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4 Electricity and Magnetism 4 Chapter 4 Electromagnetism Electromagnetism 9 (a) Practice 4.1 (p.166) 1 A 2 C not. B 4 D 5 C S N S N S S N (b) Iron is a magnetic material and aluminium is 3 N 10 N A The field lines are in the same direction and are uniformly spaced. (1) is correct. B P is a north pole and R is a south pole. (2) is incorrect. 11 The field lines between P and Q are more closely packed together. (3) is correct. 6 (a) S (b) The reading drops. Practice 4.2 (p.183) Iron is a magnetic material. The iron bar 1 and the magnet attract each other. A By right-hand grip rule, a magnetic field 7 pointing downwards is produced in the N electromagnet. It repels the bar magnet and S the reading of the scale increases. 2 A 3 D The fields due to the two currents cancel out each other. 8 (a) It points towards north. (b) The bar magnet will rotate and its north pole will point downwards as shown. S 4 C B = BP + BQ = 5 0 I 2πr 0 2I 2πr = 3 0 I 2πr B The directions of the fields due to the four wires are as shown below. N New Senior Secondary Physics at Work (Second Edition) Oxford University Press 2015 1 4 Electricity and Magnetism Chapter 4 Electromagnetism BP = BS = 4 × 10–5 T (out of page) I BQ = 0 2 πr 4 π 10 7 4 = 2 π 0.03 45° BR BQ The vertical components cancel each other. The horizontal components are all equal to 0 I cos 45 2πr I 2 0 I Resultant field = 0 cos 45 4 = 2πr πr 6 4 π 10 7 4 2 π 0.02 = 2.67 × 10–5 T (into page) 10 (a) X (b) C The two identical solenoids can be regarded as = 11 (a) B remains unchanged. 7 (a) N S N 4π 10 7 3 4π 10 7 3 2π 0.025 2π 0.075 = 3.2 × 10–5 T turns per unit length and carries the same current as the separate solenoids. By B = μ0nI, Z Magnitude of magnetic field at X I I = 0 1 0 2 2πr1 2πr2 one when they are brought end-to-end. The combined solenoid has the same number of Y P: to the left Q: to the right (b) S 12 The magnitude of the field increases. In the region between X and Y, the magnetic field due to X and that due to Y both points (b) 8 (a) S NN S into the page, i.e. no neutral point in this region. (top view) In region above X, the field due to X is always stronger than that due to Y, i.e. no neutral point in this region. P must be below Y. (b) Let x be the distance between P and X. 0 I X 0 I Y 2πrX 2πrY (top view) 5 2 x x 0.3 x 0.5 m ( 50 cm) 13 9 2 BP = 0 I 2 πr (a) Magnitude of magnetic field NI = 0 l 4π 10 7 10 5 = = 4.19 × 10–3 T 0.015 New Senior Secondary Physics at Work (Second Edition) Oxford University Press 2015 4 Electricity and Magnetism (b) Chapter 4 Electromagnetism FS Since the number of turns per unit length FR increases, the magnitude of the field increases. 14 (a) (b) 15 (a) By B = FQ 0 I , 2 πr 2πrB 2π 0.02 5 10 6 I 0.5 A 0 4π 10 7 Consider the resultant magnetic force per unit If the current is doubled, the magnetic Let I be the current and d be the side length of field is also doubled. the square. Taking the direction to the right as Magnetic field = 5 × 2 = 10 μT positive, the horizontal component is length. The vertical component is zero. FR – FS cos 45° – FQ cos45° 0 I 2 I 2 1 0 I 2 1 0 = 2π 2 d 2πd 2 2πd 2 Soft iron (Or other reasonable answers) (b) X: south = Y: north (c) (i) The current in the circuit decreases 6 For (2), A is doubled. For (1) and (3), there is and hence the strength of the no change in B, I, A or N. electromagnet increases. decreases, so the current increases A τ = BIAN sin θ (ii) The current in the circuit increases (iii) The total resistance of the circuit 7 (a) Zero (b) F = BIl = 1.2 × 4 × 0.2 and hence the strength of the = 0.96 N (into page) electromagnet increases. (c) B 2 D = 4.5 N (upwards) (d) = 1.2 N (towards upper left and the same as that of the current in X. A 4 B 5 C perpendicular to the wire) (e) F = BIl sin θ = 2 × 5 × 0.2 × sin 30° = 1 N (into page) The forces acting on P by the other three wires are as shown below. F = BIl = 0.5 × 12 × 0.2 The direction of the magnetic field due to Y is 3 F = BIl = 1.5 × 15 × 0.2 Practice 4.3 (p.204) 1 2 2 πd <0 and hence the strength of the electromagnet decreases. 0 I 2 (f) F = BIl sin θ = 1.1 × 6 × sin (90 40°) = 5.06 N (into page) New Senior Secondary Physics at Work (Second Edition) Oxford University Press 2015 3 4 Electricity and Magnetism 8 (a) Chapter 4 Electromagnetism The magnetic force applied on the wire the coil and zero when the forces are must be upwards to balance the weight along the plane of the coil in this case. of the rider. By Fleming’s left-hand rule, (i) the blue face is the south pole. (b) Take moment about the support. BIl × rm = mg × rr mgrr B= Ilrm 0.002 9.81 0.01 = 5 0.06 0.2 S N S N (ii) = 3.27 × 10–3 T The magnitude of the magnetic field is 3.27 × 10–3 T. 9 (a) (c) Any one of the following: Use stronger magnets. Increase the current. (b) Force acted on the wire Increase the area of the coil within the = BIl = 0.4 × 3 × 0.05 = 0.06 N magnetic field By Newton’s third law, a downward Increase the number of turns of the coil. force of 0.06 N is applied on the magnet Use curved magnets. by the wire, which would cause the Use a soft-iron armature with several balance reading to increase by 0.06 N = 6.12 × 10–3 kg = 6.12 g. 9.81 m s 2 coils. 11 (a) and hence no or only a small current Mass of magnet = 120 – 6.12 = 114 g (c) (i) pass through the coil. The magnitude increases. (ii) The direction reverses. 10 (a) The coil is short-circuited by the ring 12 (b) Replace the ring with a split-ring. (c) The turning effect increases. (a) The directions of the magnetic forces are shown below. S N FP Q (b) The moment of a couple is proportional to the perpendicular distance between the forces, which is maximum when the forces are perpendicular to the plane of 4 FR For each metre, 0 I P I Q FP = 2 π rP New Senior Secondary Physics at Work (Second Edition) Oxford University Press 2015 4 Electricity and Magnetism = Chapter 4 Electromagnetism 4π 10 7 2 2 2π 0.01 = 8 × 10–5 N 0 I R I Q FR = 2 πrR 4π 10 7 2 2 = 2π 0.01 Since the force is at right angle to the moving direction, no work is done and the particle does not change its speed. (3) is incorrect. 3 Whether a particle carries positive or negative charge, the direction of the electrostatic and = 8 × 10–5 N magnetic forces are opposite. The sign of the Resultant force charges cannot be determined. = FP2 FR2 (b) (1) is not necessarily correct. = 1.13 × 10–4 N (to the left) 0 I S I Q (i) FS = 2 πrS IS = = When a particle just enters the fields, net force = QE – BQv = Q(E – Bv) From the given information, it can only be E E sure that v X and vY . The B B 2 πrS FS 0 I Q 2 π 1 1.13 10 4 4 π 10 7 2 magnitudes of the charges cannot be determined. = 283 A (3) is correct and (2) is not necessarily (ii) S may be on the left of Q with the current flowing out of the page, or B correct. 4 (a) it may be on the right of Q with the 13 (a) (i) Magnitude of force = Bqv current flowing into the page. = 0.2 × 2 × 1.60 × 10–19 × 4 × 105 The motor may be overheated. = 2.56 × 10–14 N (ii) The motor requires a large voltage to drive. (iii) The motor is very large in size. (Or other reasonable answers) (b) A.c. power source can be used. (Or other reasonable answers) Practice 4.4 (p.214) 1 D 2 B There is magnetic force as long as the moving (b) Magnitude of force = Bqv sin θ = 0.2 × 1.60 × 10–19 × 4 × 105 × sin 30° = 6.4 × 10–15 N direction of the charge and the field are not parallel. (1) is incorrect. New Senior Secondary Physics at Work (Second Edition) Oxford University Press 2015 5 4 Electricity and Magnetism (c) Chapter 4 Electromagnetism The magnetic force is zero since the V Bd 3000 = 0.015 0.1 direction of motion is along the field. = 2 × 106 m s–1 The magnetic force is zero since the = charge is not moving. (d) 5 (a) The magnetic force provides the centripetal force for circular motion. mv 2 By Bqv = , r mv r= Bq 31 = 9.11 10 8.0 10 2 10 3 1.60 10 19 6 The speed of the particle is 2 × 106 m s–1. (c) Change the strength of the electric or magnetic field. Revision exercise 4 Concept traps (p.218) 1 F = 0.022775 m The magnetic field at the centre of a long 0.0228 m solenoid depends on the number of turns per N unit length . l The radius of the path is 0.0228 m. (b) 2 T 3 F v and B are not required to be perpendicular. Multiple-choice questions (p.218) (c) Distance travelled = π × 0.022775 = 0.0715 m 4 D 5 B As charge flows from one place to another, a Time of staying in the field distance = speed 0.0715 = 8.0 10 6 current is formed and this creates a magnetic field. (1) and (2) are correct. Light travels outwards from the lightning bolt –9 6 (a) = 8.94 × 10 s in straight lines. No, whether the particle carries positive (3) is incorrect. or negative charge, the electrostatic and (b) D magnetic forces it experiences are The electromagnet applies an attractive force opposite. Both cases are possible. on the bar magnet. To increase the scale The magnetic force and the electrostatic reading, the force should be reduced. force balance each other. Bqv = qE E v= B 6 6 7 A Copper is a non-magnetic material and does not affect the magnetic field strength. (1) is incorrect. New Senior Secondary Physics at Work (Second Edition) Oxford University Press 2015 4 Electricity and Magnetism 8 Chapter 4 Electromagnetism By B = μ0nI, the field strength can be raised by 14 (HKDSE 2013 Paper 1A Q26) increasing the number of turns per unit length. 15 (HKDSE 2013 Paper 1A Q28) (2) is correct and (3) is incorrect. 16 (HKDSE 2014 Paper 1A Q26) D In (1) and (2), the current through coil Y Conventional questions (p.220) increases, while in (3) magnetic field strength 17 (a) through coil Y increases. 9 D 10 A The magnetic field applies a force on the (i) electron perpendicular to its motion, so it accelerates but has no change in kinetic (b) energy. (1) is incorrect and (2) is correct. Since the magnetic force is always (Current out of the page) 1A (ii) (Correct field lines) I Magnetic field = 0 2 πr 4π 10 7 2 = 2π 0.05 1A = 8 × 10–6 T perpendicular to the magnetic field, the 1A magnetic field will never change and thus its Q: Into page 1A direction of motion will not become R: No field 1A perpendicular to the field. S: Out of page I By B = 0 , 2 πr 1A (3) is incorrect. 11 (a) (b) D Magnitude of B-field at Q 4π 10 7 0.5 2 = 2π 0.04 The force between the wires is attractive. = 5 × 10–6 T 19 12 B (a) (i) 1M 1A Electromagnet A and resistor R1 are connected to the battery in parallel. If the charge is positive, both the electrostatic 1A and magnetic forces are in the upward Therefore, the voltage across the direction. If the charge is negative, both forces electromagnet never changes and it are downwards. keeps attracting the door. (2) is correct. 13 1A P: No field component of the electron’s velocity along the 18 1M 1A (ii) R3 should be placed in series with The magnitudes of the forces and the velocity the battery. cannot be deduced from the given When the springy metal contacts information. are closed, R1 is connected in (1) and (3) are not necessarily correct. parallel with the electromagnet. (HKDSE 2012 Paper 1A Q29) Their equivalent resistance New Senior Secondary Physics at Work (Second Edition) Oxford University Press 2015 1A 7 4 Electricity and Magnetism Chapter 4 Electromagnetism decreases and the electromagnet A higher voltage results in a larger has a smaller share of voltage. 1A current and hence stronger magnetic The current passing through the field. 1A (a) From Y to X 1A (b) By F = BIl sinθ, F I= Bl sin 0.01 = 0.4 0.05 sin 30 1M electromagnet decreases, and so the 21 strength of the electromagnet decreases. (b) 1A Make the coils in A more closely spaced. 1A By B = 0 NI l , the magnetic field strength increases with Or =1A (c) N . l 1A 1A 1A increases the magnetic field. 22 (a) Magnetic force FB = BQv = 1.8 × 10 × 1.60 × 10 1A –19 × 1.22 × 107 = 3.51 × 10–15 N 1A The direction is downwards. 1A 1A Iron core 1A 0.4 m wire with diameter 2 mm 1A (b) Downwards 9–V cell 1A (c) The electrostatic force must balance the magnetic force. (b) QE = FB F E= B Q wire coil core = (Wire wound on core and connected to battery correctly) 1A (Correct labels) 1A Iron is the most readily magnetized material among the three choices. 1A 1M 3.51 10 15 1.60 10 19 = 2.20 × 104 N C–1 1A 4 –1 The field strength is 2.20 × 10 N C . (d) The 0.4 m wire with diameter 2 mm has (i) The electron deflects due to the magnetic force. 1A (ii) It performs uniform circular motion the lowest resistance among the three, in the clockwise direction. which results in the largest current and The magnetic force provides the hence the strongest magnetic field. 1A centripetal force. Therefore, mv 2 BQv = r mv r= BQ (thickness of the wire does not affect the number of turns per unit length as the wire can be wound in multiple layers) 8 1M –3 (Or other reasonable answers) (c) 1A and the direction remains unchanged. This increases the current through A and (a) The magnitude increases Replace R2 with a resistor of lower resistance. 20 1A 1A 1M New Senior Secondary Physics at Work (Second Edition) Oxford University Press 2015 4 Electricity and Magnetism = Chapter 4 Electromagnetism 9.11 10 31 1.22 10 7 1.8 10 3 1.60 10 19 25 (a) magnetic field is set up. = 0.0386 m 1A (a) the coil and the frame are attracted The poles of the electromagnet which towards the soft-iron block. have the greatest attractive force are not The scale reading thus increases. facing the materials to be sorted. 1A (b) closer to each other. of the following): Use a thicker wire to reduce the 1A Turning the electromagnet by 90 so that resistance of the coil. one end of the coil faces the materials to (Or other reasonable answers) (c) Replace the soft-iron block by a permanent magnet. By doing so, the coil is attracted towards materials to be sorted. the magnet when current flow in one Steel retains part of its magnetic direction 1A 1A property after the external magnetic field and is repelled by the magnet when is removed. current flow in the other direction. 1A The rider should be placed on XY. 1A 1A 26 (a) core by an iron core. 1A By Fleming’s left-hand rule, the wire XY Increase the current. 1A experiences a magnetic force pointing upwards. length in the coil. 1A (a) The rod rolls towards A. 1A (b) (i) (b) = mg 1M = 0.002 × 9.81 1M = 0.04 × 5 × 0.035 = 0.007 N (towards A) = 0.0196 N 0.035 sin30 1M 1A (c) By F = BIl, 1M F 0.0196 d=l= = = 0.0327 m 1A BI 0.05 12 (d) (i) 1A (ii) Magnetic force = BIl = 0.04 5 1A Magnetic force = weight of rider Magnetic force = BIl The rider should be put on WZ 1A because the direction of the = 0.014 N magnetic force is reversed. The direction is at 60° to AB. 1A (c) 1A electromagnet with both ends facing the Increase the number of turns per unit 24 1A Using a U-shape core in the This can be solved by replacing the steel (c) 1A Compress the coil so that the turns are The design can be improved by (anyone be sorted. (b) 1A The coil becomes an electromagnet and The radius of the path is 0.0386 m. 23 When a current flows through the coil, a 1A (ii) The rider should be put between Y For a higher resistance, the current will and the support. / A lighter rider be smaller should be used. 1A and hence the magnetic force will be smaller. New Senior Secondary Physics at Work (Second Edition) Oxford University Press 2015 1A Since the magnetic field is not 1A perpendicular to the current, by 9 4 Electricity and Magnetism 27 Chapter 4 Electromagnetism F BIl sin , the magnetic force Moment of coil decreases. = Fl (a) Same as I1 (b) (i) By B = 1A 0 I , 2 πr I B1 = 0 1 (r constant) 2πr 0 I1 r= 2πB1 1A = 3.6 × 10 × 0.04 1M = 1.44 × 10–4 N m the magnetic field, (ii) Into page (c) 1A moment of coil = 1.44 × 10–4 N m 4 π 10 3.5 2 π 1.4 10 4 = 0.005 m 1A (ii) Since the coil is always parallel to 1M 7 = 1M –3 (d) Any two of the following: 1A 1A + 1A Use stronger magnets. 1A Increase the current. 1A Increase the area of the coil within the At P, the magnetic field produced by I1 magnetic field. has the same magnitude as that produced Increase the number of turns of the coil. by I2. 0 I 1 0 I 2 = 1M 2 πr1 2πr2 I1 r1 r 0.005 1 = = = = I 2 r2 0.03 r 0.03 0.005 5 Use a soft-iron armature with several coils. 29 (a) The magnetic forces act along the plane of the coil and have no turning effect. 1A 1A (d) I2 = 5I1 = 5 2 = 10 A Magnetic force per unit length II = 0 1 2 2πr 4π 10 7 2 10 = 2π 0.03 28 (a) The coil should not be vertical before the 1M switch is closed. (b) 1M of the current through the coil every half It vibrates a few times 1A direction. (c) and comes to rest lying along the vertical. (c) (i) 10 that the coils stay longer in parallel with the magnetic field. 1A (Or other reasonable answers) (d) 1M 1A A practical motor use curved magnets so 1A Force on one side of coil = BIl 1A so that the coil always turns in the same 1A Use a split-ring. The commutator reverses the direction turn = 1.33 × 10–4 N (b) 1A 1A Replace the permanent magnets with electromagnets that are connected to the = 0.02 × 3 × 0.06 a.c. power supply. = 3.6 × 10–3 N When the direction of the current The magnetic forces acting on the through the coil reverses, the direction of two side of the coil form a couple. the magnetic field also reverses. 1A 1A New Senior Secondary Physics at Work (Second Edition) Oxford University Press 2015 4 Electricity and Magnetism 30 (a) Chapter 4 Electromagnetism As a result, the direction of the magnetic Physics in article (p.225) force will not reverse due to the 36 alternating current. 1A No, 1A (a) Opposite 1A (b) No, 1A the currents in the live and the neutral the magnetic force is always wires always flow in opposite directions. perpendicular to the direction of motion 1A of the particle. 1A (b) No change 1A ground fault interrupter can cut off the (c) (i) 1A power immediately while an earth wire Uniform circular motion (c) (ii) The magnetic force provides the centripetal force. mv 2 BQv r BQr v m 2 πr T= v 2 πm = BQ = (d) 1M If there is a leakage of current, the may not be able to do so. 1A No. 1A As long as the current in the live and the neutral wires remains the same, the interrupter does not response. Hence, it 1M cannot detect a large current due to a short circuit. 1A 2π 6.64 10 27 1.6 2 1.60 10 19 = 8.15 × 10–8 s 1A –8 The period is 8.15 × 10 s. 31 (HKCEE 2007 Paper 1 Q11) 32 (HKALE 2007 Paper 1 Q3) 33 (HKDSE 2013 Paper 1B Q4) Experiment questions (p.225) 34 Bring them close to each other pair-by-pair. The two bars which are neither attracted nor repelled by any of the other are the copper bars. 1A Two of them may attract each other, but repel each other if one of them is reversed. They are the bar magnets. 1A The remaining two bars are the iron bars. 1A (Or other reasonable answers) 35 (HKCEE 2011 Paper 1 Q6) New Senior Secondary Physics at Work (Second Edition) Oxford University Press 2015 11