17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 195 CHAPTER ELECTROMAGNETIC 17 INDUCTION Concept Map 6HOI,QGXFWLRQ E = – L dI dt LI = N φ Solenoid: L = P0N2A l 0XWXDOLQGXFWLRQ M = P0NpNs A l p 0DJQHWLF)OX[ φ = %·$ (OHFWURPDJQHWLF,QGXFWLRQ )DUDGD\¶V/DZDQG /HQ]¶VODZ Induced e.m.f. in linear conductors plane coils (QHUJ\6WRUHG U = 1 2 LI 2 195 Bilingual Keywords (IZVS\[L KL[LYTPUH[PVU VM YLZPZ[HUJL! 7LULU[\HU T\[SHR YPUN[HUNHU (UNSLVMKPW!:\K\[JVUKVUNHU -HYHKH`»ZKPZJ!*HRLYH-HYHKH` -HYHKH`»ZSH^!/\R\T-HYHKH` -SLTPUN»ZYPNO[OHUKY\SL!7LYH[\YHU[HUNHURHUHU-SLTPUN -S\VYLZJLU[SHTW!3HTW\ILYWLUKHYÅ\VY 0UK\JLKLTM!+NLHY\OHU 3LUa»ZSH^!/\R\T3LUa 3PULHYJVUK\J[VY!2VUK\R[VYS\Y\Z 4HNUL[PJÅ\_SPURHNL!9HUNRHPHUÅ\RZTHNUL[ 4\[\HSPUK\J[HUJL!0UK\R[HUZZHSPUN 4\[\HSPUK\J[PVU!(Y\OHUZHSPUN 7LHRLTM!+NLW\UJHR 9VV[TLHUZX\HYLYTZLTM!+NLW\UJHTPUR\HZHK\H WTRK :LSMPUK\J[HUJL!0UK\R[HUZKPYP :LSMPUK\J[PVU!(Y\OHUKPYP INTRODUCTION 1. Electromagnetic induction is the process of producing an electric current when the magnetic fl ux linked with a conductor changes. 2. Research done by physicists such as Faraday and Lenz contributed greatly to the generation of electricity, a form of energy which we sometimes take for granted.
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 196 17.1 Magnetic Flux Students should be able to: GHÀQHPDJQHWLFÁX[DVφ = B·A Learning Outcome 1. In the previous chapter, magnetic fl ux density B had been defi ned in terms of s THE FORCE F on a charge moving in the magnetic fi eld, or s THE FORCE F on a current-carrying conductor in the magnetic fi eld. 2. The magnetic fl ux, φ through an area A in a magnetic fi eld of fl ux density B(Figure 17.1(a)) is given by φ = B·A. Magnitude of the magnetic fl ux, φ = BA cos θ B A B A (a) φ = BA (a) φ = BA cos θ Figure 17.1 3. Figure 17.2(a) shows a coil of N turns and cross-sectional area A with its axis parallel to a uniform magnetic fi eld of fl ux density B. Figure 17.2 (b) Magnetic fl ux linkage Φ = Nφ Φ = NBA cos T Φ = Nφ Φ = NBA (a) Magnetic fl ux linkage A B Coil Axis of coil θ A B Axis of coil Coil The magnetic fl ux through one turn of the coil is φ = BA. The magnetic fl ux linkage through the coil, Φ1 = Nφ Φ = NBA 4. Figure 17.2(b) shows the axis of the coil at an angle θ to the direction of the magnetic fi eld. The magnetic fl ux linkage Φ = Nφ, φ = BA cos θ Φ = NBA cos θ The unit for magnetic fl ux is the weber (Wb).
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 197 Quick Check 1 1. An area of 0.20 m2 is placed in a magnetic fi eld B = 0.64 T. B 0.20 m2 B 0.20 m2 30° Find the magnetic fl ux φ through the area (a) when the area is perpendicular to the magnetic fi eld, (b) when the area is tilted through an angle of 30°. 2. A circular coil of area 6.0 cm2 and 50 turns is placed with its plane perpendicular to a uniform magnetic fi eld B = 0.35 T. The coil is then rotated about an axis perpendicular to the magnetic fi eld. B = 0.35 T Axis of rotation Calculate the magnetic fl ux linked with the coil (a) when the plane of the coil is perpendicular to the magnetic fi eld, (b) after the coil is rotated through 40°. 17.2 Faraday’s Law and Lenz’s Law Students should be able to: VWDWHDQGXVH)DUDGD\ҋVODZDQG/HQ]ҋVODZ GHULYHDQGXVHWKHHTXDWLRQIRULQGXFHGHPILQOLQHDUFRQGXFWRUVDQGSODQHFRLOVLQXQLIRUPPDJQHWLFÀHOGV Learning Outcomes 1. In 1931, Michael Faraday carried out an experiments on electromagnet induction, and the results of his experiments were summarized as Faraday’s law of electromagnet induction. 2. Faraday’s law, state that when the magnetic flux linked with a circuit changes, an electromotive force (e.m.f.) which is directly proportional to the rate of change of magnetic fl ux is induced. 3. The direction of the induced current is given by Lenz’s law: The direction of the induced current is such that it opposes the change in magnetic fl ux that gives rise to it. 4. Lenz’s law is consistent with the principle of conservation of energy. When an e.m.f. is induced in a circuit, the electric energy generated is from the conversion of mechanical work done in opposing the force produced by the effect of the induced current. 5. Faraday’s law and Lenz’s law are combined in the equation Induced e.m.f., E = – —– dΦ dt where Φ is the magnetic fl ux linkage in the circuit. Lenz’s law is represented by the negative sign in the above equation. 2011/P1/Q31, 2012/P1/Q32, 2014/P2/Q20(a), 2015/P2/Q13, 2016/P2/Q14, 2017/P2/Q17 VIDEO ,SLJ[YVTHNUL[PJ ;YPJRZ
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 198 Example 1 The diagram shows two coils P and Q arranged coaxially. The coil Q is connected in series with a battery, rheostat and switch K. Initially the switch K is closed. For each of the following changes, show on a diagram the direction of the induced current in coil P. (a) Coil P is moved closer to coil Q. (b) Coil Q is moved closer to coil P. (c) Switch K is opened. (d) The slider of the rheostat is moved to the right. Exam Tips When applying Lenz’s law, identify 1. the change 2. what is the opposite of the change? 3. what is the direction of the induced current to produce the desired effect in (2)? Solution: (a) The change is: Coil P moving closer to coil Q. Opposite of this change is: Coil P is moved away from Q. In order for coil P to be repelled from Q, the direction of the induced current in P must be opposite to that of Q. P Q Current from battery Induced current (b) The change: Coil Q moving towards coil P. Opposite to this change is: Coil Q repelled from coil P. In order for coil P to be repelled, the induced current in P must be opposite to the direction of the current in Q. P Q Current from battery Induced current (c) The change: Switch K is opened. Magnetic fl ux produced by current in Q decreases to zero. Opposite to this change: Maintain magnetic fl ux in the same direction. To achieve this, the induced current in P must be in the same direction as that in Q. P B Initial flux linkage Q P Flux linkage produced by induced current Q Induced current P Q S K
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 199 (d) The change: Slider of rheostat moved to the right. Resistance increases, current decreases and magnetic fl ux decreases. Opposite to this change: Maintain magnetic flux in the same direction. To achieve this, the induced current in P must be in the same direction as in Q. Example 2 The diagram shows a coil with its plane perpendicular to a magnetic fi eld B directed into the plane of the paper. Use Faraday’s law and Lenz’s law to discuss the production of an induced current, and the direction of the induced current in the resistor R when the magnetic fi eld strength is reduced. Solution: A current is induced in the circuit because the magnetic fl ux linked with the coil decreases. To oppose the decrease in the magnetic fl ux linkage, the induced current must fl ow through the coil such that the direction of the magnetic fi eld inside the coil is the same as the original fi eld. Hence, the direction of the induced current in the coil and the resistor is as shown in the diagram. Quick Check 2 P Induced current R B Induced current R I 1. The e.m.f. E induced in a coil where the magnetic fl ux Φ changes is given by E = – —– dΦ dt . Hence the unit for magnetic fl ux in terms of the base units is A m s–2 A–1 C m s–1 A B kg m s2 A–1 D kg m2 s–2 A–1 2. X and Y are two solenoids arranged coaxially. X carries a constant current I as shown in the diagram and moves with a constant speed away from Y. As a result of electromagnetic induction, an induced current fl ows in the straight wire MN, and there is a force between X and Y. X Y M N I Which of the following is correct? Current in MN Force between X and Y A From N to M and diminishing Attraction B From M to N and diminishing Repulsion C From N to M and diminishing Repulsion D M to N and constant Attraction
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 200 3. A copper ring is fixed to a long, light rod which is freely pivoted at X so that it may swing as a pendulum as shown in the figure. An electromagnet is mounted so that the ring passes over it as it swings. X K Copper ring Electromagnet Light rod The ring is set into oscillation with the switch K open. What happens to the motion of the copper ring after switch K is closed? A The period of oscillation decreases. B The oscillation is heavily damped. C The amplitude of oscillation increases. D The ring stops with the rod is inclined to the vertical. 4. In the circuit shown, the switch S is closed and the current in the coil is increased by adjusting the rheostat. S R G Soft iron Coil When the current is maximum, the switch S is opened. The deflection of the galvanometer is the largest A at the instant the switch S is closed. B when the current is increased. C when the current is maximum. D at the instant the switch S is opened. 5. An aluminium ring hangs vertically from a thread. A solenoid is fixed near the ring and coaxial with it. Coil Ring What is the initial motion of the aluminium ring when the current in the coil is switched on? A Remains at rest B Moves towards the coil C Moves away from the coil D Rotates about the thread E.m.f. Induced in a Linear Conductor 2010/P1/Q33, 2013/P2/Q12, 2014/P2/Q13, 2017/P2/Q12 (a) (b) (c) Fleming’s right-hand rule (d) Right-hand rule for product B v l x B E v E B Right hand v v x B Figure 17.3 1. When a linear conductor is moved across a uniform magnetic field, an e.m.f. is induced in the conductor. By the principle of conservation of energy, the work done in moving the conductor is converted into electrical energy. 2. The expression of the induced e.m.f. can be deduced using Faraday’s law of electromagnetic induction.
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 201 3. Figure 17.3(a) shows a conductor of length l being moved through a distance x in time t. The magnetic fl ux cut by the conductor φ = BA (A = lx) = Blx ............................. { Using the equation, E = – —– dφ dt , differentiating equation {, E = –Bl —– dx dt = –Blv Magnitude of induced e.m.f., E = Blv 4. In general, the e.m.f. induced in a linear conductor of length l moving with a velocity v at right angles to a uniform magnetic fi eld of fl ux density B is given by E = l (v × B) The magnitude of the induced e.m.f. is E = Blv sin θ where θ = angle between v and B. The direction of the induced e.m.f. E can be deduced using Fleming’s right-hand rule, or the right hand rule for vector product. 5. From the equation E = Blv sin θ, when θ = 90°, E = Blv 6. The expression E = Blv forms the basis for an alternative defi nition of the magnetic fl ux density B. Magnetic fl ux density, B = E lv The magnetic fl ux density B of a magnetic fi eld is equal to the e.m.f. induced in a linear conductor of unit length (l = 1 m) moving with unit velocity (v = 1 m s–1) at right angles to the fi eld. Example 3 (a) Write an equation for the force on a conductor of length, l carrying a current I at right angles to a uniform magnetic fi eld of fl ux density B. (b) Hence, deduce an expression for the e.m.f. induced in a conductor of length l which is moved with a uniform velocity v at right angles to a magnetic fi eld of fl ux density B. Solution: (a) Force F = BIl (b) When a conductor of length l is moved across a uniform magnetic fi eld a current I is induced in the conductor. According to Lenz’s law, the direction of the induced current, I would produce a force FM = BIl against the direction of motion. In order to move the conductor with uniform velocity v, the force F applied to the conductor must be of the same magnitude as FM but in the opposite direction (fi gure on the right). Using the principle of conservation of energy, electrical power output = mechanical power input EI = Fv = (BIl)v E = Blv I F v F M
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 202 Example 4 A piece of wire of mass m slides down freely from rest along two parallel rails inclined at an angle θ to the horizontal. A uniform magnetic fi eld of fl ux density B acts vertically upwards as shown in the diagram. s d θ θ Y X B Derive an expression for the instantaneous e.m.f. induced in the wire when the wire slides down a distance s along the rails, in terms of B, d, θ and the acceleration due to gravity. Solution: When the wire slides a distance s, the magnetic fl ux cut by the wire φ = A(B cos θ) (A = sd) = sdB cos θ Using Faraday’s law, induced e.m.f. E = – —– dφ dt = – —d – dt (sdB cos θ) = –Bd cos θ —ds – dt —ds dt = v = –Bdv cos θ From the equation v2 = u2 + 2as, (a = g sin θ) v2 = 0 + 2(g sin θ)s v = 2gs sin θ Hence induced e.m.f. E = –Bdv cos θ = –Bd cos θ 2gs sin θ Example 5 In the diagram shown, a rectangular coil of wire PQRS is moved with a constant speed from left to right across a uniform magnetic field. Discuss whether there is any current induced in the coil in (a) position 1, (b) position 2, (c) position 3. If there is a current induced, give the direction of the induced current. P Q R Position 1 Position 2 Position 3 Uniform magnetic field S
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 203 Solution: (a) In position 1, there is no magnetic fl ux linked with the coil. Hence, no current is induced. (b) In position 2, the side QR is moving at right angles to a uniform magnetic fi eld. Hence, a current is induced in the direction R to Q. No current is induced in the side PS since it is outside the magnetic fi eld. (c) In position 3, a current is induced in the side QR from R to Q, and an equal current is induced in the side PS from S to P. In the closed loop PQRS, the induced currents from S to P and from R to Q are equal and opposite. Hence, the resultant induced current is zero. Example 6 The diagram below shows a rigid wire being moved by a mechanical force F with a constant velocity u along two parallel, horizontal, smooth conductors placed in a uniform magnetic fi eld of fl ux density B acting vertically downwards. The rigid wire has a mass m, cross-sectional area A and resistivity ρ. The resistance of the parallel conductors is negligible. (a) Write an expression for E, the magnitude of the e.m.f. induced in the rigid wire. (b) Deduce an expression for the induced current I in terms of B, u, ρ and A. (c) The mechanical force F on the rigid wire is suddenly removed. Explain why the rigid wire subsequently decelerates. (d) When the speed of the rigid wire reduces to v, after the removal of the mechanical force F, what is the resultant force experienced by the rigid wire in terms of A, B, d, v and ρ? Solution: (a) Induced e.m.f. E = Bdu (b) Induced current, I = induced e.m.f. ——————– resistance = ——Bdu ρd —–A (R = ρd —–A ) = ——BuA ρ (c) According to Lenz’s law the direction of the induced current opposes the motion of the rigid wire. When the mechanical force F is removed, the force F = BId due to the induced current acts in the opposite direction to the motion of the wire. This causes the wire to decelerate. (d) Resultant force retarding the wire is F = BId I = ——BvA ρ = B —— BvA ρ d = B2 ———vAd ρ F d Rigid wire Uniform magnetic field
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 204 Quick Check 3 1. A metal rod is moved from rest across a uniform magnetic field with a constant acceleration perpendicular to the magnetic fi eld. Which graph shows the variation of the e.m.f. induced E in the rod with time t? A E 0 t C E 0 t B E 0 t D E 0 t 2. A metal rod of length 0.15 m is moved by a constant force F along conducting rails that are connected to a 0.40 : resistor in a region of uniform magnetic fi eld B = 1.2 T. F 0.40 B If the rod moves with a constant velocity of 4.5 m s–1, what is the magnitude of the force F? A 0.32 N C 0.45 N B 0.36 N D 0.81 N 3. The diagram shows two identical coils of wire X and Y which are arranged coaxially. A constant current I fl ows in the coil Y. X Y I Which of the following changes would cause a current to be induced in the coil X in the same direction as the current I in coil Y? A Moving coil X away from coil Y. B Moving coil Y towards coil X. C Moving both coils X and Y towards each other. D Moving both coils towards the right at the same speed. 4. Figure (a) shows a circular coil with its plane perpendicular to a uniform magnetic fi eld of fl ux density B. Figure (a) The fl ux density B varies with time t as shown in Figure (b). Figure (b) B t / s 0 2468 Which of the following shows the variation of magnitude of the e.m.f. induced E in the coil with time t? A t/s E 2 4 6 8 0
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 205 B t/s E 2 4 6 8 0 C t/s E 2 4 6 8 0 D t/s E 2 4 6 8 0 5. The diagram shows a copper rod of length l being rotated in a circle of radius r with constant angular velocity ω round a long straight wire carrying a current I. l I ω r If μ0 is the permeability of free space, what is the e.m.f. induced in the copper rod? A Zero C μ0 Ilω 2 Sr B μ0 Ilω 2 Sr D μ0 Ilω 2 Sr 6. A straight conductor is moved at right angles across a uniform magnetic field. The graph below shows the variation of the displacement s of the conductor from its initial position with time t. s t 0 Which graph best represents the variation of the e.m.f. E induced in the conductor with time t? A t E 0 B t E 0 C t E 0 D t E 0 7. A uniform magnetic field of flux density B passes normally through a square aluminium plate of area A. On top of the metal plate is a coil of 20 turns and area 1 4 A as shown below. Square metal plate of area A Coil of 20 turns, area Uniform magnetic field B 4 A What is the magnetic flux linkage for the coil? A Zero C 5BA B BA D 20BA 8. A conductor of length l is moved with constant velocity v along conducting rails placed in a magnetic field B by an external force F as shown in the diagram. The total resistance in the circuit is R. B l F Which of the following is the correct expression for the external force F ? A F = Blv C Blv R B F =B2 l 2 v D B2 l 2 v R
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 206 9. An aeroplane flies horizontal at 250 m s–1 from east to west. Calculate the potential difference between the tips of the wings if the vertical component of the Earth’s magnetic field is 4.0 × 10–5 T and the distance between the wingtips is 25 m. If the wing-tips are connected by a wire, discuss whether a current flows in the wire. 10. Initially, a rectangular coil of wire PQRS of length a, width b, resistance R is outside a uniform magnetic field as shown in the diagram. The flux B x S R b P a Q l density of the magnetic field is B, and its width l > a. The coil is then pulled with a constant speed v along the x-direction for a distance a, and then with a constant speed of 2v. (a) (i) Sketch a graph to show how the magnetic flux φ linked with the coil varies with the distance x moved by the side QR. Mark on your graph, x = a and x = l. What is the maximum magnetic flux φm linked with the coil? (ii) Deduce an expression for the induced e.m.f. E in terms of the gradient of the graph in (a)(i) and v, the velocity of the coil. Hence, write the equations for the induced e.m.f. E in terms of B, l and v for the various stages of the coil’s motion across the magnetic field. (iii) Derive an expression, in terms of B, a, v, and R for the total heat dissipated from the coil when it moves across the magnetic field for x = 0 to x = a + l. Neglect the effect of self-induction. (b) If the coil PQRS is pulled with a constant speed v throughout its motion, discuss how the answers to (a)(i), (ii), and (iii) above would change, if any. 11. An aeroplane flies southwards with a speed of 120 m s–1 with its wings tilted at an angle of 30° from the horizontal. An e.m.f. of 240 mV is induced across the wings of total length 50 m. Calculate the vertical component of the Earth’s magnetic field. 12. (a) The diagram shows a conductor XY being moved at a constant speed v by a mechanical force F along two parallel, smooth, horizontal conductors in a uniform vertical magnetic field of flux density B. The distance between the two parallel conductors is l, and a resistor of resistance R is connected across the ends W and Z. l F R W Z Y X (i) Write the equation for the e.m.f. E induced in the conductor XY. (ii) Explain why the speed of the conductor remains constant, although it is moved by a force F. Hence deduce an expression for the current I induced in the conductor. Give the direction of the induced current. (iii) If v = 1.6 m s–1, F = 2.0 N, and R = 12 :, what is the magnitude of the induced current I? (b) The conductor is now moved by a force F c which causes the conductor to move from rest with a uniform acceleration a. Derive an expression for the magnitude of the e.m.f. E induced in the rod XY at time t = T. 13. A bar magnet is placed inside a solenoid which is connected to a resistor. The magnet is pulled out of the solenoid at a constant speed. R P Solenoid N S (a) Explain why a current is induced in the solenoid when the S-pole of the magnet moves out from the solenoid. (b) Use Lenz's law to deduce the direction of the induced current as the (i) S-pole, and (ii) the N-pole leaves the solenoid. (c) Sketch a graph to show how the variation of the induced current as the magnet leaves the solenoid.
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 207 E.m.f. Induced in a Rotating Coil 2007/P2/Q6, 2009/P2/Q11 1. Figure 17.4(a) shows a coil of N turns and area A initially with its plane at right angles to a uniform magnetic fi eld of fl ux density B. Figure 17.4 (a) (b) B Axis of rotation ω B ω ω ωt /`KYVLSLJ[YPJWV^LYZ[H[PVU >H[LYÅV^Z[OYV\NO[OLKHTHUK[\YUZ H SHYNL ^OLLS JHSSLK H [\YIPUL ;OL [\YIPUL [\YUZ H ZOHM[^OPJO YV[H[LZ H ZLYPLZ VMTHNUL[Z WHZZ JVWWLY JVPSZ [VWYVK\JLLSLJ[YPJP[` Info Physics 2. The coil is then rotated about an axis perpendicular to the magnetic fi eld with a uniform angular velocity ω. 3. Figure 17.4(b) shows the plane of the coil at a time t. The angle turned through by the coil is ωt. Magnetic fl ux linked with the coil, Φ = NBA cos ωt Using E = – —– dΦ dt , e.m.f. induced in the coil, E = – — d dt (NBA cos ωt) = NBA ω sin ωt which is a sinusoidal alternating e.m.f. of peak value, E0 = NBAω and it occurs when ωt = 90° or 270°, that is when the plane of the coil is parallel to the magnetic fi eld B. 4. Figure 17.5(a) is the graph of the induced e.m.f. against time, and Figure 17.5(b) shows the positions of the coil relative to the magnetic fi eld B. Example 7 A coil rotates at a constant speed inside a uniform magnetic fi eld. The peak value of the e.m.f. induced in the coil is 10.0 V. Deduce (a) the root-mean-square (r.m.s.) value of the induced e.m.f., (b) the instantaneous value of the induced e.m.f., one quarter of its period of rotation after the peak e.m.f. Figure 17.5 T B P Q Q P P NBAω – NBAω 0 (a) (b) Plane of coil Induced e.m.f. Time Q Q P P Q T 4 T 2 T 3 4
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 208 Solution: (a) R.m.s. value of e.m.f. = ————— Peak e.m.f. 2 = —— 10.0 2 = 7.07 V (b) From the graph, the e.m.f. is at its peak when t = —T 4 . —T 4 after the e.m.f. is peak, t = —T 4 + —T 4 = —T 2 From the graph, when t = —T 2 , induced e.m.f = 0. Example 8 The diagram shows a square coil of 546 turns, length of each side 6.0 cm rotating at 55 revolution per second in a uniform magnetic fi eld of fl ux density 0.50 T. (a) Write an expression for the magnetic fl ux linked with the coil as a function of time t. (b) Hence, deduce the expression for the induced e.m.f. (c) Sketch a graph of induced e.m.f. against time. (d) The resistance of the lamp connected to the carbon brushes A and B is 12.0 :. Calculate the r.m.s. current in the lamp. Solution: (a) B ωt After a time = t, angle turned through by the coil = ωt. Magnetic fl ux linkage, Φ = NBA sin ωt = 546 × 0.50 × (6 × 10–2)2 sin (2S × 55t) Φ = 0.983 sin (110 St) (b) Using E = – —– dΦ dt , induced e.m.f. E = – —d dt (0.983 sin (110 St)) = –(0.983)(110 S) cos (110 St) E = –340 cos (110 St) Induced e.m.f. T 4 T T T = Period of rotation t 3 4 0 T 2 S A B Lamp N 55 r.p.s.
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 209 (c) 1.8 t / x 10–2 0 s –340 340 E / V T = —1 f = —1 55 = 0.018 s (d) R.m.s. voltage = E —–0 2 = —— 340 2 = 240 V R.m.s. current in the lamp = —–– 240 12.0 = 20 A Neglect the resistance of the coil. Example 9 To provide for a pulse of electricity to the heart of a patient, a small circular coil is inserted into the patient’s chest. Outside the patient’s chest is a larger coil with is connected to a battery. The two coils are coaxial and in the same plane. Small coil = 150 turns, area 1.0 cm2 Large coil = 500 turns, radius 10.0 cm and resistance 10 : Battery = E.m.f. 60 V, internal resistance 10 : Time for current in large coil to drop to zero = 10–4 s From the above data, estimate the potential difference across the small coil when the external circuit is opened. Explain why the induced e.m.f. is larger when the current in the external circuit is switched off, than when the current is switched on. Solution: R E, r Battery Switch Large coil Small coil I Current in the large coil, I = E R + r = 60 10 + 10 = 3.0 A
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 210 The magnetic fl ux density at the centre of the large coil, B = μ0 NI 2r = (4S × 10–7)(500) × 3.0 2 × 0.10 = 3S × 10–3 T Initial magnetic fl ux linkage in small coil, ) = NcBA = (150)(3S × 10–3)(1.0 × 10–4) = 1.41 × 10–4 Wb After the switch is opened, fi nal fl ux linkage, )1 = 0 Using induced e.m.f. = rate of change of magnetic fl ux linkage = 1.41 × 10–4 – 0 10–4 = 1.41 V When the current is switched on, self-induction in the large coil induces a back-e.m.f. which retards the increase of the current to its maximum value. The e.m.f. induced in the small coil is small because the time taken for the change of magnetic fl ux is longer. When the switch is opened, the time taken for the current to drop to zero is very much shorter compared to the time taken for the current to increase from zero to its maximum value. The rate of change of magnetic fl ux is greater and a larger e.m.f. is induced. Quick Check 4 1. A rectangular coil of wire, initially in the position shown in Figure (a) is rotated with constant angular velocity in a uniform magnetic fi eld in the direction XXc. The e.m.f. induced is sinusoidal as shown in Figure (b). Figure (a) Figure (b) Z Y X X` Y` Z` E.m.f. 0 Time To obtain the e.m.f. induced as shown in Figure (b), the coil must be rotated through A a quarter revolution about the axis YYc B a half revolution about the axis XXc C a half revolution about the axis ZZc D a quarter revolution about the axis XXc 2. A magnetic fi eld B = B0 sin 2St T as shown in the fi gure is applied perpendicular to the plane of a fl at coil of copper wire. B T 0 8 T 4 T T t 2 3T 8 5T 8 8 3T 4 7T B0 –B0 At which of the following values of t is the magnitude of the e.m.f. induced in the coil a maximum? A —T 8 C 3 —T 8 B — T 4 D — T 2
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 211 3. The magnetic flux through a coil varies with time t as shown in the diagram. Φ 0 t Which graph best represents the variation of the e.m.f. E induced in the coil with time t? A E 0 t B E 0 t C E 0 t D E 0 t 4. A plane circular coil of 50 turns, each of diameter 0.01 m, rotates 25 times each second about a diameter which is perpendicular to a uniform magnetic field of flux density 5.0 × 10–5 T. (a) Calculate (i) the maximum instantaneous value of the induced e.m.f., (ii) the r.m.s. value of the e.m.f. (b) If the coil were fixed and the field were rotated at the same rate and in the same direction, what difference would this make to the induced e.m.f.? 5. Calculate the r.m.s. value of the e.m.f. induced in a coil of 50 turns, each of area 30 cm2 , when it is rotated at a uniform speed of 2 100 revolutions per minute about an axis perpendicular to the magnetic field of flux density 0.80 T. 6. A small square coil has its plane set at right angles to the uniform magnetic field between the pole pieces of a horseshoe magnet as shown in the diagram. A B D C Coil (a) The magnet is now rotated at constant angular velocity about the axis AB. Draw sketch graphs, on the same time axis, to show the variation of (i) the magnetic flux through the coil, (ii) the e.m.f. induced in the coil. (b) Draw a second set of graphs for the case where the magnet rotates at constant angular velocity about the axis CD. Assume that the magnetic flux density in the region outside the pole pieces is zero.
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 212 17.3 Self-Induction Students should be able to: H[SODLQWKHSKHQRPHQRQRIVHOILQGXFWLRQDQGGHÀQHVHOILQGXFWDQFH XVHWKHIRUPXODHE = –L dI dt and LI = N φ GHULYHDQGXVHWKHHTXDWLRQIRUWKHVHOILQGXFWDQFHRIDVROHQRLGL = μ0 N2 A l Learning Outcomes 1. Self-induction is the phenomenon when an e.m.f. is induced in a conductor itself due to changes in the magnetic fl ux linked with the conductor. 2. Figure 17.6 shows a solenoid connected in series with a battery and a switch K. Before the switch K is closed, the magnetic fl ux linked with the coil is zero. When the switch K is closed, a current starts to fl ow in the solenoid. The magnetic fl ux linkage in the solenoid increases. Figure 17.6 K Back-e.m.f. I 3. This changing magnetic fl ux induces an e.m.f. in the solenoid. According to Lenz’s law, the direction of the induced e.m.f. opposes the increase in the current, and therefore is in the direction shown in Figure 17.6. This induced e.m.f. is known as the back-e.m.f. 4. The magnetic fl ux linkage Φ in the solenoid is directly proportional to the current I in the coil. Φ v I or Φ = LI where L is a constant known as the self-inductance of the solenoid. Unit for self-inductance L is the henry (H). 5. Using E = – —– dΦ dt , Back-e.m.f., E = – —d dt (LI) E = –L—– dI dt The back-e.m.f. is directly proportional to the rate of change of current, —– dI dt . 6. From the equation back-e.m.f E = –L—– dI dt , we have, self-inductance, L = –—– E —– dI dt Defi nition: The self-inductance L of a conductor is the back-e.m.f. in the conductor when the current in the conductor changes at a rate of 1 ampere per second. 2013/P2/Q16, 2014/P2/Q12, Q20(b), 2016/P2/Q11, 2017/P2/Q20, 2018/P2/Q20
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 213 7. The self-inductance of a conductor depends on its shape and linear dimension. We will derive the expression for the self-inductance L of a solenoid of length l, radius a and n turns per unit length. When the current in the solenoid is I, the magnetic flux density in the solenoid is B = μ0 nI = μ01 N l 2I Magnetic flux linkage in the solenoid Φ = NBA (N = nb) = N3μ01 N l 2I4A = LI self inductance, L = μ0 N2 A l = μ0 (n2 l)(Sa2 ) 8. Self-induction in the solenoid causes the current to increase slowly after the switch K is closed. Figure 17.7(a) shows the variation of for the current I in the solenoid with time t. The back-e.m.f. at the instant the switch K is closed (t = 0) is about the same value as the e.m.f. of the battery, but in the opposite direction. 9. No back-e.m.f. is produced when the current is steady. 10. At the instant that the switch K is opened, the current takes a time t 2 which is smaller than t 1 to drop to zero. Hence, the back-e.m.f. is of greater magnitude than the back-e.m.f. at the instant the switch is closed. 11. This large back-e.m.f. may cause sparking at a switch of an electrical appliance due to the insulation of the air breaking down. For this reason, a relay is used for electrical appliance that consumes a large current. 12. A back-e.m.f. eback is produced in the armature coil of a motor when it rotates. If V is the voltage applied across the motor, R the armature resistance and I the current in the armature when the speed to rotation is Z, then back e.m.f eback = V – IR, and eback v Z 13. The output power = I eback = IV – I2 R = (input power) – (rate of heat loss) 14. At the instant the motor is switched on, t = 0 initial Z = 0 and eback = 0 eback = V – IR = 0 Current, I = —V R Figure 17.7 Time Time Current, I Back-e.m.f K closed K closed K closed t2 t2 t1 t1 0 (a) (b) 0 K
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 214 Example 10 The current in a coil changes at a rate of 2 A in 50 ms. The back-e.m.f. in the coil is 4.0 V. What is the self-inductance of the coil? Solution: Using back-e.m.f. E = –L—– dI dt , 4.0 = L ———— 2.0 50 × 10–3 Self-inductance, L = 0.10 H Example 11 A coil which has 250 turns carries a current of 7.5 A. The magnetic fl ux linked with each turn of the coil is 0.055 Wb. Calculate the self-inductance of the coil. Solution: Magnetic fl ux linkage in 1 turn = 0.055 Wb Magnetic fl ux linkage in 250 turns, Φ = 250 (0.055) Wb Also, Φ = LI Self-inductance, L = Φ —I = 250 × 0.055 7.5 = 1.83 H Example 12 (a) The armature resistance of a 3.0 V d.c. motor is 1.5 :. What is the initial current in the armature when the motor is switched on? (b) When the motor rotates at 200 revolution per minute (rpm) the current in the armature is 0.50 A. What is the back e.m.f? (c) When the motor is loaded, its speed is 120 rpm. (i) What is the back e.m.f? (ii) What is the output power? (iii) What is the rate of heat produced? Solution: (a) Initial current, I = — V R = ——3.0 1.5 A = 2.0 A (b) Back e.m.f. eback = V – IR = 3.0 – (0.50)(1.5) V = 2.25 V Induction cooker $ KLJK IUHTXHQF\ DOWHUQDWLQJ FXUUHQW IORZV LQ D FRLO SODFHG EHORZWKH FRRNLQJ VXUIDFH7KH RVFLOODWLQJPDJQHWLFÀHOGLQGXFHV D FXUUHQW LQ WKH PHWDO XWHQVLO +HDWLVSURGXFHGLQWKHXWHQVLO Info Physics
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 215 (c) (i) eback v Z New back e.m.f. = 1—— 120 200 2(2.25)V = 1.35 V (ii) Back e.m.f. eback = V – IR 1.35 = 3.0 – I(1.5) I = 1.10 A Output power = I eback = (1.10)(1.35) W = 1.49 W (iii) Rate of heat loss = I2 R = (1.10)2 (1.5) W = 1.82 W Quick Check 5 1. A fl at coil of wire has 250 turns. When the current in the coil is 3.5 A, the magnetic fl ux linked with each turn is 0.11 Wb. Calculate the self-inductance of the coil. 2. A solenoid of length 20.0 cm has 100 turns per cm. Its cross-sectional area is 1.8 cm2 . A current of 2.0 A fl ows in the solenoid. (a) Calculate the magnetic fl ux linkage in the solenoid. (b) Calculate the self-inductance of the solenoid. 3. A coil of self-inductance 57 mH is connected to a battery. The current takes 60 ms to reach the maximum steady value of 0.50 A. What is the back-e.m.f. induced? 17.4 Energy Stored in an Inductor Students should be able to: XVHWKHIRUPXODIRUWKHHQHUJ\VWRUHGLQDQLQGXFWRUU = 1 2 LI 2 Learning Outcome 1. Self-induction in an inductor opposes the change of current in the inductor. Work is done by a battery to drive a current through an inductor. The work done is stored in the inductor. 2. The back e.m.f. in an inductor E = –L —– dI dt When the instantaneous current is I, the instantaneous power = IE —– dU dt = IL—– dI dt dU = IL(dI) Intergrating ³ dU = ³LI dI Energy stored in inductor, U = 1 –– 2 LI2
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 216 Quick Check 6 1. The current in an inductor of self-inductance 80 mH is 0.40 A. Calculate the energy stored in the inductor. What is the form of energy stored in the inductor? 2. An inductor has self-inductance of 25 mH and can carry a current of up to 5.0 A. (a) What is the energy stored in the magnetic fi eld of the inductor when the steady current in it is maximum? (b) The energy in the inductor is completely discharged in 5.0 ms. Calculate the mean back-e.m.f. induced in the inductor. 17.5 Mutual Induction Students should be able to: H[SODLQWKHSKHQRPHQRQRIPXWXDOLQGXFWLRQDQGGHÀQHPXWXDOLQGXFWDQFH GHULYHDQH[SUHVVLRQIRUWKHPXWXDOLQGXFWDQFHEHWZHHQWZRFRD[LDOVROHQRLGVRIWKHVDPHFURVVVHFWLRQDODUHDM = μ0 NS Ns A l S Learning Outcomes 1. Figure 17.8 shows two solenoids P and Q arranged coaxially close to each other. When a current I fl ows in the coil P, a magnetic fi eld is produced by the current in P. There is a magnetic fl ux linked with the solenoid Q. 2. When the current in P changes, the magnetic fl ux linkage in Q changes, and an e.m.f. is induced in coil Q. This is called mutual induction. 3. Mutual induction in the phenomenon where an e.m.f. is induced in a conductor when the current in a neighbouring conductor is changing. 4. The magnetic fl ux linkage Φ in solenoid Q is directly proportional to the current I in the solenoid P. Φ v I Φ = MI where M is a constant known as the mutual inductance of the two solenoids P and Q. Unit for mutual inductance M is the henry (H). 5. When the current I in solenoid P changes, the induced e.m.f. in solenoid Q is E = – —– d) dt = – —– d dt (MI) E = –M—– dI dt 6. For a system of two conductors, the value of M, the mutual inductance depends on the shape and physical dimensions of the conductors. 7. In Figure 17.8, solenoid P is the primary coil, and Q the secondary coil. A change of current in P induces an e.m.f. in Q. The reversed happens when the positions of the battery and galvanometer are interchanged. Solenoid Q then becomes the primary coil, and P the secondary coil. Figure 17.8 Mutual induction P Q I G 2015/P2/Q16, Q19(c), 2017/P2/Q13, ,2018/P2/Q13
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 217 8. We will now derive an expression for the mutual inductance between two coaxial coils show in Figure 17.9. 9. When a current I fl ows in the primary coil which is a solenoid of length l, radius r, and has Np turns, the magnetic fl ux density in the solenoid is B = μ0 nI Np n = —– lp = μ0 Np I —– l p 10. The secondary coil is a solenoid that has N2 turns and of the same radius r. Magnetic fl ux linkage in the secondary coil Φ = BA = Ns ƫ0 Np I —–—lp A Also, Φ = MI 11. A transformer transforms alternating voltages. It uses the principle of mutual induction. 12. Figure 17.10 shows a transformer which consists of a primary coil and a secondary coil wound on a soft iron core. 13. An alternating voltage VP is connected to the primary coil. The alternating current in the primary coil produces a magnetic fl ux linkage with the secondary coil which constantly changes. This induces an e.m.f. in the secondary coil. Example 13 A small coil is placed near a solenoid carrying a current as shown in Figure (a). The variation of the current I in the solenoid with time t is shown in Figure (b). Figure (b) I 0 Time Figure (a) Solenoid Coil (a) Sketch the following graphs on the same time axis. (i) The variation of the magnetic fl ux ) in the small coil with time t. (ii) The variation of the e.m.f. E induced in the small coil with time t. (b) Describe and explain the effect on the amplitude and frequency of the e.m.f. E induced in the coil if, (i) a ferrous core is slowly introduced into the solenoid, (ii) the frequency of the current in the solenoid is increased, whilst maintaining the same amplitude. Figure 17.9 l I I Primary coil, Np turns Secondary coil, Ns turns Figure 17.10 Vp Np Ns Vs Primary coil Soft iron core Secondary coil Mutual inductance, M = — Φ I M = μ0 NpNs A l p
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 218 Solution: (a) (i) From the graph of I against t, the variation of the current I with t may be represented by I = I0 cos ωt The magnetic fl ux linkage in the coil is ) = MI M = mutual inductance = MI0 cos ωt Hence the graph of ) against time t is as shown. (ii) E.m.f. induced in the coil, E = –M—– dI dt = –M—– d dt (I0 cos ωt) = MI0 ω sin ωt Graph of E against t is as shown below. E 0 t (b) (i) When a ferrous core is introduced into the solenoid the magnetic fl ux density increases. The mutual inductance M increases. Hence the amplitude of the e.m.f. E induced in the coil increases, but the frequency remains unchanged. (ii) The e.m.f. induced, E = MI0 ω sin ωt (from (a)(ii) ω = 2Sf ) E = 2SfMI0 sin ωt Hence, when the frequency f of the current I is increased, the amplitude of the induced e.m.f. E increases. Quick Check 7 0 t 1. The self-inductance of a solenoid can be increased by introducing a ferrous core. The ferrous core A reduces the electrical resistance. B increases the fl ux linkage in the solenoid. C reduces the effect of the Earth’s magnetic fi eld. D increases the mutual inductance of the solenoid and the core. 2. The magnetic flux density in a solenoid which has 200 turns and cross-sectional area 2 × 10–4 m2 is 5 × 10–5 T when the current in the solenoid is 4 A. What is the value of the self-inductance of the solenoid? A 0.5 PH C 2.0 PH B 1.0 PH D 2.5 PH 3. In the circuit shown, the inductors P and Q both have the same resistance but the selfinductance of P is smaller than that of Q. Switch P Q Which of the following graphs shows correctly the variations of the current in P, and Q after the switch is closed with time t?
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 219 A C 0 t Q P Current 0 t Q P Current B D 0 t P Q Current 0 t P Q Current 4. Two coils X and Y are arranged as shown in the diagram below. X Y One quarter of the magnetic flux produced by the current in X is linked with Y. The mutual inductance between the coils is M. What is the mutual inductance when Y is the primary coil? A —M 4 C M B — M 2 D 4M 5. The mutual inductance of two coils is 2.0 H. Due to mutual induction, an e.m.f. of 3.0 V is induced in one of the coils. Which of the following is true about the other coil? A A current of 1.5 A passes through it. B The current in it changes at 1.5 A s–1. C The potential difference across it is 1.5 V. D The potential difference across it changes at 1.5 V s–1. 6. Two solenoids P and Q of 10 turns and 100 turns respectively are arranged as shown. In Figure (a), when the current IP in P changes at 5 A s–1, an e.m.f. of 2.0 mV is induced in Q. Figure (a) Figure (b) P IP Q P IQ Q The current IP is switched off and a current IQ that changes at 2 A s–1 flows in Q. What is the e.m.f. induced in P? A 0.8 mV B 2.0 mV C 5.0 mV D 20.0 mV 7. When the current in a coil is reduced uniformly from 20 A to zero in 0.50 s, an e.m.f. of 4.0 V is induced in a neighbouring coil. What is the value of the mutual inductance between the two coils? A 0.10 H B 0.40 H C 2.5 H D 40 H 8. A short coil is wound over the middle part of a long solenoid as shown in Figure (a). Figure (a) Coil Solenoid 0 t I Figure (b) The current I in the solenoid varies with time t as shown in Figure (b). The flux density in the solenoid is given by B = μ0 nI, n = number of turns per metre. Which graph shows the variation the induced e.m.f. E in the coil with time t?
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 220 A E 0 t C t E 0 B E 0 t D t E 0 9. The diagram shows a battery E of e.m.f. 6.0 V, whose internal resistance is negligible, connected in series with an inductor of selfinductance 2 H and resistance 3.0 :. E S L Sketch a graph to show the variation of the current I in the circuit with time t if the switch S is closed at time t = 0, and open at time t = T when the current had achieved the maximum value. Calculate the maximum rate of change of current during the period T. 10. Two identical coils A and B are wound on a soft iron core as shown in the diagram. A B When the coils are connected in series, the resultant self- inductance is either twice that of one coil, or is negligible, depending on the manner the coils are connected. Explain this observation. 11. (a) Explain the meaning of self-inductance and mutual inductance. (b) (i) Calculate the value of the selfinductance of a solenoid which has length 40.0 cm, diameter 4.0 cm and 100 turns. (ii) Another solenoid which is shorter is coaxial and inside the solenoid in (i) above but has 200 turns and diameter 2.0 cm. Find the mutual inductance of the two solenoids. 12. An inductor has inductance 200 ƫH and can carry currents up to 1.0 × 106 A. (a) What is the energy stored in the inductor when the current in it is the maximum allowed? (b) The energy stored in (a) may be transferred to an uncharged capacitor in 10 ms. (i) Calculate the mean power delivered by the inductor. (ii) What is the mean back-e.m.f. in the inductor? Important Formulae 1. Magnetic flux, φ = B·A = BA cos θ 2. Maximum magnetic flux linked with a coil, Φ = NBA 3. Faraday’s law of electromagnetic induction: Induced e.m.f., E = – —– dΦ dt 4. E.m.f. induced (a) Straight conductor, E = l(v u B), E = Blv sin θ (b) Rotating coil, E = NBAω sin θ or E = NBAω cos θ 5. Self-induction: Induced e.m.f., E = –L—dI dt , and Φ = LI Energy stored, U = —1 2 LI2 6. Mutual induction: Induced e.m.f., E = –M—dI dt , and Φ = MI
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 221 1. The current in a loop of conducting wire changes at a rate of 18 A s–1, and an e.m.f. of 6.0 V is induced. What is the self inductance of the loop? A 3.0 H C 3.0 mH B 0.33 H D 33 mH 2. A circular coil with 20 turns and area 15.0 cm2 is placed in a uniform magnetic fi eld of 0.60 T which is perpendicular to the plane of the coil as shown in the diagram. The coil has a resistance of 8.0 :. 20 turns A = 15.0 cm2 The magnetic fi eld is reduced to zero in 50 ms. What is the magnitude and direction of the current induced in the coil? A 45 mA, clockwise direction B 45 mA, anticlockwise direction C 0.36 A, clockwise direction D 0.36 A, anticlockwise direction 3. A solenoid has N turns, a length l, and a cross-sectional area A. The current through the solenoid is reduced uniformly from I 0 to zero in time T. Which expression represents the back-em.f. produced in the solenoid? A μ0 NI0 A T C μ0 N2 I0 A T B μ0 NI0 A Tl D μ0 N2 I0 A Tl 4. A coil of N turns each of area A is placed in a uniform magnetic fi eld B. The angle between the normal to the plane of the coil and the magnetic fi eld is θ. Coil Normal B The magnetic fl ux linked with the coil is A BA C NBA sin θ B NBA D NBA cos θ 5. A magnet is positioned with its N-pole at the centre of a coil. Which movement of the coil will not produce an induced e.m.f. in the coil? A Down N S C Left to right N S B Rotate about its axis N S D To the left N S 6. A rectangular coil of length L is moved through a uniform magnetic field B at a constant speed v. The length of the magnetic fi eld is 3L. Coil Coil P L v x B Q 3L P Q Which graph shows correctly the variation of the e.m.f. E induced in the coil with the displacement x of the side PQ of the coil? A x E 0 2L 4L STPM PRACTICE 17
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 222 B x E 0 2L 4L C x E 0 2L 4L D x E 0 2L 4L 7. A magnetic field has a flux density of B. Which of the statements is incorrect? A Magnetic flux through an area A perpendicular to the field is BA. B Force on a stationary electron in the field is eB. C Force on a charge q moving with a velocity v perpendicular to the field is qvB. D Force on a conductor of length 1 m, carrying a current of 1 A and perpendicular to the field is B. 8. A flat circular coil has N turns each of radius r. The current in the coil is reduced from I to zero in time T. What is the magnitude of the average e.m.f. induced in the coil? A μ0 Sr2 NI 2T C μ0 Sr2 N2 I T B μ0 SrNI T D μ0 SrN2 I 2T 9. A conductor of length L is placed on two rails that are separated by a distance x in a uniform magnetic field B. A resistor of resistance R is connected between the rails at one end. The conductor is moved with a constant speed v along the rails. x L v R B What is the current in the resistor? A Bxv R C BxR v B BLv R D BLR v 10. A flat coil of 240 turns, each of area 0.070 m2 , is placed with its axis parallel to a uniform magnetic field. The flux density of the field is decreased steadily from 100 mT to 40 mT over a period of 4.0 s. What is the e.m.f. induced in the coil? A 0 B 170 mV C 250 mV D 420 mV 11. A plane coil of wire with N turns each of area A is placed in a uniform magnetic field of flux density B. The plane of the coil makes an angle θ with the direction of the magnetic field. The coil is then moved through a distance x in time t to the dotted position shown in the diagram below. x θ B Magnetic field Plane of coil What is the e.m.f. induced in the coil? A Zero C NABx t B NAB t D NABx cos θ t
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 223 12. Which statement about an ideal transformer is true? A Equal number of turns in the primary and secondary coils B Same current in both the primary and secondary coils C Output voltage and input voltage are in phase D Same magnetic flux through both the primary and secondary coil 13. A constant voltage V is applied across an inductor of inductance L and resistance R. The variation of the current in the inductor with time is as shown in the graph. Current 0 Time V R Which of the following explains for the shape of the graph? A The value of L decreases as the current increases. B The current is directly proportional to the rate of change of back-e.m.f. C The back-e.m.f. is directly proportional to the rate of change of current. D The back-e.m.f. is directly proportional to the current. 14. The diagram (a) shows two coils P and Q wound on a soft iron core. V0 Power supply P Q t1 t2 0 t I (a) (b) The graph of current I against time t in the coil P is shown in diagram (b). Which graph represents the variation of the output voltage V0 with time t? A C t1 t2 VO 0 t t1 t2 VO 0 t B D t1 t2 VO 0 t t1 t2 VO 0 t 15. A short bar magnet moves at a uniform speed along the axis of a solenoid. A galvanometer is connected across the solenoid. N S Which graph best represents the variation of the galvanometer deflection θ with time t? A 0 e t B θ 0 t C θ 0 t D θ 0 t
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 224 16. An aircraft of length l, wing span x, flies horizontal at speed v from south to north in a region where the Earth’s magnetic field of flux density B is inclined at an angle θ to the horizontal. What is the magnitude of the e.m.f. induced between the wingtips of the aircraft? A Blv C Blv cos θ B Bxv sin θ D Bxv 17. Large alternating currents in a cable can be measured by monitoring the e.m.f. induced in a small coil held near the cable. In which arrangement of coil and cable is the e.m.f. induced in the coil the maximum? A Cable Coil B Cable Coil C Cable Coil D Cable Coil 18. An inductor of inductance 6.0 H and resistance 24 : is connected in series with a battery of e.m.f. 12 V and negligible resistance. What is the maximum energy stored in the inductor? A 0.75 J C 3.00 J B 1.50 J D 6.00 J 19. A plane circular coil is arranged with its plane perpendicular to the magnetic field between the poles of a horseshoe magnet. The coil is then pulled out of the magnetic field. The charge induced does not depend on A area of the circular coil. B resistance of the circular coil. C magnetic flux density of the magnetic field. D time taken for the coil to move out of the field. 20. A flat coil is placed inside a solenoid with its axis parallel to the axis of the solenoid. An alternating current I = I0 sin 100 St, where I0 is the peak current and t is the time flows in the solenoid. Which of the following graphs shows the variation of the induced e.m.f. E in the coil with time t? A 0 0.02 E t /s B 0 0.02 E t /s C 0 0.02 E t /s D 0 0.02 E t /s 21. A copper ring falls vertically through a bar magnet as shown in the diagram. N S Compare the acceleration a of the copper ring when it is level with the N-pole, and the S-pole of the magnet with the acceleration due to gravity g. Neglect air resistance.
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 225 Level with N-pole Level with S-pole A a > g a < g B a < g a > g C a < g a < g D a > g a > g 22. In the diagram shown below, the switch S is closed and a steady current flows in the primary coil of the transformer. S A B C Y X D Soft iron core Resistor Which of the following describes correctly the magnitude of the magnetic flux linkage, the direction of the magnetic flux and the direction of the induced current in the secondary coil at the instant the switch S is opened? Magnitude of Direction of Direction magnetic flux magnetic of induced linkage flux current A Increases ABCD X to Y B Increases DCBA Y to X C Decreases ABCD X to Y D Decreases DCBA Y to X 23. A circular coil has 200 turns each of radius 5.0 cm and carries a current of 6.0 A. Calculate (a) the magnetic flux through the coil, (b) the self-inductance of the coil. 24. The diagram shows a metal rod P of mass 4.0 g initially stationary inside a uniform magnetic field B = 0.60 T. Another metal rod Q of mass 5.0 g enters the magnetic field with an initial velocity u = 0.20 m s–1. The rods are free to move on conducting rails separated by a distance of 8.0 cm. 8.0 cm B Q P u (a) Calculate the e.m.f. induced in the rod Q when it enters the magnetic field. State the direction of the induced current. (b) The rod P is observed to move to the right when rod Q moves towards it. Explain why. (c) The rods P and Q can be considered be collide elastically. Determine the velocities of the rods after collision. 25. (a) The plane of a coil of wire is initially parallel to a uniform magnetic field B. The area of the coil is A. The coil is then rotated at a constant angular speed ω about the axial which is perpendicular to B. B Axial (i) Write an expression for the magnetic flux through the coil after a time t. (ii) Hence deduce the expression for the e.m.f. induced in the coil. (iii) Calculate the maximum e.m.f. induced for a coil of area 20.0 cm2 and rotated at 2.0 rotations per second in a magnetic field of 0.80 T. (iv) On the same time-axis, sketch the graph of 1. magnetic flux against time, and 2. induced e.m.f. against time. (b) A solenoid is of length l and has n turns per unit length. The cross-sectional area of the solenoid is A. (i) Derive an expression for the selfinductance of the solenoid.
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 226 (ii) Discuss how the self-inductance of the solenoid is affected by (1) changing its radius, and (2) changing its length. 26. (a) State the laws of electromagnetic induction. Write an equation to summarise the two laws. (b) The diagram shows a circular coil of wire of radius 15.0 cm in a uniform magnetic field of flux density 0.80 T. The radius of the coil is reduced to 12.0 cm in 0.16 s. B R What is the magnitude and direction of the e.m.f. induced in the coil? (c) A d.c. motor is connected to a 6.0 V battery. The initial current is 2.4 A, and when the armature is rotating at 45 revolutions per second, the armature current is 1.7 A. Calculate (i) the resistance of the armature, (ii) the back e.m.f. when the speed of rotation is 45 revolutions per second, (iii) the output power at 45 revolutions per second. (iv) the back e.m.f. when the speed of rotation is 20 revolutions per second, 27. (a) (i) What is meant by electromagnetic induction. What is the effect opposite to electromagnetic induction? (ii) A conductor of length l is moved with a velocity v across a uniform magnetic field B. Derive the expression: e.m.f. induced E = Blv (b) A copper rod of mass 0.025 kg slides down freely from rest along two parallel metal rails inclined at an angle of 20o to the horizontal. The rails are separated by a distance of 0.20 m. A uniform vertical magnetic field B = 1.20 T is applied between the rails. Across one end of the rails is connected a 3.0 : resistor. (i) What is magnitude of the e.m.f. when the speed of the rod is 2.2 m s–1. (ii) Determine the terminal velocity of the rod, and the direction of the magnetic field. (iii) Describe the transformation of energy when the rod moves with its terminal velocity. 28. A plane coil has 100 turns each of radius 10.0 cm and carries a current of 10 A. At the centre of the coil and coaxial with it is a small coil which has 16 turns each of radius 0.25 cm. Calculate the e.m.f. induced in the small coil if the current in the large coil is reduced uniformly to zero in 2.0 s. Sketch a diagram to show the direction of the current in the large coil, and the direction of the induced e.m.f. in the small coil. Explain how you deduce the direction of the induced e.m.f. 29. A long straight vertical wire carries a current 100 A downwards. The current is reduced to zero in 2.0 s. A small coil of radius 0.25 cm, of 16 turns is placed with its centre 10.0 cm from the wire. (a) Calculate the maximum e.m.f. induced in the coil. (b) Draw a diagram to show the direction of the current in the wire, and the direction of the induced e.m.f. (c) Explain how you deduce the direction of the induced e.m.f. 30. (a) The energy stored in an inductor of inductance L carrying a current I is —1 2 LI2 . In what form is the energy stored? How may the energy stored be converted into other forms?
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 227 (b) A plane coil of 50 turns each of area 10 cm2 rotates about a horizontal axis which is parallel to the horizontal component of the Earth’s magnetic field. If the vertical component of the Earth’s magnetic field is 3.0 × 10–5 T, calculate (i) the maximum magnetic flux linkage with the coil, (ii) the mean e.m.f. induced in the coil if the flux decreases from the maximum to zero in 10–2 s. 31. (a) What is meant by electromagnetic induction? (b) Describe how you would determine the magnetic flux density of a uniform magnetic field using a search coil, an a.c. motor to turn the search coil and an a.c. voltmeter. Show how the flux density may be calculated from the readings taken. (c) The diagram shows a long metal sheet of width l placed in a uniform magnetic field of flux density B in the Oz direction. x l y z V A voltmeter is connected to two sliding contacts on opposite edges of the metal sheet. What is the reading of the voltmeter, in terms of B, l, and v if (i) the metal sheet is moved along the y-axis with constant velocity v? (ii) both the metal sheet and voltmeter are moved at the same velocity along the y-axis? (d) The metal sheet and voltmeter is removed from the magnetic field. A plane circular coil of N turns, each of area A is placed with its plane parallel to the magnetic field. (i) The coil is turned through 90°. Find an expression for the e.m.f. induced in the coil, in terms of B, A, R and t, where R is the resistance of the coil and t is the time taken to turn the coil. (ii) If the coil is stationary and N = 100, R = 6 :, and A = 80 cm2 , at what rate must the magnetic field change in order to induce a current of 1.0 mA in the coil? 32. (a) The figure below shows two long, straight, parallel wires P and Q carrying currents I 1 and I 2 respectively in opposite directions perpendicular to the plane of the page. P Q r (i) Write the expression for the magnetic flux density at distance r from wire P. Hence derive an expression for the force per unit length on the wire Q. Sketch a diagram to show the directions of the current in Q, the magnetic field and the force on Q. (ii) Explain how the expression in (a)(i) is used to define the ampere. (b) The diagram below shows a bar magnet hanging from a helical spring above a solenoid which is connected to a battery and switch. Solenoid N S Spring Bar magnet Switch Battery Explain what happens to the magnet when the switch is closed.
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 228 (c) The battery and switch are then removed, and the solenoid connected to a signal generator whose output is a sinusoidal alternating voltage. The frequency of the signal generator is increased slowly from zero to a high value. The magnet is observed to vibrate. Solenoid N S Signal ~generator (i) Sketch a graph to show the variation of the amplitude of vibration of the magnet with the frequency of signal generator. Explain the shape of the graph. (ii) If the mass of the magnet is 0.020 kg and the force constant k of the spring is 5.00 N cm–1, calculate the frequency of the signal generator when the amplitude of vibration of the magnet is maximum. 33. (a) What is meant by self induction? (b) An inductor of inductance 1.5 H and resistance R is connected in series with a 12.0 V battery and a switch. A neon lamp which requires a minimum voltage of 150 V to light up is connected across the inductor. The switch is closed and after 1.0 s it is opened. The variation of current I in the inductor with time t is as shown in the graph below. 0 0.5 1.0 1.02 t/s I/A 3.0 (i) Explain the shape of the graph. (ii) Sketch a graph to show the variation of the back e.m.f. with time. (iii) Determine the resistance R of the inductor. (iv) Find the maximum energy stored in the inductor. (v) Discuss when the neon lamp is lighted. 1 1. (a) φ = BA = (0.64)(0.20) = 0.128 Wb (b) φ = BA cos 30° = (0.128) cos 30° = 0.111 Wb 2. (a) Φ = NBA = (50)(0.35)(6 u 10–4)= 1.05 u 10–2 Wb (b) Φ = NBA cos 40° = 8.04 u 10–3 Wb 2 1. D 2. A 3. B 4. D 5. C 3 1. B 2. B 3. A 4. C 5. A 6. D 7. C 8. D: Fv = E2 R = (Blv) 2 R 9. 0.25 V No current E = Blv 10. (a) (i) 0 φ a l a + l x Bab ANSWERS
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 229 (ii) E = – dΦ dt = –vdΦ dx E = –Bbv 0 x < a = 0 a x < l = 2Bbv l x a + l (iii) Power = B2 b2 v2 R 0 x < a 1Use P = E2 R + = 0 a x < l = 4B2 b2 v2 R l x < a + l Total energy dissipated = 3B2 b2 va R Energy = Pt (b) (i) No change (ii) E = Bbv 0 x < a + l (iii) Power = B2 b2 v2 R 0 x a + l Energy = 2B2 b2 va R 11. 4.62 × 10–5 T, E = Blv cos 30° 12. (a) (i) E = Blv (ii) Electromagnetic force FM = BIl due to induced current in conductor equal and opposite to mechanical force F. I = Blv R , direction: Y to X (iii) 0.516 A Use power = Fv = I 2 R (b) E = BlaT E = Blv, v = u + at 13 D 0DJQHWLFÁX[OLQNHGZLWKWKHVROHQRLG decreases when the magnetic is pulled out. According to Faraday's law, an e.m.f. which is proportional to the rate of change of PDJQHWLFÁX[LVLQGXFHGLQWKHVROHQRLG$Q LQGXFHGFXUUHQWÁRZV (b) Lenz's law: the direction of the induced current opposes the S-pole leaving the solenoid. The induced current forms a N-pole at the right end of the solenoid. Induced FXUUHQWÁRZVIURPULJKWWROHIWLQWKHUHVLVWRU When the N-pole leaves the solenoid, the induced current forms S-pole at the right end RIWKHVROHQRLG,QGXFHGFXUUHQWÁRZVLQWKH opposite direction. (c) 0 Time Induced current 4 1. A 2. D 3. C 4. (a) (i) 3.08 × 10–5 V E0 = NBAω (ii) 2.18 × 10–5 V Er.m.s. = E0 Ƽ 2 (b) In the opposite direction. 5. 18.7 V E0 = NBAω, Er.m.s. = E0 Ƽ 2 6. (a) 0 t t E 0 = NBA cos ωt dΦ dt E = – 5 1. (a) Using LI = Φ,L = Φ I = (250)(0.11) 3.5 = 7.86 H 2. (a) B = μ0 nI = (4S u 10–7)(100)2 (2.0) = 2.51 u 10–2 T Φ = NBA = (2000)(2.51 u 10–2)(1.8 u 10–4) = 0.009 Wb (b) Using Φ = LI, L = 0.0045 H 3. E = –L dI dt = (57 u 10–3) 1 0.50 60 u 10–32 = 0.475 V 6 1. U = 1 2 LI2 = 1 2 (80 u 10–3)(0.40)2 = 6.4 u 10–3 J 2. (a) U = 1 2 LI2 = 1 2 (25 u 10–3)(5)2 = 0.3125 J (b) E = –L dI dt = (25 u 10–3) 1 5.0 5 u 10–32 = 25 V 7 1. B 2. A 3. A 4. C 5. B 6. A 7. A 8. B 9. t T 0 2.0 Current /A 3.0 A s–1
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 230 10. L QHJOLJLEOH EHFDXVHPDJQHWLF ÀHOGVLQ WKH WZR solenoids are in opposite directions. LWZLFHEHFDXVHPDJQHWLFÀHOGVLQWKHWZRVROHQRLGV are in the same direction. 11. (a) Refer page 212 and 216 (b) (i) 39.5 PH (ii) 19.7 PH 12. (a) 1.0 × 108 J Energy = 1 2 LI2 (b) (i) 1.0 × 1010 W Power = Energy Time (ii) 2.0 × 104 V E = –L dI dt STPM PRACTICE 17 1. C: E = –6.0 V = –L dI dt gives L = 0.33 H 2. A: Direction: Use Lenz’s law. 3. 'ffl %DFNHPI ² Φ1 – Φ0 T = 0 – NP0 (N/l)I0 A T 4. D 5. B 6. C: L < x < 3LfflQRFKDQJHLQÁX[E = 0. 3L < x < 4L: E induced in the opposite direction to the E induced for 0 < x < L 7. B: No force on a stationary electron in B. 8. D: Magnitude of induced e.m.f. = 'Φ 't = NBA – 0 T = N(μ0 NI/2r)(Sr2 ) T = μ0 SrN2 I 2T 9. A: E = Bxv, I = E R = Bxv R 10. C 11. A 12. 'ffl1RORVVRIPDJQHWLFÁX[ $OOPDJQHWLFÁX[SURGXFHG E\SULPDU\ FRLOLV OLQNHG with the secondary coil. 13. C 14. C: Use E = –M dI dt 15. A 16. B 17. B 18. A 19. D 20. B 21. C 22. C 23. (a) B = μ0 NI 2r )OX[ NBA = μ0 N2 I 2r (Sr2 ) = (4S u 10–7)(200)2 (6.0)(5.0 u 10–2)(S) 2 = 0.0237 Wb (b) Φ = LI , L = 395 mH 24. (a) E = Blv = (0.60)(0.080)(0.20) V = 9.6 mV 7KHGLUHFWLRQLVFORFNZLVH E ,QGXFHGFXUUHQWÁRZVLQRSSRVLWHGLUHFWLRQVLQ the rods. Repulsive magnetic force between the rods. (c) Momentum is conserved: (5.0)(0.20) + (4.0)(0) = (5.0)v1 + (4.0)v2 v1 – 0.20 = –(0.80)v2 ...................... Kinetic energy is conserved: 1 2 (5.0)v1 2 + 1 2 (4.0)v2 2 = 1 2 (5.0)(0.20)2 + 0 v1 2 – (0.20)2 = –0.80v2 2 ........................ /: v1 + 0.20 = v2 ......................................... Solving and : v2 = 0.22 m s–1 v1 = 0.020 m s-1 25. (a) (i) Φ = BA sin ωt (ii) E = – d dt (BA sin ωt) = (–BA ω cos ωt) (iii) EPD[ = BAω = (0.80)(20.0 u 10–4)(4S) = 20.1 mV (iv) t E 0 (b) (i) Φ = LI = NBA = N(μ0 nI)A L = (nl)(μ0 nA) = μ0 n2 l A (ii) (1) L increases if radius is greater (A = Sr2 ). (2) L increases if the length l is increased. 26. (a) Refer page 197 (b) E = – dΦ dt = (0.80)S(0.152 – 0.122 ) ———––—————– 0.16 V = 0.13 V (c) (i) R = V —– I0 = 2.5 ۙ (ii) eEDFN = V – IR = 1.75 V (iii) Pout = I = eEDFN = 3.0 W (iv) eEDFN v 20 rps, 1.75V v 45 rps eEDFN = 0.78 V 27. (a) (i) Electromagnetic induction: E.m.f. induced LQDFRQGXFWRUZKHQPDJQHWLFÁX[OLQNHG with it changes. (ii) Opposite of electromagnetic induction is PDJQHWLFHͿHFWRIFXUUHQW (b) (i) E = Blv cos 20o = 0.496 V (ii) At terminal velocity. I = E R F = BIl cos 20o = mg sin 20o B1 Blv cos 20° R 2 l cos 20o = mg sin 20o v = mgR sin 20° (Bl cos 20°)2 = 4.94 m s–1 B vertically downwards
17 Physics Term 2 STPM Chapter 17 Electromagnetic Induction 231 (iii) Loss of gravitational potential energy JDLQNLQHWLFHQHUJ\KHDWGLVVLSDWHG from resistor. 28. 9.87 × 10–7 V B = μ0 NI 2r , E = – dΦ dt E = – N2 BA2 t I 10 A Induced current opposes the decrease in the PDJQHWLFÀHOGLQWHQVLW\ZKHQWKHFXUUHQWLVUHGXFHG 29. (a) 3.14 × 10–8 V B = μ0 I 2r , E = – dΦ dt (b) I I B Induced (c) Induced current opposes decrease in magnetic ÁX[GHQVLW\ 30. D (QHUJ\LQPDJQHWLFÀHOG Energy stored can be converted into heat, by connecting a resistor to inductor, or into electric field by connecting capacitor to inductor. (b) (i) 1.50 × 10–6 Wb ) = NBA (ii) 1.50 × 10–4 V E = – d) dt 31. (a) Connect a.c. voltmeter to search coil. E 5RWDWHVHDUFKLQPDJQHWLFÀHOGXVLQJPRWRU 3HDNHPI E0 = NBAω B = E0 NAω E0 = Ƽ 2 V, V = voltmeter reading = Ƽ 2 V NAω (c) (i) V = Blv (ii) V = 0 (d) (i) E = NBA t (ii) 7.5 × 10–3 T s–1 I = E R 32. (a) (i) B = μ0 I1 2Sr F = BI2 l F l = μ0 I 1 I2 2Sr I Q B F (ii) I1 = I2 = 1 A, r = 1 m, μ0 = 4S × 10–7 H m–1 F l = 2 × 10–7 N m–1 (b) Magnet attracted into solenoid. Current in solenoid produces an S-pole at the top end of the solenoid. S-pole attracts N-pole of magnet. (c) (i) Amplitude Frequency Resonant frequency 0 (ii) f = 1 2S k m , natural frequency of loaded spring = 25 Hz 33. (a) Refer page 212 (b) (i) 0 to 0.5 s, current increases slowly. 5HDVRQffl%DFNHPIRSSRVHVWKHJURZWK of the current. 0.5 s to 1.0 s, current = 3.0 A (constant), QREDFNHPI t = 1.0 s, switch opened but currents WDNHVVWRGURSWR]HUR5HDVRQffl e.m.f. induced delayed the decrease of the current. (ii) 0 t/s Back e.m.f 0.5 1.0 1.02 (iii) R = 12.0 V 3.0 A ۙ LY 0D[LPXPHQHUJ\ 1 2 LI2 = 6.75 J Y WRVDYHUDJHEDFNHPI = L dI dt = (1.5)( 3.0 0.50 ) V = 9.0 V < 150 V, lamp not lighted. 1.0 s to 1.02 s, EEDFN = (1.5)( 3.0 0.02 ) V = 225 V, lamp lighted
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 232 CHAPTER ALTERNATING CURRENT 18 CIRCUITS Concept Map $OWHUQDWLQJ&XUUHQW $FWKURXJKD5HVLVWRU $FWKURXJKDQ,QGXFWRU 5/&LQ6HULHV $FWKURXJKD&DSDFLWRU 232 Bilingual Keywords 0TWLKHUJL!0TWLKHUZ 0UZ[HU[HULV\ZWV^LY!2\HZHZLRL[PRH 7OHZLKPMMLYLUJL!)LaHMHZH 7\YLPUK\J[VY!0UK\R[VY[\SLU 9LHJ[HUJL!9LHR[HUZ :PU\ZVPKHSHS[LYUH[PUNJ\YYLU[!(Y\Z\SHUNHSPRZPU\ZVPKHS INTRODUCTION 1. The electricity supplied to our homes, schools and industries is in the form of alternating current (a.c). 2. Audio and video signals in electrical circuits are in the form of alternating currents. 3. Principles that you learned in direct current such as Ohm’s law and Kirchhoff’s laws are applicable to alternating current circuits. 4. An alternating current is a current whose direction changes periodically.
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 233 5.Figure 18.1 shows three forms of alternating current. Figure 18.1 (a) Sinusoidal alternating current (b) Saw-tooth alternating current (b) Rectangular-wave alternating current Current, I 0 Time, t + – Current, I 0 Time, t + – Current, I 0 Time, t + – 6. The alternating current supplied to our home is the sinusoidal alternating current which we will study in detail in this chapter. 18.1 Alternating Current Through a Resistor Students should be able to: H[SODLQWKHFRQFHSWRIWKHUPVYDOXHRIDQDOWHUQDWLQJFXUUHQWDQGFDOFXODWHLWVYDOXHIRUWKHVLQXVRLGDOFDVHRQO\ GHULYHDQH[SUHVVLRQIRUWKHFXUUHQWIURPV = V0 VLQωt H[SODLQWKHSKDVHGLIIHUHQFHEHWZHHQWKHFXUUHQWDQGYROWDJHIRUDSXUHUHVLVWRU GHULYHDQGXVHWKHIRUPXODIRUWKHSRZHULQDQDOWHUQDWLQJFXUUHQWFLUFXLWZKLFKFRQVLVWVRQO\RIDSXUHUHVLVWRU Learning Outcomes 1. A pure resistor has no capacitive or inductive effect. 2. A sinusiodal a.c. may be represented by the equation I = I0 sin ωt where I0 is known as the peak current. I0 is the maximum value of the a.c. and ωt is known as the phase of the a.c. ω = 2Sf where f = frequency of a.c. = number of complete cycles per second The frequency of the a.c. in the household supply is 50 Hz. 3. The period T of an a.c. is the time for a complete cycle. period, T = 1 frequency T = 1 f If f = 50 Hz, period T = 1 50 s = 0.02 s Figure 18.2 R V Exam Tips Terms and formula for a.c. are identical to those in simple harmonic motion and waves. 2016/P2/Q13, 2018/P2/Q14
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 234 4. When a sinusoidal a.c. flows through a pure resistor of resistance R, the potential difference across the resistor, V = IR = (I0 sin ωt)R = I0 R sin ωt = V0 sin ωt where V0 = I0R is the peak voltage. 5. Figure 18.3 shows the graph of current I against time t, and voltage V against time t plotted on the same time axis. From the graphs, and also from the equations I = I0 sin ωt V = V0 sin ωt, we can conclude that the a.c. in a pure resistor is in phase with the voltage V across the resistor. Both the current I and voltage V attain their respective peaks at the same instant. Power Dissipated from a Pure Resistor 1. Suppose that the alternating current in a pure resistor is represented by the equation I = I0 sin ωt Then the potential difference V across the pure resistor may be represented by the equation V = V0 sin ωt 2. Since the value of I and V are not constant, the power P dissipated from a pure resistor varies with time t according to the equation P = IV = I0 V0 sin2 ωt 3. Figure 18.4 shows how the power P varies with time t. Mean power = mean of (I0 V0 sin2 ωt) = I0 V0 mean of 1 2 (1 – cos 2ωt) Mean power = I0 V0 2 (Mean of cos 2ωt = 0 = 1 2 I0 2 R over a complete cycle) = 1 2 V0 2 R (V0 = I0 R) Figure 18.3 V0 I 0 I V 0 t Figure 18.4 V0 I 0V0 I 0 I P V Power, P 0 Time, t T 2 T 3T 2T 2 2010/P1/Q35, 2011/P1/Q36, 2014/P2/Q14, Q15, 2016/P2/Q15
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 235 Root-Mean-Square Current (Ir.m.s.) 1. The root-mean-square (r.m.s) value of an alternating current is the value of the steady direct current which produces the same power in a resistor as the mean power produced by the alternating current. The r.m.s. current is the effective value of the a.c. 2. If Ir.m.s. is the r.m.s. value of an a.c., then I2 r.m.s. R = mean power = I0 2 R 2 Hence, Ir.m.s. = I0 Ƽ2 3. Similarly the r.m.s. value of an alternating voltage is the value of the steady direct voltage which when applied across a resistor, produces the same power as the mean power produced by the alternating voltage across the same resistor. If Vr.m.s. is the r.m.s. voltage, then V2 r.m.s R = mean power = V0 2 2R Vr.m.s. = V0 Ƽ2 ;OLHJ]VS[HNL\ZLKPU4HSH`ZPHPZ¶fl= *V\U[YPLZSPRL<:HUK*HUHKH\ZL¶= Info Physics Example 1 A sinusoidal a.c. of r.m.s. value 4.00 A and frequency 50 Hz fl ows in a pure resistor. (a) What is the peak current? (b) Find the magnitude of the instantaneous value of the current 6 u 10–4 s after it changes direction. Solution: (a) Peak current, I0 = Ƽ2 Ir.m.s. = Ƽ2 u (4.00) = 5.66 A (b) Use the equation I = I0 sin ωt, (ω = 2Sf) I = 5.66 sin (2S u 50 u 6 u 10–4) = 5.66 sin (0.189) = 1.06 A Note that the angle 0.189 is in radians. Exam Tips I 0V0 I 0V0 P Time 2 0 = Mean power T Area under the power-time graph = energy Area shaded = area shaded Mean power = 1 0 0 =0 = 0 r.m.s. =r.m.s. = 0 2 r.m.s.9 = V2 UPV R Exam Tips I UPV = 2 I 0 and VUPV = 2 V0 is only applicable for ZPU\ZVPKHS a.c.
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 236 Example 2 An alternating current of r.m.s. value 2.0 A when passing through a pure resistor dissipates power at the same mean rate as a steady direct current I. What is the value of I? Solution: Ir.m.s. = 2.0 A means that the mean power dissipated by a.c. is equal to the power dissipated by a steady direct current 2.0 A. Hence I = 2.0 A Example 3 A fuse is used to protect an electrical appliance which is normally connected to a 240 V a.c. supply. The circuit is broken if the current exceeds 13 A. For what value of the current would the fuse be broken if the electrical appliance is connected to a 120 V d.c. supply? Explain your answer. Solution: A fuse works on the principle of the heating effect of a current. A 13 A fuse would break if the current in it exceeds 13 A irrespective of the voltage supply. Hence when the appliance is connected to a 120 V d.c. supply the fuse would break when the current exceeds 13 A. Example 4 An ammeter uses the heating effect of the current to produce the defl ection of the pointer. Discuss whether the same calibration can be used for measuring direct currents and alternating currents. Solution: The same calibration can be used for measuring direct currents and alternating currents because the r.m.s. value of the alternating current is equal to the value of the direct current that produces the same heating effect. Hence the ammeter measures r.m.s. value of the alternating current. Example 5 When the sinusoidal potential difference V1 (Figure (a)) is applied across a resistor of resistance R the mean rate of heat dissipated is W. Figure (a) Figure (b) V1 V0 –V0 0 Time V2 V0 –V0 0 Time What is the rate of heat dissipated from the resistor when the potential difference V2 (Figure (b)) is applied across it?
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 237 Solution: The mean power dissipated by the sinusoidal alternating voltage V1 is W = 1 2 V0 2 R The alternating rectangular voltage V2 has constant magnitude V0 . Power = V0 2 R = 2W Example 6 An electric heater is rated 240 V a.c. 200 W. (a) What is the r.m.s. value of the current in the heater when it is connected to a 240 V a.c. supply? (b) What is the peak current in the heater? Fuses rated as 6 A, 10 A and 13 A are available. Which fuse is the most suitable for the heater. Explain your answer. Solution: (a) Vr.m.s. = 240 V Using power = Ir.m.s. Vr.m.s., Ir.m.s. = 2 000 240 = 8.33 A (b) Peak current, I0 = Ƽ2 Ir.m.s. = 11.8 A The effective current (r.m.s.) current = 8.33 A. Hence the suitable fuse that is rated slightly more than the r.m.s. value is 10 A. Quick Check 1 Exam Tips For a.c. the values of the voltage and current mentioned without further FODVVLðFDWLRQPHDQVWKHUPVYDOXHV e.g. 240 V a.c. means =r.m.s. = 240 V 1. When an alternating current flows in a resistor of resistance 16 :, the power P dissipated varies with time t as shown in the diagram. 200 P / W t / s 0 0.01 0.02 0.03 What is the r.m.s. value of the alternating current? A 1.5 A C 3.5 A B 2.5 A D 10 A 2. An ammeter that uses the heating effect of a current to produce a defl ection of the pointer records 5.0 A when used to measure a direct current. What is the ammeter reading when used to measure an alternating current of 5.0 A r.m.s.? A 1.5 A C 5.0 A B 3.5 A D 7.1 A 3. An ammeter uses the heating effect of a current to produce the deflection of the pointer. The reading, which is proportional to the heating, is X when a direct current I fl ows through the meter. When connected to a circuit in which an alternating current of r.m.s. value I fl ows, the reading is
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 238 A X Ƽ2 , because the average heating effect is half that of the direct current B X Ƽ2 , because it measures the r.m.s. current which is equal to the direct current divided by Ƽ2 C X, because it measures the r.m.s. current which has the same heating effect as the direct current D Ƽ2 X, because it measures the peak current which is Ƽ2 times the direct current 4. A sinusoidal current is represent by the equation I = I0 sin ωt Which of the following equations represents a sinusoidal current of twice the amplitude and frequency? A I = I0 sin (2ωt) C I = I0 sin 1 2 ωt B I = 2I0 sin (2ωt) D I = 2I0 sin 1 2 ωt 5. A sinusoidal voltage is connected across a pure resistor, and the frequency of the voltage is varied, keeping the r.m.s. voltage constant. The power dissipated from the resistor is A proportional to (frequency)2 . B proportional to frequency. C proportional to (frequency) 1 —2 . D independent of frequency. 6. The current in a resistor of resistance R is I = I0 sin ωt Which of the following best represents the variation of the power P dissipated in the resistor with time t? A C P I 0 2 R 0 t P I 0 2 R 0 t B D P I 0 2 R 0 t P 0 t I 0 2 R 2 7. When an alternating voltage of peak value V0 is connected across a resistor of resistance R, the rate of heat loss from the resistor is A V0 2 R C V0 2 2R B V0 2 Ƽ2R D V0 2 4R 8. An alternating current is represented by the equation I = 2 sin (50 St) where t is in s. What is the mean power developed by the current in a resistor of resistance 10 :? A 20 W C 100 W B 40 W D 200 W 9. An a.c. voltage is connected to a resistor of resistance R. The mean power dissipated in the resistor is P. If the same a.c. voltage is connected to a resistor of resistance 2R, the mean power dissipated in the resistor is A 1 4 P C 2P B 1 2 P D 4P 10. A sinusoidal alternating voltage is connected across a resistor. Which comparison between the current in the resistor and the voltage is correct? A When the voltage is zero, the current is maximum. B When the voltage is maximum, the current is zero. C When the voltage is zero, the current is zero. D When the voltage is positive, the current is negative.
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 239 11. The sinusoidal current is a resistor of resistance 8.0 :as shown in the graph below. I / A 0 20 40 60 –5 5 t / ms Find (a) the peak current, (b) the r.m.s. value of the current, (c) the frequency of the current, (d) the change in phase in a time of 15 ms. (e) the mean power dissipated 12. When an electric heater is operated from a 240 V a.c. supply, an r.m.s. current of 12.0 A fl ows. Assuming that the heater is purely resistive, calculate (a) its resistance, (b) the mean power, (c) the maximum instantaneous power. 18.2 Alternating Current Through an Inductor Students should be able to: GHULYHDQH[SUHVVLRQIRUWKHFXUUHQWIURPV = V0 VLQωt H[SODLQWKHSKDVHGLIIHUHQFHEHWZHHQWKHFXUUHQWDQGYROWDJHIRUDSXUHLQGXFWRU GHÀQHWKHUHDFWDQFHRIDSXUHLQGXFWRU XVHWKHIRUPXODXL = ωL GHULYHDQGXVHWKHIRUPXODIRUWKHSRZHULQDQDOWHUQDWLQJFXUUHQWFLUFXLWZKLFKFRQVLVWVRQO\RIDSXUHLQGXFWRU Learning Outcomes 1. A pure inductor has only inductive effect. Its resistance is zero and it has no capacitive effect. 2. Since an inductor usually consists of a coil of wire which has a fi nite resistance, pure inductors are quite diffi cult to come by in practice. 3. Figure 18.5 shows a pure inductor connected to an alternating voltage V. Suppose the alternating current in the inductor is I = I0 sin ωt The magnetic fl ux linkage in the inductor changes sinusoidally as the current changes. The back-e.m.f. E induced in the inductor is given by E = –L—dI – dt L = inductance = –L—d – dt (I o sin ωt) = –ωLI0 cos ωt In order that the current continues to fl ow in the inductor, the applied voltage V must be of the same magnitude but opposes the back e.m.f. Hence, V = –E = ωLI0 cos ωt V = ωLI0 sin ωt + S 2 ............................. { V = V0 sin ωt + S 2 V0 = ωLI0 ~V L A.c. Inductor Figure 18.5 2010/P1/Q34, 2011/P1/Q35, 2011/P2/Q12, 2013/P2/Q14, Q17, 2017/P2/Q25, 2018/P2/Q15
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 240 4. Comparing the equations I = I0 sin Zt with V = V0 sin ωt + S 2 and from the graphs of I and V in Figure 18.6, we concluded that I and V for a pure inductor are not in phase. The voltage V is leading the current I by T 2 or S 2 radians or 90°. 5. From the equation for the peak voltage V0 = ωLI0 , XL = V0 I0 = ωL = 2SfL (ω = 2Sf) The ratio of V0 I0 = Vr.m.s. Ir.m.s. is known as the reactance of the inductor. Reactance of inductor, XL = V0 I0 = Vr.m.s. Ir.m.s. XL = 2SfL The reactance of the inductor is a measure of the opposition of the inductor to alternating current. 6. Figure 18.7 shows how the reactance XL of an inductor varies with frequency f of the alternating current. Exam Tips Figure 18.7 XL = 2SfL 0 Reactance, XL Frequency, f VUPV IUPV )RUDQLQGXFWRU ;ɧL = 2SfL = = V0 I 0 VUPV = IUPV;ɧL Example 7 Calculate the potential difference across an inductor of inductance 5.0 H when an alternating current of 20 mA and frequency 50 Hz passes through it. Solution: V = IXL , XL = 2SfL = 2SfLI = 2S u 50 u 5.0 u (20 u 10–3) = 31.4 V Figure 18.6 T 0 4 + – Time, t T T I V 2 T4 3
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 241 Power in a Pure Inductor Circuit 1. For a pure inductor, the potential difference V leads the current I by S 2 radians. If current I = I0 sin ωt, then potential difference, V = V0 sin ωt + S 2 = V0 cos ωt 2. The instantaneous power, P = IV = I0 V0 sin ωt cos ωt = 1 2 I0 V0 sin 2ωt = Ir.m.s.Vr.m.s. sin 2ωt Mean power over a complete cycle P = mean of (Ir.m.s.V r.m.s. sin ωt) = 0 3. Figure 18.8 shows the variation of power P with time t for a complete cycle. Figure 18.8 V T P Time, t P 4 1 2 T T 2 3T 4 – + + 0 – – I –I0 I0 –V0 V0 I0V0 1 2I0V0 Figure 18.8 may be used to explain why the mean power over a complete cycle is zero for a pure inductor. The area under the power-time graph represents energy. 4. For the first quarter of the cycle, from time t = 0 to t = T 4 , the area under the graph is positive. This area represents the energy from the a.c. supply to the inductor. The energy stored in the inductor is maximum at time t = T 4 when the current I is maximum. Recall the equation energy = 1 2 LI2 . 5. During the second quarter of the cycle, t = T 4 to t = T 2 , the area under the graph is negative. This area represents the energy from the inductor back to the a.c. supply, which is equal to the energy received from the a.c. supply during the first quarter cycle. Hence, there is no energy dissipated in the inductor for the first half of the cycle. 6. Similarly, during the third quarter cycle, energy is supplied to the inductor. This energy is returned to the supply during the last quarter cycle. Hence, over a complete cycle, the mean power dissipated is zero.
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 242 Quick Check 2 Which of the following graphs correctly represents the variation of the voltage V and power P in the inductor with time t? A 0 t V P V, P B 0 t V P V, P C 0 t V P V, P D 0 t V P V, P 4. An alternating current I = I0 sin ωt fl ows in a pure inductor. On the same axes, sketch graphs to show (a) the variation of potential difference V with time, (b) the variation of power in the inductor with time. 1. Which of the following statements about an alternating current circuit containing a pure inductor is true? A The applied voltage leads the current by S 2 radians. B The applied voltage leads the current by S radians. C The current leads the applied voltage by S 2 radians. D The current leads the applied voltage by S radians. 2. The graph shows the alternating voltage V and current I in an electrical component. 0 t / s V I 0.2 0.4 0.6 0.8 1.0 The current I leads the voltage V by A S 2 radians B S radians C – S 2 radians D –S radians 3. The diagram shows a pure inductor L connected to an a.c. supply. L A.c. ~
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 243 18.3 Alternating Current Through a Capacitor Students should be able to: GHULYHDQH[SUHVVLRQIRUWKHFXUUHQWIURPV = V0 VLQωt H[SODLQWKHSKDVHGLIIHUHQFHEHWZHHQWKHFXUUHQWDQGYROWDJHIRUDSXUHFDSDFLWRU GHÀQHWKHUHDFWDQFHRIDSXUHFDSDFLWRU 8VHWKHIRUPXODXC = 1 ωC GHULYHDQGXVHWKHIRUPXODIRUWKHSRZHULQDQDOWHUQDWLQJFXUUHQWFLUFXLWZKLFKFRQVLVWVRQO\RIDSXUHFDSDFLWRU Learning Outcomes 1. A pure capacitor has no resistive or inductive effect. 2. When a capacitor is connected to an a.c. supply, electric charges fl ow into and out of the capacitor plates. 3. Suppose that the alternating voltage applied across a capacitor of capacitance C is given by V = V0 sin ωt then charge in capacitor, Q = CV = CV0 sin ωt Using I = —– dQ dt , current in capacitive circuit, I = —– d dt (CV0 sin ωt) = CV0 ω cos ωt I = I0 sin ωt + S 2 where I0 = CV0 ω 4. Comparing the equations I = I0 sin ωt + S 2 and V = V0 sin (ωt), we conclude that in a capacitive circuit, the current I leads the voltage by S 2 radians. The graph in Figure 18.10 shows the variation of I and V with time t. 5. From the equation I0 = CV0 ω, the reactance of a capacitor in an a.c. circuit is defi ned as XC = V0 I0 = Vr.m.s. Ir.m.s. = 1 ωC ω = 2Sf XC = 1 2SfC Figure 18.11 shows the variation of XC with frequency f of the a.c. Figure 18.9 ~ V C Figure 18.11 XC 0 f Figure 18.10 T 0 4 + – Time, t T T I V 2 T4 3 2009/P1/Q34, 2011/P1/Q35, 2012/P1/Q34, 2013/P2/Q15, 2015/P2/Q14, 2017/P2/Q14
18 Physics Term 2 STPM Chapter 18 Alternating Current Circuits 244 Example 8 A 100 μF capacitor is connected to an alternating voltage of 12 V of frequency 50 Hz. Calculate (a) the reactance of the capacitor, (b) the current in the circuit. Solution: (a) Reactance, XC = 1 2SfC = 1 2S u 50 u (100 u 10–6) = 31.8 : (b) Using XC = Vr.m.s. Ir.m.s. , Ir.m.s. = 12 31.8 = 0.377A Power in a Pure Capacitor Circuit 1. For an a.c. circuit containing a capacitor, the current I leads the voltage by S 2 radians. V = V0 sin ωt I = I0 sin ωt + S 2 = I0 cos ωt 2. The variation of power P in the capacitor varies with time t according to the equation P = IV = (I0 cos ωt)(V0 sin ωt) = I0 V0 sin ωt cos ωt = 1 2 I0 V0 sin 2ωt = Ir.m.s.Vr.m.s. sin 2ωt Figure 18.12 shows the variation of power P with time t for a pure capacitor. 3. From t = 0 to T 4 , the power is positive. This means that energy is supplied to the capacitor by the voltage supply. The energy stored in the capacitor is maximum at time t = T 4 when the voltage V is maximum. This happens when the charge in the capacitor is maximum. Energy is stored in the electric fi eld between the plates of the capacitor. 4. From t = T 4 to t = T 2 , the power is negative. Energy stored in the capacitor is now returned to the voltage supplied. The net energy dissipated from the capacitor during the half cycle is zero. 5. Similarly the energy supplied to the capacitor from time t = T 2 to t = 3 4 T is returned to the voltage supply from time t = 3 4 T to t = T. Hence no energy is dissipated from the pure capacitor for a complete cycle. The mean power = 0. V T P Time, t T 4 T 2 3T 4 – + + 0 – I 1 2 I0V0 I0 V0 Figure 18.12