d) 0.065 H 9µF 2µF 23. Arrange the following electromagnetic radiations in the order of increasing +– energy. (2016 ) (a) 240 N/C 8V A. Blue light B. Yellow light (c) 420 N/C (b) 360 N/C C. X-ray D. Radio wave (d) 480 N/C (a) D, B, A, C (b) A, B, D, C (c) C, A, B, D (d) B, A, D, C 20. Two identical wires A and B, each of 24. When 5V potential difference is applied length l, carry the same current I. Wire A across a wire of length 0.1m, the drift speed of electrons is 2.5 × 10− 4 ms−1 . If the is bent into a circle of radius R and wire B electron density in the wire is 8 × 1028 m− 3 the resistivity of the material is close to is bent to form a square of side a. If BA and BB are the values of magnetic field at the centres of the circle and square (2015) respectively, then the ratio BA is (2016 ) (a) 1.6 × 10−8 Ω - m BB (b) 1.6 × 10−7 Ω - m (c) 1.6 × 10− 5 Ω - m (a) π2 (b) π2 (c) π2 (d) π2 (d) 1.6 × 10− 6 Ω - m 8 16 2 16 82

4 Electricity & Magnetism 25. In the circuit shown below, the current in (a) Charge (b) Charge the 1Ω resistor is Q2 Q2 6V P 2Ω 3 µF C 3 µF C 1 µF 1 µF 1Ω 9V Charge Charge (c) Q2 (d) Q2 3Ω Q 3Ω (2015 ) 1 µF 3 µF C 1 µF 3 µF C (a) 1.3 A, from P to Q (b) 0.13 A, from Q to P 29. Two coaxial solenoids of different radii (c) 0 A (d) 0.13 A, from P to Q carry current I in the same direction. Let 26. A uniformly charged solid sphere of radius F1 be the magnetic force on the inner solenoid due to the outer one and F2 be the R has potential V0 (measured with respect magnetic force on the outer solenoid due to ∞) on its surface. For this sphere, the to the inner one. Then, equipotential surfaces with potentials (2015 ) 3V0 , 5V0 , 3V0 and V0 have radius R1 , (a) F1 is radially outwards and F2 = 0 2 44 4 (b) F1 is radially inwards and F2 is radially outwards R2, R3, and R4 respectively. Then, (2015 ) (c) F1 is radially inwards and F2 = 0 (a) R1 ≠ 0 and (R 2 − R 1) > (R 4 − R 3) (d) F21 = F2 = 0 (b) R1 = 0 and R 2 > (R 4 − R 3) (c) 2R < R 4 30. Two long current carrying (d) R1 = 0 and R 2 < (R 4 − R 3) thin wires, both with L current I, are held by θ insulating threads of 27. A long cylindrical shell carries positive length L and are in surface charge σ in the upper half and equilibrium as shown in negative surface charge − σ in the lower the figure, with threads I I half. The electric field lines around the making an angle θ with cylinder will look like figure given in the vertical. If wires have mass λ per unit (figures are schematic and not drawn to length then, the value of I is scale) (2015 ) (g = gravitational acceleration) (2015 ) (a) +–+––+–+––+–+– (b) +––+–+–+––+––+ (a) 2 sinθ πλgL (b) sinθ πλgL µ 0 cos θ µ 0 cos θ (c) 2 πgL tan θ (d) πλgL tan θ µ0 µ0 31. A rectangular loop of sides 10 cm and 5 cm (c) +––+–+–+––+––+ (d) +–+––+–+––+––+ carrying a current I of 12 A is placed in different orientations as shown in the figures below. (2015 ) zz 28. In the given circuit, 1µF I B 2µF BI I charge Q2 on the 2µF C (a) y (b) I I I y capacitor changes as I x I C is varied from 1 µF z x to 3 µF. Q2 as a z function of C is given E IB B properly by (figures I y (d) I I y are drawn schematically and are not to (c) I I I xI scale) (2015 ) x

Previous Years’ Questions (2018-13) 5 If there is a uniform magnetic field of 0.3 T 36. The coercivity of a small magnet where in the positive z-direction, in which orientations the loop would be in the ferromagnet gets demagnetised is (i) stable equilibrium and (ii) unstable equilibrium? 3 × 103 Am−1. The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet (a) (a) and (b) respectively gets demagnetised when inside the (b) (b) and (d) respectively solenoid is (2014 ) (c) (a) and (c) respectively (a) 30 mA (b) 60 mA (c) 3 A (d) 6 A (d) (b) and (c) respectively 32. An inductor (L = 0.03 H) and a resistor 37. In the circuit shown here, the point C is (R = 0.15 kΩ ) are connected in series to a kept connected to point A till the current battery of 15V EMF in a circuit shown flowing through the circuit becomes below. The key K1 has been kept closed constant. Afterward, suddenly point C is for a long time. Then at t = 0, K1 is opened disconnected from point A and connected and key K 2 is closed simultaneously. At to point B at time t = 0. Ratio of the t = 1 ms, the current in the circuit will be voltage across resistance and the inductor (e5 =~ 150) (2015 ) at t = L / R will be equal to (2014 ) 0.03H 0.15 kΩ ACR K2 K1 L B 15V (a) e (d) 1 − e (a) 100 mA (b) 67 mA (c) 0.67mA (d) 6.7 mA e 33. In a large building, there are 15 bulbs of 1− e (b) 1 (c) −1 40 W, 5 bulbs of 100 W, 5 fans of 80 W 38. A conductor lies along the z-axis at −1.5 ≤ z < 1.5 m and carries a fixed and 1 heater of 1 kW. The voltage of the current of 10.0 A in −az direction (see figure). electric mains is 220 V. The minimum For a field B = 3.0 × 10−4 e−0.2xay T, find the power required to move the conductor at capacity of the main fuse of the building constant speed to x = 2.0 m, y = 0 in 5 × 10−3s. will be (2014 ) Assume parallel motion along the x-axis. (a) 8 A (b) 10 A (c) 12 A (d) 14 A (2014 ) 34. Assume that an electric field E = 30x2 $i z 1.5 exists in space. Then, the potential difference VA − VO, where VO is the (2014 ) potential at the origin and VA the potential at x = 2 m is (a) 120 J (b) −120 J (c) −80 J (d) 80 J 35. A parallel plate capacitor is made of two l By circular plates separated by a distance of –1.5 5 mm and with a dielectric of dielectric 2.0 x constant 2.2 between them. When the electric field in the dielectric is 3 × 104 V/m, the charge density of the positive plate will be close to (2014 ) (a) 1.57 W (b) 2.97 W (c) 14.85 W (d) 29.7 W (a) 6 × 10−7 C/m2 (b) 3 × 10−7 C/m2 (c) 3 × 104 C/m2 (d) 6 × 104 C/m2

6 Electricity & Magnetism 39. This question has Statement I and 43. Two capacitors C1 and C2 are charged to Statement II. Of the four choices given after the statements, choose the one that 120 V and 200 V respectively. It is found best describes the two statements. (2013 ) that by connecting them together the potential on each one can be made zero. Statement I Higher the range, greater Then (2013 ) is the resistance of ammeter. (a) 5C1 = 3C2 (b) 3C1 = 5C2 Statement II To increase the range of (c) 3C1 + 5C2 = 0 (d) 9C1 = 4C2 ammeter, additional shunt needs to be used across it. 44. Two short bar magnets of length 1 cm (a) If Statement I is true, Statement II is true; each have magnetic moments 1.20 Am2 Statement II is the correct explanation for Statement I and 1.00 Am2, respectively. They are (b) If Statement I is true, Statement II is true; placed on a horizontal table parallel to Statement II is not a correct explanation for Statement I each other with their N poles pointing (c) If Statement I is true; Statement II is false towards the South. They have a common (d) If Statement I is false; Statement II is true magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the mid-point O of the line joining their 40. The supply voltage in a room is 120 V. centres is close to (Horizontal component The resistance of the lead wires is 6 Ω. A of the earth’s magnetic induction is 60 W bulb is already switched on. What is 3.6 × 10−5 Wb/m2) (2013 ) the decrease of voltage across the bulb, (a) 3.6 × 10−5 Wb /m2 (b) 2.56 × 10−4 Wb /m2 (c) 3.50 × 10−4 Wb /m2 (d) 5.80 × 10−4 Wb /m2 when a 240 W heater is switched on in parallel to the bulb? (2013 ) 45. A circular loop of radius 0.3 cm lies (a) zero (b) 2.9 V (c) 13.3 V (d) 10.4V parallel to a much bigger circular loop of 41. Two charges, each equal to q, are kept at radius 20 cm. The centre of the smaller x = − a and x = a on the x-axis. A particle loop is on the axis of the bigger loop. The of mass m and charge q0 = q is placed at distance between their centres is 15 cm. If 2 a current of 2.0 A flows through the the origin. If charge q0 is given a small bigger loop, then the flux linked with displacement y ( y << a) along the y-axis, smaller loop is (2013 ) the net force acting on the particle is (a) 9.1 × 10−11 Wb (b) 6 × 10−11 Wb proportional to (2013 ) (c) 3.3 × 10−11 Wb (d) 6.6 × 10−9 Wb (a) y (b) − y (c) 1 (d) − 1 46. A metallic rod of length l y y is tied to a string of length 2l and made to 2l l 42. A charge Q is uniformly distributed over a ω long rod AB of length L as shown in the figure. The electric potential at the point rotate with angular speed O lying at distance L from the end A is ω on a horizontal table with one end of the string fixed. If there is (2013 ) a vertical magnetic field B in the region, OA B the emf induced across the ends of the L L rod is (2013 ) (a) Q (b) 3 Q (a) 2B ωl 3 (b) 3B ωl 3 8 π ε0L 4 π ε0L 2 2 (c) Q (d) Q ln (2) (c) 4Bωl 2 (d) 5Bωl 2 4 π ε0L ln (2) 4 π ε0L 2 2

Previous Years’ Questions (2018-13) 7 47. In a L-C-R circuit as shown below, both (a) Work done by the battery is half of the energy dissipated in the resistor switches are open initially. Now, switch (b) At t = τ, q = CV / 2 S1 and S2, are closed. (q is charge on the (c) At t = 2τ, q = CV (1 − e−2 ) capacitor and τ = RC is capacitance time (d) At t = τ / 2, q = CV (1 − e−1) constant). Which of the following statement is correct? (2013 ) 48. The amplitude of a damped oscillator V decreases to 0.9 times its original C S1 magnitude is 5 s. In another 10 s, it will R S2 decrease to α times its original L magnitude, where α equals (2013 ) (a) 0.7 (b) 0.81 (c) 0.729 (d) 0.6 Answer with Explanations 1. (b) Potential of B = Potential due to charge on Wattless current is, A + Potential due to charge on B + Potential due to I = Irms sin φ charge on C. = 20 × sin π 24 C = 20 = 10A BA 2 +σ ∴ Pav = 1000 watt –σ 2 +σ and Iwattless = 10A 10 Ω 4. (b) k(QA + QB) + kQC ∴ VB = b c = 1  σ 4 πa2 − σ 4 πb2 + σ 4 πc 2  + 12V 4 πε0  b b c  – 1Ω = σε  a2 − b2 + c2  = σ  a2 − b2 + c  – ε0  b c  ε0  b  + 2Ω  a2 − b2 13V σ  b  VB = ε0 + c  For parallel combination of cells, 2. (a) Magnitude of induced charge is given by E1 + E2 r1 r2 Q′ = (K − 1) CV0 Eeq = 1+ 1 =  5 − 1 90 × 10−12 × 20 = 1.2 × 10−9C r1 r2 3 12 + 13 ⇒ Q′ = 1.2 nC ∴ Eeq = 1 = 37 V 1+ 2 3 3. (b) Given, e = 100sin 30 t 1 and i = 20sin  30 t − π  12 4 Potential drop across 10 Ω resistance, ∴ Average power , 37 Pav = VrmsIrms cos φ V =  E × 10 = 3 2  × 10 = 11.56V  Rtotal  10 + 3 = 100 × 20 × cos π = 1000 watt 22 42 ∴ V = 11.56 V

8 Electricity & Magnetism Alternative Method 7. (a) Sharpness of resonance of a resonant L-C-R circuit E I2 13 V, 2 Ω is determined by the ratio of resonant frequency with the selectivity of circuit. This ratio is also called F I1 12V, 1Ω D ‘‘Quality Factor’’ or Q-factor. C Q-factor = ω0 = ω0L = 1 A B 2 ∆ω R ω0CR 10 Ω 8. (b) With only the cell, Applying KVL, in loop ABCFA, E′ −12 + 10 (I1 + I2 ) + 1 × I1 = 0 …(i) ⇒ 12 = 11I1 + 10 I2 52 cm Similarly, in loop ABDEA, G E, r −13 + 10 (I1 + I2 ) + 2 × I2 = 0 ⇒ 13 = 10 I1 + 12 I2 …(ii) Solving Eqs. (i) and (ii), we get On balancing, E = 52 × x …(i) where, x is the potential gradient of the wire. I1 = 7 A, I2 = 23 A 16 32 When the cell is shunted, ∴Voltage drop across 10 Ω resistance is, E′ V = 10  7 + 23  = 11.56 V 16 32 40 cm 5. (b) From Bqv = mv 2 , we have r r = mv = 2 mK G Bq Bq E, r where, K is the kinetic energy. As, kinetic energies of particles are same; R=5 Ω r∝ m ⇒ re : rp : rα = me : mp : 4mp Similarly, on balancing, q e e 2e V = E − Er = 40 × x (R + r) …(ii) Clearly, rp = rα and re is least [Q me < mp] …(i) So, rp = rα > re Solving Eqs. (i) and (ii), we get 6. (c) As m = IA, so to change dipole moment (current is E = 1 = 52 V r 40 kept constant), we have to change radius of loop. 1− R+ r Initially, m = IπR 2 and B1 = µ 0I 2 R1 ⇒ E = R + r = 52 ⇒ 5 + r = 52 V R 40 5 40 Finally, m′ = 2m = IπR 2 2 ⇒ r = 3 Ω ⇒ r = 1.5 Ω 2 ⇒ 2 IπR12 = IπR 2 2 or R2 = 2 R1 9. (c) We have, X + Y = 1000 Ω So, B2 = µ 0I = µ 0I X Y=1000 – X 2(R2 ) 2 2 R1 Hence, ratio B1 =  µ 0I  = 2 G  2 R1  100 – l B2  µ 0I  l    2 2 R1  X = 1000 − X Initially, l 100 − l ∴ Ratio B1 = 2 B2 When X and Y are interchanged, then