Title 6.ELECTROMAGNETIC INDUCTION - Explanations.pdf ✅

1 (d) During decay of current i = i0e − Rt L = E R e − Rt L = 100 100 e − 100×10−3 100×10−3 = 1 e A 2 (d) Whenever a magnet is moved either towards or away from a conducting coil, the magnetic flux linked with the coil changes and therefore, an emf is induced in the coil. The magnitude of induced emf e = −N dφ dt e = −N d(BA) dt Time interval dt, depends on the speed with which the magnet is moved. Therefore, the induced emf is independent of the resistance of the coil. 3 (a) H = V 2t R and V = N(B2−B1 )A cos θ t V = 1 × (1 − 2) × 0.01 × cos 0° 10−3 = 10 V So, H = (10) 2×10−3 0.01 = 10 J 4 (c) e = NBAω; ω = 2πf = 2π × 2000 60 ∴ e = 50 × 0.05 × 80 × 10−4 × 2π × 2000 60 = 4π 3 5 (c) When frequency is high, the galvanometer will not show deflection 6 (c) At t = 0 inductor behaves as broken wire then i = V R2 At t = ∞ Inductor behaves as conducting wire i = V R2R2/(R1 + R2) = V(R1 + R2) R1R2 8 (a) φ = BA = 10 weber 9 (b) V = 200V; r = 10Ω R ′ = 10 + 100Ω = 110Ω I = V R′ = 220 110 = 2A P = I 2R = 4 × 100 = 400W 10 (d) By Fleming’s right hand rule 11 (d) Mutual inductance between two coil in the same plane with their centers coinciding is given by M = μ0 4π ( 2π 2R2 2N1N2 R1 ) henry 12 (a) φ = Li ⇒ NBA = Li Since magnetic field at the centre of circular coil carrying current is given by B = μ0 4π . 2πNi r ∴ N. μ0 4π . 2πNi r . πr 2 = Li ⇒ L = μ0N 2πr 2 Hence self inductance of a coil = 4π × 10−7 × 500 × 500 × π × 0.05 2 = 25 mH 13 (b) Induced emf is given by e = − dφ dt If the radius of loop is r at a time t, then the instantaneous magnetic flux is given by φ = πr 2B ∴ e = − d dt (πr 2B) e = −πB ( 2r dr dt ) e = −2π Br dr dt Numerically, e = 2πBr ( dr dt) 14 (a) When a north pole of a bar magnet moves towards the coil, the induced current in the coil flows in a direction such that the coil presents its north pole to the bar magnet as shown in figure (a). Therefore, the induced current flows in the coil in the anticlockwise direction. When a north pole of a bar magnet moves away from the coil, the induced current in the coil flows in a direction such that the coil presents its such pole to the bar magnet as shown in figure (b) Therefore induced current flows in the coil in the clockwise direction 15 (b) Polarity of emf will be opposite in the two cases while entering and while leaving the coil. Only in option (b) polarity is changing. 16 (c) From Faraday’s law of electromagnetic induction e = − dφ dt = −BAN Given, B = 0.1 T, N = 20, A = πr 2 = π(0.1) 2 ∴ e = −0.1 × 20 × π(0.1) 2 = 20π mV 17 (b)
We know that for step down transformer Vp > Vs but Vp Vs = is ip ⇒ is > ip Current in the secondary coil is greater than the primary 19 (c) e = Bvl ⇒ e ∝ v ∝ gt 22 (a) ∵ L ∝ N 2 r; L1 L2 = ( N1 N2 ) 2 × r1 r2 ⇒ L L2 = ( 1 2 ) 2 × ( r r/2 ) = 1 2 ; L2 = 2L 23 (b) Induced e.m.f. = Blv = 0.3 × 10−4 × 10 × 5 = 1.5 × 10−3V = 1.5 mV 24 (b) e ∝ dφ dt ; if φ → maximum then e → minimum 25 (b) According to Lenz’s law of electromagnetic induction, the relative motion between the coil and magnet produces change in magnetic flux. 29 (c) Since the rod is moving in transverse magnetic field, so it will cut no flux passing through the field and hence no induced emf is produced. So, no current will flow through the rod. 30 (a) h = L − L cos θ ⇒ h = L(1 − cos θ) ....(i) ∴ v 2 = 2gh − 2g L(1 − cos θ) = 2g L (2 sin2 θ 2 ) ⇒ v = 2√gL sin θ 2 Thus, maximum potential difference Vmax = BvL = B × 2√gL sin θ 2 L = 2BL sin θ 2 (gL) 1/2 32 (a) Current in the inner coil i = e R = A1 R1 dB dt Length of the inner coil = 2πa So it’s resistance R1 = 50 × 10−3 × 2π(a) ∴ i1 = πa 2 50 × 10−3 × 2π(a) × 0.1 × 10−3 = 10−4A According to lenz’s law direction of i1 is clockwise Induced current in outer coil i2 = e2 R2 = A2 R2 dB dt ⇒ i2 = πb 2 50 × 10−3 × (2πb) × 0.1 × 10−3 = 2 × 10−4A(CW) 33 (a) Back emf ∝ speed of motor 34 (c) E.m.f. or current induces only when flux linked with the coil changes 36 (c) L = μ0N 2A l = 4π × 10−7 × (1000) 2 × 10 × 10−4 1 = 1.256 mH 37 (d) Rod is moving towards east, so induced emf across it’s end will be e = BVvl = (BH tan φ)vl ∴ e = 3 × 10−4 × 4 3 × (10 × 10−2) × 0.25 = 10−5V = 10μV 38 (b) The flux associated with coil of area A and magnetic induction B is φ = BA cos θ = 1 2 Bπr 2 cos ωt [∵ A = 1 2 πr 2 ] ∴ einduced = − dφ dt = − d dt ( 1 2 Bπr 2 cos ωt) = 1 2 Bπr 2ω sin ωt ∴ power p = einduced 2 R = B 2π 2 r 4ω 2 sin2 ωt 4R Hence,Pmean =< p > = B 2π 2 r 4ω 2 4R . 1 2 (∵< sinωt >= 1 2 ) = (Bπr 2ω) 2 8R 39 (d) By Faraday’s second law, induced emf e = − Ndφ dt which gives e = −L dI dt ∴ |e|=2 × 10−3 × 20 × 10−3V = 40μV 40 (a) As the north pole approaches, a north pole is developed at the face, i. e., the current flows anticlockwise. Finally when it completes the oscillation, no emf is present. Now south pole approaches the other side, i. e., RHS, the current flows clockwise to repel the south pole. This means the current is anticlockwise at the LHS a
before. The break occurs when the pendulum is at the extreme and momentarily stationary 41 (b) ΔQ = NBA R (cos θ1 − cos θ2 ) = 500 × 0.2 × 0.1(cos 0 − cos 180) 50 = 0.4 C 42 (c) |e| = L | di dt| = 0.5 × 10 2 = 2.5V 43 (a) M = K√L1L2 For perfect coupling K = 1 M12 = M21 44 (c) Given;NP NS = 1 25 , VP = 230V,Is = 2 A For an ideal transformer NP NS = VP VS = IS IP or NP NS = IS IP Or IP = IS × NP NS = 2A × 25 1 = 50A 45 (c) According to Lenz’s Law 46 (b) At t = 0 current through L is zero so it acts as open circuit. The given figures can be redrawn as follow i1 = 0 i2 = E R i3 = E 2R Hence i2 > i3 > i1 47 (c) e = −N ( ΔB Δt ) . A cos θ = −100 × (6 − 1) 2 × (40 × 10−4) cos 0 ⇒ |e| = 1 V 48 (b) When ring enters and leaves the field polarity of induced emf is opposite. Also during the stay of ring completely in the field there is no induction 49 (d) |e| = dφ dt = 8 × 10−4 0.4 = 2 × 10−3V 51 (b) e0 = ωNBA = (2πv)NB(πr 2) = 2 × π 2v NBr 2 = 2 × (3.14) 2 × 1800 60 × 4000 × 0.5 × 10−4 × (7 × 10−2) 2 = 0.58 V 52 (c) Horizontal conductor intercepts vertical component = B0 sin δ ∴ e = (B0 sin δ)lv 53 (a) |e| = L di dt ⇒ 1 = L × [10 − (−10)] 0.5 ⇒ L = 25mH 54 (d) e = Bl 2πv = 0.4 × 10−4 × (0.5) 2 × (3.14) × 120 60 = 6.28 × 10−5V 55 (b) ( dφ dtr) In first case = e ( dφ dt ) relative velocity 2v = 2 ( dφ dt ) I case = 2e 56 (b) e = Bvl ⇒ e = 0.7 × 2 × (10 × 10−2) = 0.14 V 59 (a) In step-up transformer, number of turns in primary coil is less than the number of turns in secondary coil. ie, Ns Np > 1 60 (b) e ∝ ω 61 (a) Mutual inductance for two concentric coplanar circular coils, M = πμN1N2r 2 2R Here, N1 = N2 = 1 ∴ M = πμ0r 2 2R 62 (b) If resistance is constant (10Ω) then steady current in the circuit i = 5 10 = 0.5 A. But resistance is (i) S R E (ii) E R S R (iii) E R S R N S v v r O R A B I
increasing it means current through the circuit starts decreasing. Hence inductance comes in picture which induces a current in the circuit in the same direction of main current. So i > 0.5 A 63 (b) This is the case of periodic EMI 64 (b) Magnetic induction depends upon the magnetic permeability of medium between the coils (μr) or nature of material on which two coils are wound. 65 (c) As per the phenomenon of mutual induction when two coils are placed near each other and current is passed through one of them then due to the phenomenon of electromagnetic induction current is induced in the other coil, in this case since, current in loop A increases with time, hence direction of current induced in loop B will be same as direction of current in loop A. 67 (a) When the secondary coil circuit is open, the magnetic flux in the core is produced by the primary current only. When the secondary circuit is closed, the currents in the secondary coil also produce magnetic flux in the core but in opposite direction. This decreases the core flux and hence reduces the back emf more current is drawn in the primary coil. Hence, power factor is no longer zero. The power factor has increased or the phase difference is no longer 90°, i. e., phase difference has decreased. Thus, dynamic resistance has increased 68 (c) In a transformer 1. Iron losses In actual iron cores, inspite of lamination, eddy current are produced. The magnitude of eddy current may however be small and a part of energy is lost as the heat produced in the iron core. 2. Copper losses In practice, the coils of the transformer possess resistance. So, a part of the energy is lost the due to the heat produced in the resistance of the coil. 3. Flux leakage The coupling between the coils is seldom perfect. So whole of the magnetic flux produced by the primary coil is not linked up with the secondary coil. And hysteresis loss, humming losses also occur in the transformer. 70 (d) Magnetic field, φB = BA cos θ Where θ is the angle between normal to the plane of the coil and magnetic field Induced emf, ε = BA sin θ Here, θ = 0° ∴ Magnetic flux is maximum and induced emf is zero 71 (d) Transformer doesn’t work on dc 72 (d) Emf induced in the wire is given by e = Blv Given, l = 50 cm = 0.5 m v = 300 m −min-1= 300 60 = 5 ms−1 e = 2 V Magnetic field, B = e lv = 2 0.5×5 = 0.8 T 73 (a) L ∝ n (Number of turns). For straight conductor n = 0, hence L = 0 74 (a) The current flows through the coil 1 is I1 = I0 sinωt Where I0 is the peak value of current Magnetic flux linked with the coil 2 is φ2 = MI1 = MI0 sin ωt Where M is the mutual inductance between the two coils The magnitude of induced emf in coil 2 is |ε2 | = dφ2 dt = d dt (MI0 sin ωt) = MI0ω cos ωt ∴ Peak value of voltage induced in the coil 2 is = MI0ω = 150 × 10−3 × 2 × 2π × 50 = 30π V 76 (b) The emf developed between the centre and the rim is e = 1 2 Bωl 2 = 1 2 × 0.05 × 60[1] 2 = 1.5 V 77 (c) According to Faraday’s law, “the induced emf in a closed loop equals the time rate of change of magnetic flux through the loop.” ie., e = − dφB dt Hence, induced emf in a coil depends on rate of change of flux. 78 (a) Induced current in the circuit i = Bvl R Magnetic force acting on the wire Fm = Bil = B ( Bvl R ) l ⇒ Fm = B 2vl 2 R . External force needed to move the rod with constant velocity (Fm) = B 2vl 2 R = (0.15) 2 × (2) × (0.5) 2 3 = 3.75 × 10−3N 79 (a)