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NUMERICALS-ANSWERS QUESTIONS A coil of area 20 cm2 is placed so that a magnetic field of 0.1 T is perpendicular to the plane of the coil If the coil is turned through 180° in 0.1 s, what is the e.m.f induced ? A coil of 100 turns and area of cross section 0.01 m2 is placed in a magnetic field of 1.0 T perpendicular to the plane of the coil If the coil is withdrawn from the field in 0.05 s, what is the induced e.m.f ? A train is moving at 90 km h–1 The axle of the wheel of the carriage is 1.8 m long If the vertical component of earth’s magnetic field is 0.5 G, what is the potential difference developed across the ends of the axle !" A circular coil of radius 10 cm is placed in a uniform horizontal magnetic field B !"of strength –2 0.5 × 10 T Find the magnetic !" flux crossing the coil if its plane is !(i) " parallal to B, (ii) inclined at an angle of 60° to B, (iii) inclined at an angle of 45° to B !" A rectangular coil cm × 3!"cm is placed in a uniform magnetic!" field B such that its plane makes an angle of 90º with B What should be the strength of B to have a magnetic flux of 1.5 × 10–6 Wb through the coil Magnetic flux of microweber is linked with a coil when a current of mA flows through it What is the self inductance of the coil (AISSCE 1993) The current in a 0.5 mH solenoid varies from to 3A in ms Calculate the induced e.m.f produced An induced e.m.f of 200 V is developed across a solenoid when the current through it is changed from A to in 0.01 s Calculate the self inductance of the solenoid The mutual inductance between the primary and the secondary of a transformer is 10 H Compute the induced e.m.f in the secondary when 0.5 A current in primary is cut off in 0.1 second S Chand & Company Limited 10 How much current is drawn by the primary of an ideal transformer if it steps down 110 V to 11V, given that the output resistance is 220 Ω 11 The two rails of a railway track, insulated from each other and the ground, are connected to a voltmeter What is the reading of the voltmeter when a train travels at 108 kmh–1 along the track Vertical component of earth’s magnetic field is 0.20 G and the rails are separated by a distance of m 12 A wheel with 10 metallic spokes, each 0.50 m long, is rotated with a speed of 120 r.p.m in a plane normal to the earth’s magnetic field at that place If the magnitude of the field is 0.40 G, What is the induced e.m.f between the axle and the rim of the wheel ? (NCERT Book) 13 A short solenoid of length 4.0 cm, radius 2.0 cm and 100 turns is placed inside on the axis of a long solenoid of length 80.0 cm and 1500 turns A current of 3.0 A flows through the short solenoid Find (a) the mutual inductance of the two solenoids, (b) the magnetic flux through the long solenoid (NCERT Book) 14 A toroidal solenoid with an air core has an average radius of 15 cm, area of cross section 12 cm2 and 1200 turns Compute its self inductance µ0 = 4π × 10–7 Hm–1 15 In a car spark coil, when the current in the primary is reduced from 4.0 A to zero in 10 µs, an e.m.f of 40,000 V is induced in the secondary Find the mutual inductance M of the primary and the secondary windings of the spark coil 16 In a 0.4 mH coil, the current rises uniformly from zero to 250 mA in 0.1 s Find the self induced e.m.f in the coil 17 A closed coil consists of 500 turns wound on a rectangular frame of area 0.4 cm2 and has a resistance of 50 Ω It is kept with its plane perpendicular to a uniform magnetic field of 0.2 Wb m–2 Calculate the amount of charge flowing through the coil if it is turned over (rotated through 180°) Will this answer depend on the speed with which the coil is rotated ? S Chand & Company Limited 18 A 28 turn coil with average diameter of 0.02 m is placed perpendicular to a magnetic field of 8000 T If the field changes to 3000 T in 4s, what is the magnitude of the induced e.m.f ? (π = 22/7) 19 A Jet plane is travelling west at a speed of 1800 km h–1 What is the voltage developed between the ends of the wing 25 m long, if the earth’s magnetic field at that location has a magnitude of 5.0 × 10–4 T ? Dip angle is 30° 20 A long solenoid of 10 tuns/cm has a small loop of area cm2 placed inside with the normal of the loop parallal to the axis Calculate the voltage across the small loop if the current in the solenoid is changed at a steady rate from A to A in 0.1 s (NCERT Book) 21 A wheel with 10 metallic spokes each 0.50 m long is rotated with a speed of 120 rev/min in a plane normal to the earth’s magnetic field at the place If the magnitude of the field is 0.40 G what is the induced e.m.f between the axle and the rim of the wheel ? (NCERT Book) 22 What is the self inductance of a solenoid of length 40 cm, area of cross section 20 cm2 , and total number of turns 800 ? (NCERT Book) 23 A coil consists of 2000 tuns of wire How many volts of e.m.f will be developed if the flux linked to it is changed from 5.0 × 10–4 Wb to 0.5 × 10–4 Wb in 0.1 s If the resistance of the coil is 100 Ω, calculate the strength of the induced current ANSWERS ε=– Here ∆φ ∆t B = 0.1 T, A = 20 cm2 = 20 × 10–4 m2 φ1 = BA cos 0° = 0.1× 20 × 10–4 = × 10–4 Wb φ2 = BA cos 180º = –BA = – × 10–4 Wb ∆ φ=φ2 – φ1 = – × 10–4 Wb S Chand & Company Limited ∴ ε= × 10 –4 = ×10 –3 V 0.1 ∆φ ∆t N = 100, A = 0.01 m2 , t = 0.5 s B=1T φ1 = BA = × 0.01 = 0.01 Wb φ2 = Zero ∆ φ = – 0.01 Wb – 0.01 = 20V ∴ ε = – 100× 0.05 Induced emf ε = Bvl Here ε = 0.5 G = 0.5 × 10–4 T, l = 1.8 m v = 90 × = 25 m/s 18 ∴ ε = 0.5 × 10–4 × 25 × 1.8 ε = –N = 22.5 × 10–4 = 2.25 × 10–3 V Magnetic flux is given by φ = BA cos θ, where θ is the angle between the normal to the plane of the coil and the direction of the magnetic field Here r = 10 cm = 0.1 m A = πr2 = π × (0.1)2 m2 S Chand & Company Limited B = 0.5 × 10–2 T (i) When the plane is parallal to B, θ = 90° – 0° = 90° ∴ φ = 0.5 × 10–2 × π × (0.1)2 (ii) In this case θ = 90º – 60º = 30º φ = BA cos θ × cos 90º = 0.5 × 10–2 × π × (0.1)2 × cos 30º = 1.4 × 10–4 Wb (iii) θ = 45º φ = BA cos θ = 0.5 × 10–2 × π × (0.1)2 × cos 45º = 1.1 × 10–4 Wb Magnetic flux φ = BA Here φ = 1.5 × 10–6 Wb A = × 10–2 × × 10–2 = 15 × 10–4 m2 φ 1.5 × 10 –6 B= = 1×10 –3 T = –4 A 15 × 10 φ = LI Here φ I φ = × 10–6 Wb, I = × 10–3 A ∴ L= or L= × 10 –6 10 –3 = × 10 –3 H S Chand & Company Limited =L Self induced e.m.f dI dt = 0.5 × 10–3 × Self induced e.m.f ε = L or 10–3 = 1.5V di dt ε di / dt 200 0.01 = = 200 × = 2H 1.0/0.01 1.0 L= Induced e.m.f in the secondary dI | εS | = M P dt 10×0.5 = = 50V 0.1 10 εS = 11 V, R = 220 Ω eS 11 = = A = 0.05 A 220 20 R εS i = P εP iS iS = Now Here ε P = 110 V S Chand & Company Limited i 11 = P 110 0.05 ∴ or iP = 11 × 0.05 = × 10 –3 A 110 11 v = 108 km/h = 108 × = 30 m/s BH = 0.20 × 10–4 T, l = 1m ε = BH vl = 0.20 × 10–4 × 30 × = × 10–4 V 12 Let r be the radius of the wheel Area of the wheel = π r Area swept by each spoke per second = π r ν Change in magnetic flux per second = B π r ν ω dφ = Bπr ν = Br dt 120 = 0.50 m, ν = = rps 60 Induced e.m.f | ε | = Here r B = 0.40 × 10 –4 T | ε | = (0.40 × 10–4 ) × (3.14) × (0.50)2 × = 6.28×10 –5 V S Chand & Company Limited 13 (a) Mutual inductance of two solenoids is given by A M = µ N1 N 2 l1 where is used for the long solenoid and is used for the short solenoid Here N1 = 1500, N2 = 100, A2 = π (2 × 10–2)2 m2 l1 = 80.0 × 10–2 m ∴ M= π × 10 –7 × 1500 ×100 × π × × 10 –4 (80 ×10 –2 ) = 2.96 × 10–4 H (b) Total flux through the long solenoid due to current i2 in the short solenoid is given by φ1 = M i = (2.96 × 10–4) × 3.0 = 8.88 × 10–4 Wb 14 The self inductance of a long solenoid is given by µ0 N A l For a toroidal solenoid l = πr L= ∴ L= µ0 N A 2πr Here r = 15 cm = 0.15 m, A = 12 × 10–4 m2, S Chand & Company Limited N = 1200 ∴ L= π × 10 –7 × (1200)2 × 12 × 10 –4 π × 0.15 = 2.3 × 10 –3 H di dt Here ε = 40,000 V, di = 4.0 A dt = 10 µs = 10 × 10–6 s 40, 000 40, 000 × 10 –5 ∴ M= = 4 /(10 × 10 –6 ) 15 We have ε = M = 0.1H di 16 ε = L dt Here L = 0.4 × 10–3 H, di = 250 × 10–3 A dt = 0.1 s ∴ ε= 0.4 × 10 –3 × 250 × 10 –3 = 10 –3 V 0.1 N × ∆φ R Here N = 500, R = 50 Ω When the coil is turned through 180º, change in the magnetic flux ∆φ = 2BA 17 Charge q = = × 0.2 × × 10–4 = 1.6 × 10–4 Wb S Chand & Company Limited 500 × 1.6 × 10 –4 = 1.6 × 10 –3 C 50 Answer does not depend on the speed of rotation of the coil 18 r = 0.01 m A = π r2 = π × (0.01)2 m2 ∴ ∴ ∴ q= φ1= B1A = 8000 × π × (0.01)2 Wb φ2= B2A = 3000 × π × (0.01)2 Wb ∆ φ = 5000 × π × (0.01)2 Wb ε =N dφ dt 28 × 5000 × π × (0.01)2 = 2.2 V = 500 ms –1 19 v = 1800 kmh–1 = 1800 × 18 l = 25 m = B = × 10–4 T, Induced e.m.f θ = 30º ε = Bvl sin θ ε = × 10–4 × 500 × 25 × sin 30º 3.125 V 20 Magnetic field inside the solenoid B = µ0ni Rate of change of magnetic field dB di = µ0 n dt dt S Chand & Company Limited = π × 10 –7 × (10 × 100) × 0.1 = 12.57 × 10 –3 T/s Therefore, induced e.m.f is dB ε =A dt = (12.57 × 10–3) × (10–4) = 12.57 × 10–7 V 21 l = 0.50 m, B = 0.4 × 10–4 T Induced e.m.f ε = ν= 120 = rev/s 60 (ωBL2 ) (2πν BL2 ) = × (2 × 3.14 × × 0.40 × 10 –4 × 0.5 × 0.5) = 6.28 × 10 –5 V 22 L = 40 cm = 0.40 m or ε= A = 20 × 10–4 m2,N = 800 Self inductance L = µ0 n2 lA = µ0 N2 A/l S Chand & Company Limited 4π × 10 –7 × (800) × 20 × 10 –4 0.40 –3 = 4.02 × 10 H = 4.02mH | ∆φ | 23 Induced e.m.f | ε | = n ∆ t = Here n = 2000, | ∆φ | = | φ – φ1 | = | (0.5 × 10–4 – 50 × 10–4)| = 4.5 × 10–4 Wb ∆t = 0.1 s ∴  4.5 ×10 –4  | ε | = 2000 ×   = 9.0 V  0.1  Induced current i = ε / R = 9.0 = 0.09 A 100 S Chand & Company Limited