15 145 29. In the circuit shown, the voltmeter has a resistance of 500 : and reads 1.0 V. A C B 1 500 8 V 500 8 Calculate (a) the potential difference across AB, (b) the potential difference across CB when the voltmeter is removed. 30. Two identical resistors each of resistance 5 k: are connected in series with a 6.0 V battery. A voltmeter of resistance 5 k: is connected across one of the resistors. Calculate the percentage error in the voltmeter reading. 31. (a) A cell of e.m.f. E and internal resistance r are connected to a load of resistance R. (i) Write an expression for the potential difference V across the load. (ii) Deduce the expression for V if R is large compared to r. (b) Two identical cells, each of e.m.f. E and internal resistance r are connected to the load of resistance R as shown in the diagram below. R E r E r Derive the expression for the potential difference V across the load of resistance R. 32. A cell is connected in series with an ammeter and a variable resistor. A voltmeter is connected across the terminals of the cell. When the resistance of the variable resistor is changed, the readings of the ammeter and voltmeter are a shown below. Current/A 0.10 0.20 0.30 0.40 0.50 Potential 1.36 1.25 1.15 1.07 0.97 difference/ V Plot a graph of potential difference against current, and deduce from the graph the e.m.f. of the cell and its internal resistance. 33. Measurements using a potentiometer gives the e.m.f. of a battery as 2.80 V, but a voltmeter gives a value of 2.50 V. The voltmeter consists of a moving-coil galvanometer of resistance 50 : connected in series with a large resistor X. The voltmeter reads 3.00 V when a current of 10 mA flows in it. Calculate (a) the internal resistance of the battery, (b) the value of the resistance of the resistor X. 34. (a) Define e.m.f. of a battery. (b) Show that when a battery of e.m.f. E and internal resistance r, is connected to a resistor of resistance R, the relation between E and the potential difference V across the resistor is given by —–— E – V – V = —r – R In the circuit shown, V is a high resistance voltmeter and the resistance of the resistor R is 2.0 :. R K V The table below shows the voltmeter reading V at time T. The switch K is closed 15 seconds after the start of the experiment and opened 45 seconds from the start. T/s 0 10 15 20 30 40 45 V/V 1.5 1.5 K close 0.9 0.8 0.75 K open T/s 50 60 70 V/V 1.1 1.2 1.25 (i) Plot a graph of potential difference against time. Physics Term 2 STPM Chapter 15 Direct Current Circuits
15 146 (ii) From the graph, determine the voltmeter reading immediately after the switch K is closed. (iii) Explain the shape of the graph. (iv) Estimate the internal resistance of the battery at time T = 15 s, and at time T = 45 s. 35. (a) Explain why an ammeter should have a low resistance, and a voltmeter should have a high resistance. (b) The resistance R of a resistor is determined using circuit P, and circuit Q. R Circuit P V A R Circuit Q V A Discuss which of the circuit would give a more accurate result if the resistance R is small. (c) A potentiometer is used to determine the e.m.f. E and internal resistance r of a cell as shown in the diagram. The slide wire PQ is 100.0 cm long and has a resistance of 5.0 :. The drive cell D has an e.m.f. of 2.0 V and negligible internal resistance. When the switch S is open, the centrezero galvanometer is balanced when PJ = 87.5 cm. When the switch S is closed the balanced length becomes 78.2 cm. E, r S ȍ ȍ 2.0 V D P Q J G Determine (i) the e.m.f. and (ii) the internal resistance of the cell. 36. (a) (i) Define the resistance of an electric conductor. (ii) Differentiate between ohmic c o n d u c t o r a n d n o n - o h m i c conductor. (iii) Sketch a graph to show the I-V characteristic of a filament lamp. Hence discuss to what extend the filament lamp obeys Ohm’s law. (b) A galvanometer of resistance 25 : has a full-scale deflection of 5.0 mA. (i) With the aid of a labelled diagram, show how you would convert the galvanometer into an ammeter with full-scale deflection of 1.5 A. (ii) Calculate the reading of the ammeter in (b) (i) and the percentage error of the reading when the ammeter is used to measure the current in a 2.0 : resistor that is connected in series with a cell of e.m.f. 3.0 V and negligible internal resistance. (iii) If the galvanometer is converted to an ammeter with full-scale deflection 3.0 A, and again used to measure the current in (b) (ii), explain whether the reading would be more accurate. 37. (a) Define the ohm, and hence express the ohm in terms of the base S.I. units. (b) A car headlamp is marked 12 V, 72 W. Calculate (i) its resistance, (ii) the energy dissipated in 20 minutes. (c) Two of the headlamps each marked 12 V, 72 W, are connected into the circuit shown in the figure below, in which one source of e.m.f. is the generator of the car. The generator is connected in parallel with the car battery and the two headlamps. Both lamps are working normally. Switch 12 V Battery 15.0 V Generator R Physics Term 2 STPM Chapter 15 Direct Current Circuits
15 147 1 1. B 2. B 3. B 4. —ε V = R r ——– R hence, V = ( R ——– R r )պ 5. A 6. (a) Cell in open circuit, I = 0 (b) Cell is charged, charging voltage > e.m.f. of cell 7. (a) 0.625 W (b) R = r = 0.5 : 8. (a) I = E ——– R + r (b) P = E2 R ———– (R + r)2 P = I 2 R (c) R = r 9. 9.0 I = E ——– R + r, P = I 2 R 10. (a) Refer page 103 (b) (i) V = E – Ir (ii) r = –gradient of graph = 4.54 : (iii) 4.00 : (iv) 1.21 : Power is maximum when R = r 2 1. C 2. C 3. D 4. A: —1 R = —1 9 + 1 9 + 9 R ۙ 6.0 V 3 1 0 V 6 1 Y X Z VXZ = 6.0 V, VXZ = VY – VX = 2.0 V 5. (a) Refer page 108 (b) (i) Ia = 17 —– 33 A = 0.52 A Ib = 4 —– 11 A = 0.36 A (ii) Ic = Id = 0.25 A 6. (a) R Љ 7.2 = 0.3(1.5 + 5.0) + 0.40R (b) I = 0.10 A (c) 6.44 V 3 1. C 2. D 3. B 4. C 5. B 6. B 7. D 8. B 9. A 10. B 11. D 12. A 4 1. D 2. (a) 0 (b) 2.0 V (c) 2.0 V (d) 0 (e) 1.25 V 3. —V 2 , 10 —11V 5 1. B 2. A 3. C 4. A: Voltmeter is connected in parallel with the resistor. 5. D 6. (a) (R1 + R2 ) = 50 : 1 mA u (R1 + R2 ) = 1 mA u 50 (b) R2 = 5 : 19 mA u R2 = 1 mA u (50 + R1 ) 6 1. C 2. C 3. B 4. C 5. B The e.m.f. of the battery is 12.0 V and its internal resistance is negligible. The generator has an e.m.f. of 15.0 V and negligible internal resistance. The generator is in series with a variable resistor R. The value of R is adjusted so that there is no current from the battery when the lamps are on. Calculate (i) the current in the generator, (ii) the value of the resistance of R. The lamps are then switch off, the value of R remains unchanged. (iii) Calculate the current in the battery. (iv) Suggest two advantages which the circuit shown has over a single power source. ANSWERS 15 Physics Term 2 STPM Chapter 15 Direct Current Circuits
15 148 6. (a) 0.50 V I1 = E ——–– R + 10 (b) 7.50 : I2 = E —– 10 (c) 25.0 cm I2 = E —– 10 (d) 25.0 cm I1 = E ——–– R + 10 7 1. D 2. C 3. C 4. ffiffiۙLQVHULHVZLWKVOLGHZLUH 5. 0.252 V 1.018 = I(10.0 + 726 ——100 × 5.00) EQ = I( 67.4 ——100 × 5.00) 8 1. D 2. (a) S1 at X : 55.0 cm Potentiometer measures e.m.f. E2 S1 at Y : 55.0 cm (b) S1 at X : 27.5 cm VR1 measured S1 at Y : 41.25 cm VR1 + R2 measured 9 1. C 2. 'ffl(ͿHFWLYHUHVLVWDQFHRI: and R is ( 20 10 )(15 :) = 30 : 1 30 = 1 60 + 1 R hence, R = 60 : 3. Zero STPM PRACTICE 15 1. A: 60 W = 2402 —–—R , P’ = 1102 —–—R = 12.6 W 2. D: Power = IV = (0.5 A)(2.5 V) = 1.25 W 3. A: Current in 3.0 :, I = (3.2 – 1.4) A = 1.8 A V = IR = (1.8 A)( 3.0 :) = 5.4 V 4. A: 20 + R 20 = 63.0 35.0, R = 16 : 5. B 6. D 7. C 8. C: 1 R = 1 1 + 1 2 = 3 2 R = 2 3 Љ 3.0 = I12 + 2 3 2, I = 1.125 A Also, 3.0 V = 2I + I1 = 2I + 2I2 , solving, I1 = 0.750 A 9. D: 8.0 : v 4.0 V, therefore 6.0 : v 3.0 V, VX = 12.0 V – 3.0 V = 9.0 V, and (9.0 – 4.0) V = 5.0 V v 10.0 : 10. A: I + 1.5 A = (0.5 + 2.8 – 0.8) A I = 1.0 A 11. B: V E = R R + r , V = 1 10.0 10.0 + 2.0 2(6.0 V) = 5.0 V Power lost = I 2 r = 1 5.0 10.0 2 2 (2.0) W = 0.50 W 12. B: (5.0 – 0.020)(R1 + R2 ) = (0.020)(500), R1 + R2 = 2.008 : (15.0 – 0.020)(R1 ) = (0.020)(500+ R2 ) Solving, R1 = 0.669 :, R2 = 1.34 : 13. A: Equivalet circuits are shown below X Y R1 ȍ R2 R3 R4 X Y R1 ȍ R34 ȍ R2 = R3 = R4 ȍ R2 1 —— RXY = 1 —— 2.0 + 1 ————– 1.0 + 1.0 RXY = 1.0 : 14 &ffl8VH.LUFKRͿ·VVHFRQGODZ (6.0 – 1.5)V = I(9.0 + 3.0 + 3.0) I = 0.30 A VPQ = 6.0 – (0.30)(9.0) V = 3.3 V 15. A: 3.00 V = I(1200) 12.0 V = I (1200 + R) R = 3600 : 16. A: E = SRWHQWLDOGLͿHUHQFHDFURVV: = 1 ———— 4.0 8.0 + 4.0 2(2.0 V) = 0.67 V 17. B 18. A 19. B 20. A 21. B 22. B: Current, I = 6.0 —— 2.0 A = 3.0 A :KHQFXUUHQWÁRZVIURP$WR%VA VB. Potential drops across 2.0 :, 0.5 : and 1.0 : resistor = (3.0)(2.0) V, 1.5 V and 3.0 V respectively Total potential drop = (6.0 + 1.5 + 3.0) V = 10.5 V Cell raise the potential by 1.5 V VA – VB = resultant drop in V = (10.5 – 1.5) V = 9.0 V VB = VA – 9.0 V = (9.0 – 9.0) V = 0 V V / V VA = ȍ 1.5 V ȍ ȍ A 0 3.0 6.0 9.0 B Physics Term 2 STPM Chapter 15 Direct Current Circuits
15 149 23. B 24. B 25. C 26. No current in 5.0 : resistor when capacitor is fully charged. 1 R = 1 4.0 + 1 3.5 + 2.5 I = V R = 2.5 A 27. (a) I(360 kV) = (120 A)(2.0 kV) I = 0.667 A (b) Power lost km–1 = I2 R = (0.667 A)2 Љ = 2.22 W 28. D L .LUFKKRͿ·VODZV UHIHUWRWH[W (ii) First law based on principle of conservation of charge. Second law based on principle of conservation of energy. (b) (i) At B: I 2 = I 1 + I 3 Loop ABEF, (3.0 + 4.0) = (2.0 + 3.0)I1 + 2.0 I2 7 = 5 I1 + 2(I1 + I 3 ) = 7 I1 + 2I 3 …………… Loop CBED, (6.0 + 4.0) = (6.0 + 4.0)I 3 + 2.0 I2 10 = 10 I3 + 2(I 1 + I 3 ) = 2 I1 + 12I 3 ………… Solving, I 1 = 0.80 A, I 3 = 0.70 A, I 2 = 1.50 A (ii) VBE = E – I2 r = 4.0 – (1.5)(2.0) V = 1.0 V (iii) P = I3 2 (6.0 :) = (0.70)2 (6.0) W = 2.94 W 29. (a) 2.33 V (b) 1.75 V 30. 'V = (3.0 – 2.0) V % error = 33.3% 'V—–V u 100% 31. (a) (i) —V ε = R ——–– R + r (ii) If R o f, V | ε V = ( R ——–– R + r )ε E (ͿHFWLYHHPI ε (ͿHFWLYHLQWHUQDOUHVLVWDQFH r —2 3RWHQWLDOGLͿHUHQFHDFURVVR, V = ( ———R R + r —2 ) ε = 2Rε ——— 2R + r 32. 1.46 0 V I0 I / A x x x x x r = 0.9 : Using E = V + Ir, V = E – Ir r = –gradient E = 1.46 V 33. (a) r = 36 : E —– V = 1 + r —– R , R = (50 + X) (b) 250 : 34. (a) Refer page 102 (b) Using V = IR, E = I(R + r) (c) V / V 1.5 1.0 0.73 0 15 45 T /s (iv) r = 1.0 :, use E —– V = 1 + V —– R at T = 15 s E = 1.5 V, V = 1.0 V r = 0.74 :, use E —– V = 1 + r —– R at T = 45 s E = 1.0 V, V = 0.73 V 35. (a) Ammeter is connected in series. Low resistance so as not to increases resistance. Voltmeter is connected in parallel. High resistance so that current in voltmeter is negligible. (b) Circuit P is better. R is small, and resistance of voltmeter is high. Current in voltmeter is negligible. In circuit Q: voltmeter measures V across R and ammeter. Hence value of resistance = R + resistance of ammeter. (c) (i) E = (———–— 2.0 5.0 + 2.0 )(0.875 × 5.0) V = 1.25 V —– E V = ——– 87.5 78.2 = ———– 4.0 + r 4.0 r = 0.48 : 36. (a) (i) Refer page 103 (ii) Ohmic conductor obeys Ohm's law. Nonohmic: does not (iii) (b) (i) V I 0 R = 0.0836 : (ii) I· $ 'I = (1.500 – 1.440) A % error = 4.0% (iii) R = 0.0417 : I· = 1.469 A, 'I = 0.031 A % error = 2.1% More accurate 37. (a) Ohm (:): resistance of a conductor if when a SRWHQWLDOGLͿHUHQFHRI9LVDSSOLHGDFURVVLW the current in it is 1A. : —V A JC-1 A (kgm2 s-2)(As)-2 A kgm2 s-4A-3 (b) (i) 2.00 : P = IV, R = —V I (ii) 8.64 × 104 J Energy = Pt (c) (i) 12.0 A (ii) 0.25 : V = E – IR (iii) 12 A LY 9ROWDJHDFURVVODPSVFRQVWDQWDW9 %DWWHU\LVFKDUJHGE\WKHJHQHUDWRU when the e.m.f. < 12 V. R Physics Term 2 STPM Chapter 15 Direct Current Circuits
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 150 CHAPTER MAGNETIC FIELDS 16 0DJQHWLF)LHOGV 0DJQHWLF)LHOG GXHWR &XUUHQWV Straight wire, B = P0I 2Sd Circular coil, B = P0NI 2r Solenoid, B = P0nI 'HWHUPLQDWLRQRIWKHUDWLR +DOO(IIHFW H P Mahflhgl^qi^kbf^gm )RUFHRQD0RYLQJ&KDUJH ) = q Y u % F = qvB sin θ )RUFHEHWZHHQ7ZR&XUUHQW &DUU\LQJ&RQGXFWRUV F = P0I1I2l 2Sd Concept Map Bilingual Keywords *PYJ\SHYSVVW!.LNLS\UNI\SH[ -SLTPUN»ZSLM[OHUKY\SL!7LYH[\YHU[HUNHURPYP-SLTPUN /HSSLMMLJ[!2LZHU/HSS 3VUNZVSLUVPK!:VSLUVPKWHUQHUN 3VUNZ[YHPNO[^PYL!+H^HPWHUQHUN`HUNS\Y\Z 4HNUL[PJÄLSKZ[YLUN[O!2LR\H[HUTLKHUTHNUL[ 4HNUL[PJÄLSK!4LKHUTHNUL[ 9HKPHSTHNUL[PJÄLSK!4LKHUTHNUL[QLQHYPHU :WLJPÄJJOHYNL!*HZZWLZPÄR 150
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 151 INTRODUCTION 1. Magnets are able to attract certain materials such as iron, steel, nickel and cobalt. 2. Magnets are used as pointers of compasses which point north. This is because the Earth has a magnetic fi eld. 3. Magnetic fi elds are not only produced by magnets but also by current fl owing in conductors. 16.1 Concept of a Magnetic Field Students should be able to: H[SODLQPDJQHWLFÀHOGDVDÀHOGRIIRUFHSURGXFHGE\FXUUHQWFDUU\LQJFRQGXFWRUVRUE\SHUPDQHQWPDJQHWV Learning Outcome 1. A magnetic fi eld is a fi eld of force produced by a magnet or by current-carrying conductors. Figure 16.1 The N-pole of a magnet points north Figure 16.2 Magnetic fi eld around a bar magnet Thread S N N Magnet S N 2. When a bar magnet is freely suspended, it always comes to rest along the north-south direction. The pole of the magnet which points north is known as the north-seeking pole or N-pole. The other pole of the magnet is the S-pole (Figure 16.1). 3. A magnetic fi eld is represented graphically by lines of force. Figure 16.2 shows the magnetic fi eld around a bar magnet. 4. The magnetic fi eld lines start from the N-pole and end at the S-pole. 5. At the poles, the magnetic fi eld lines are close together, showing that the magnetic fi eld is stronger at the poles. Figure 16.3 Magnetic fi eld of the Earth N S Horizontal component Vertical component 2015/P2/Q10
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 152 6. Figure 16.3 shows the magnetic fi eld of the Earth. The direction of the fi eld is from the south pole of the Earth to the north south. It is as if there is a bar magnet in the Earth. 7. At the equator, the magnetic fi eld is almost horizontal. 8. At other points, for example P, the direction of the Earth’s magnetic fi eld is not horizontal but at an angle to it. A bar magnet freely suspended at P dips downwards (Figure 16.4). S Vertical component Horizontal component P Earth's magnetic field N θ Figure 16.4 Bar magnet dips downwards 9. Hence, the Earth’s magnetic fi eld has a vertical component and a horizontal component. Quick Check 1 1. (a) What is represented by a line of force in a magnetic fi eld? (b) What is represented by the direction of the line of force? 2. State the difference in the magnetic fi eld pattern between a region where the fi eld is strong and another region where the fi eld is weak. 3. Magnetic fi elds exist around all of the following except A the Earth B a bar magnet C a current carrying wire D a rheostat 16.2 Force on a Moving Charge Students should be able to: XVHWKHIRUPXODIRUWKHIRUFHRQDPRYLQJFKDUJHF = qv u B XVHWKHHTXDWLRQF = qvBVLQθWRGHÀQHPDJQHWLFÁX[GHQVLW\B GHVFULEHWKHPRWLRQRIDFKDUJHGSDUWLFOHSDUDOOHODQGSHUSHQGLFXODUWRDXQLIRUPPDJQHWLFÀHOG Learning Outcomes 1. When a charge +q moves with a velocity v in a magnetic fi eld of fl ux density B, a force F acts on the charge. The force F is given by the vector product below. F = q(v u B) The magnitude of the force is F = qvB sin θ where θ is the angle between the vectors v and B. 2. The direction of the force F is perpendicular to the plane containing the velocity v and the magnetic fl ux density B and obtained using the right-hand rule for vector product (Figure 16.5(a)) or Fleming’s left-hand rule (Figure 16.5(c)). 2010/P1/Q31, 2012/P2/Q12, 2014/P2/ Q11,Q19, 2015/P2/Q9, 2016/P2/Q12
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 153 (a) Right-hand rule (b) (c) Fleming’s left-hand rule ( v x B ) v x B B e +q v F B v Figure 16.5 3. The right hand rule for vector product states that if the fingers of the right hand curve from the direction of the first vector, i.e. v to the second vector, i.e. B, then the direction of the product (v u B) is shown by the thumb. 4. Fleming’s left-hand rule is shown in Figure 16.5(c). If the thumb, first and second finger of the left hand is extended so as to be mutually perpendicular to each other with the first finger pointing and in the direction of the magnetic field B, and the second finger in the direction of the velocity v, then the direction of the force F is shown by the thumb. 5. If the moving particle is negatively charged then the direction of the force is opposite to that obtained by the rules mentioned above. 6. When the charged particle moves in a direction perpendicular to the magnetic field, then the angle θ between v and B is 90°. From F = qvB sin θ, θ = 90° F = qvB 7. The expression F = qvB is used to define the magnetic flux density B of a magnetic field. The magnetic flux density B of a magnetic field is the force on a unit charge (q = 1 C) moving with unit velocity (v = 1 m s–1) at right angles (θ = 90°) to the field. 8. The unit for magnetic flux density B is tesla (T). The quantity of magnetic flux through an area is known as the magnetic fluxI (pronounced as phi) and is measured in webers (Wb). 9. Since the magnetic force F = qvB acting on a particle of charge q moving with a velocity v at rightangles to the magnetic field B is perpendicular to the velocity v, the subsequent motion of the charged particle is a circle (Figure 16.6). (a) (b) Figure 16.6 r F Uniform magnetic field B v q Positive charge + F Negative charge – v 10. In Figure 16.6, the direction of the magnetic field is into the plane of the paper, shown by the ‘’ sign. The direction of F on the negative charged particle is opposite to that of the positive charge.
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 154 11. Since the motion of the charge is circular, the centripetal force is provided by the force F = qvB. Hence, mv2 ——r = qvB r = radius of circle or mrω2 = qvB m = mass of charged particle mrω2 = q(rω)B v = rω Angular velocity, ω = qB —– m Period of circular motion, T = —– 2S ω T = ——– 2Sm qB which is independent of the velocity v of the charged particle. 12. The same particle with a velocity of v 2 at right-angles to B would move in a circle of radius r 2 , and take the same time T to complete a revolution. B Stationary force F = D v q q q v Figure 16.7 13. From the equation F = qvB sin θ, there is no force (F = 0) if v = 0, that is the charged particle is stationary in the magnetic fi eld. θ = 0° or 180°, that is the charged particle moves in a direction parallel to the magnetic fi eld B (Figure 16.7). The particle continues its motion in the fi eld in a straight line with constant velocity. Example 1 The path of an electron in a vacuum under the influence of a magnetic field of flux density 1.0 u 10–2 T is a circle of radius 1.0 u 10–2 m. The specifi c charge, —e m of an electron is –1.76 u 1011 C kg–1. Calculate (a) the period of orbit, (b) its period if the kinetic energy of the electron is halved. Solution: (a) Using F = qBv = mrω2 , (v = r) qB (rω) = mr2 ω = qB —– m (q = e) = —e m B Info Physics The Northern Lights (or Aurora Borealls) 6HH YLGHR DW YLGHRQDWLRQDOJHRJUDSKLFFRP YLGHR«QRUZD\DXURUDERUHDOLVYLQ 0XOWLFRORXUHGFXUWDLQVRIOLJKWÀOOWKHVNLHVZKHQ ODUJHQXPEHUVHOHFWULFDOO\FKDUJHGSDUWLFOHVIURP WKH6XQVWUHDPDWKLJKVSHHGWRZDUGVWKH(DUWK DQGDUHGHÁHFWHGE\WKH(DUWKҋVPDJQHWLFÀHOG 7KHVHSDUWLFOHVFROOLGHZLWKWKHDLUSDUWLFOHV7KH DLUWKHQOLJKWVXSUDWKHUOLNHZKDWKDSSHQVLQD ÁXRUHVFHQWOLJKWWXEH
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 155 Period T = —– 2S ω = —– 2S B —m e = ————–– 2S (1.0 u 10–2) 1 –———— 1.76 u 1011 = 3.57 u 10–9 s (b) From the equation, period T = —– 2S B —m e The period T is independent of the velocity. Hence, when the kinetic energy, — 1 2 mv 2 is halved, the period remains the same at 3.57 u 10–9 s. Quick Check 2 1. The diagram shows four electrons A, B, C and D moving with the same speed v in the directions shown in the magnetic fi eld of a bar magnet. The magnitude of the magnetic force on which one of these electrons is the greatest? v v v v B A C D N S 2. An electron moving with a constant speed in a vacuum enters a uniform magnetic fi eld in a direction at right angles to the fi eld. Its subsequent path is A a straight line parallel to the fi eld. B a circle in a plane normal to the fi eld. C a parabola in plane parallel to the fi eld. D a helix with its axis parallel to the fi eld. 3. A particle of mass m, charge q enters a uniform magnetic fi eld of fl ux density B with speed v and describes a circular path of radius r. The radius of the circular path is directly proportional to A q C 1 –– v B 1 –– m D 1 ––B 4. The diagram below shows the circular tracks of two particles X and Y in a uniform magnetic fi eld which acts downwards into the paper. The radius of the track of X is half that of the track of Y. X Y Which of the following statements is certainly true? A The speed of Y is half that of X. B The mass of Y is half that of X. C The charge of Y is twice that of X. D Both X and Y carry a positive charge. 5. An oxygen ion (O2–) and a lithium ion (Li+) travel at the same speed perpendicular to a uniform magnetic fi eld. The ratio of the mass of oxygen ion to lithium ion is 16 : 7. What is the ratio of radius of oxygen ion’s orbit to lithium ion’s orbit? A 16 : 7 C 7 : 8 B 8 : 7 D 7 : 32 6. In a cyclotron, a beam of protons with speed 1.5 u 107 m s–1 move in a circular path of diameter 120 m inside a uniform magnetic fi eld. Calculate the magnetic fl ux density of the fi eld. Draw a sketch to show the direction of the magnetic fi eld relative to the circular path. (Mass of proton = 1.67 u 10–27 kg)
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 156 16.3 Force on a Current-Carrying Conductor Students should be able to: H[SODLQWKHH[LVWHQFHRIPDJQHWLFIRUFHRQDVWUDLJKWFXUUHQWFDUU\LQJFRQGXFWRUSODFHGLQDXQLIRUPPDJQHWLF ÀHOG GHULYHDQGXVHWKHHTXDWLRQF = IlBVLQθ Learning Outcomes 1. Figure 16.8 shows a metal rod of length l carrying a current I in a uniform magnetic fi eld of fl ux density B. Figure 16.8 l I I v Fe v Fe F Magnetic field B v Fe – – – 2. The free electrons drift with a velocity v in the direction opposite to the fl ow of current I. The force on each electron due to magnetic fi eld is Fe = Bev 3. The number of free electrons in the metal rod = (nAl) where n = number of free electrons per unit volume A = cross-sectional area of metal rod Hence, total force on the free electrons in the metal rod F = (nAl)Fe = (nAl)(Bev) = (nAve)(lB) (nAve = I) = BIl in the direction shown in Figure 16.8. Therefore, the force on a conductor of length l carrying a current in a magnetic fi eld B which is perpendicular to the conductor is F = BIl 4. In general, the force F on a conductor of length l carrying a current I in a magnetic fi eld of fl ux density B is given by the vector equation F = I (l u B) where the direction of the vector l is in the direction of the current I. The magnitude of the force F = BIl sin θ where θ = angle between the vector l and the magnetic fi eld of fl ux density B (Figure 16.9) (a) Right-hand rule for vector product (b) Fleming’s left-hand rule Figure 16.9 e B F l ( l x B ) F B l 2013/P2/Q10, 2015/P2/Q10, 2016/P2/Q10
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 157 5. The direction of the force F can be determined using the right-hand rule for vector product (Figure 16.9(a)) or Fleming’s left-hand rule (Figure 16.9(b)). 6. The equation F = BIl sin θ can be used to deduce a defi nition for B, the magnetic fl ux density. When θ = 90°, F = BIl B = —F Il The magnetic fl ux density B of a magnetic fi eld is the force on a conductor of unit length (l = 1 m) carrying a unit current (I = 1 A) at right-angles to the fi eld. Example 2 A copper wire is stretched between two fi xed points P and Q and carries an alternating current of frequency f. Describe and explain what will happen if a magnet is arranged to apply a strong magnetic fi eld perpendicular to the central portion of the wire, as shown in the diagram. Solution: I P F B Magnetic field Q I P F B Magnetic field Q Figure (a) Figure (b) When the current I is from P to Q, a force F acts downwards on the wire (Figure (a)). When the direction of the current I is reversed, from Q to P, the force F of the wire is upwards. Since the ends P and Q of the wire are fi xed, the wire vibrates between the poles of the magnet. Example 3 The diagram shows a rectangular coil carrying a current hanging from a light helical spring. The lower section XY of the coil is at right angles to a uniform horizontal magnetic fi eld. There is no magnetic fi eld for the other sections of the coil. The coil has 100 turns and a mass of 10 g. The length of XY is 5.0 cm and the current is 1.0 mA. When the current is switched off, the coil oscillates with a period of 1.0 s. Finally it comes to rest 1.0 mm above its initial position. Calculate the fl ux density of the magnetic fi eld. Solution: Force on the section XY which has 100 turns and carries a currect I = 1.0 mA in a magnetic fi eld of fl ux density B is F = 100(BIl) This force stretches the spring by a further extension x = 1.0 mm Using Hooke’s law, F = kx 100(BIl) = kx ............................. { Q N S P I I I X Spring Y
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 158 When the current is switched off, the coil oscillates with period T = 1.0 s T 2 = 4S——– 2 m k k = 100( ———– BIl) x = 4S ——–—2 mx 100(BIl) Flux density, B = 4S ——–—– 2 mx 100T 2 Il = 4S2 u (10 u 10–3)(1 u 10–3 ———————–——–—–––– ) 100 u (1.0)2 u (10–3)(5 u 10–2) = 7.9 u 10–2 T Quick Check 3 1. The diagram shows a wire frame in the form of a triangle XYZ carrying a current of 2.0 A inside a uniform magnetic fi eld of fl ux density 5.0 T. 8 cm 6 cm 2.0 A 10 cm Y Z X B = 5.0 T Calculate the magnitude and direction of the force on the sides (a) XY, (b) YZ, (c) XZ, and show that the resultant force on the wire frame is zero. 2. In the diagram shown, the metal rod PQ is of length 40 cm and mass 50 g. The rod is free to slide on metal strips CD and EF. C P D F Q E I I Cell 37° The direction of the current I is as shown. A uniform magnetic fi eld of fl ux density 0.20 T acts vertically downwards. (a) Calculate the value of the current I for the rod to remain stationary. (b) If I = 2.50 A, calculate the acceleration of the rod. 3. (a) Defi ne magnetic fl ux density. (b) Write an expression for the force F on an electron moving with a velocity v at an angle of θ to a uniform magnetic fi eld of fl ux density B. Give the unit for each of the quantities in the expression. Draw a sketch to show the directions of the vector quantities involved. (c) A wire XY of weight W can slide freely on two metal rods inclined at an angle θ to the horizontal, and separated by a distance d. A vertical uniform magnetic fi eld acts in the region shown. d Q Q X Y Metal rod Magnetic field Derive an expression, in terms of W, θ, B, and d, for the current I in the wire XY which will prevent the wire from sliding. What is the direction of the current?
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 159 16.4 Magnetic Fields due to Currents Students should be able to: VWDWH$PSHUHҋVODZDQGXVHLWWRGHULYHWKHPDJQHWLFÀHOGRIDVWUDLJKWZLUHB = μ0 I 2Sr XVHWKHIRUPXODHB = μ0 NI 2r IRUDFLUFXODUFRLODQGB = μ0 nIIRUDVROHQRLG Learning Outcomes 1. When a current fl ows in a conductor, a magnetic fi eld is produced. The magnetic fi eld pattern around the conductor depends on the shape of the conductor. 2. Figure 16.10 shows the magnetic fi eld pattern produced by the current in a long straight wire. On the plane of the cardboard, the magnetic lines of force form concentric circles along the wire. In fact these circles are the cross sections of concentric cylinders with the wire as axis. 3. The direction of the magnetic fi eld can be determined experimentally by placing a plotting compass close to the wire. 4. The right-hand grip rule gives the direction of the magnetic fi eld relative to the direction of the current. If the wire is griped with the right hand, and the thumb pointing in the direction of the current, then the fi ngers curve in the direction of the magnetic fi eld (Figure 16.10). 5. The magnitude of the magnetic fl ux density B at a perpendicular distance r from the wire is given by B = μ0 I ——2Sr where μ0 = 4Su 10–7 H m–1, the permeability of free space. Ampere’s Law 1. The expression for the magnetic fi eld due to a current I in a long straight wire, B = μ0 I 2Sr , can be derived using Ampere’s law. 2. Ampere’s law states that for any closed path, the sum of the length elements l times B//, the magnetic fi eld parallel to the length elements equals μ0 I where I is the current enclosed by the closed path. 6(B// 'l) = μ0 I μ0 = 4 u 10–7 H m–1 is the permeability of free space. 3. Ampere’s law is similar to Gauss’s law for electric fi eld. Gauss’s law: 6Q = ε0 Φ where 6Q is the charge enclosed by a closed surface, and ) is the electric fl ux. Figure 16.10 Thumb points in the direction of the current Right hand Direction of field Current Magnetic field pattern Cardboard 2014/P2/Q9, 2015/P2/Q12, 2017/P2/Q19, 2018/P2/Q10
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 160 4. Below are the steps taken to apply Ampere’s law to derive the expression for the magnetic fl ux density B at a distance r from a long straight current-carrying wire. s 2EFER TO THE MAGNETIC l ELD PATTERN SHOWN IN &IGURE TO CHOOSE A SUITABLE CLOSED PATH Closed path I I r B B B B Figure 16.11 s 3INCE THE MAGNETIC l ELD CONSISTS OF CONCENTRIC CIRCLES THE MOST SUITABLE CLOSED PATH IS A CIRCLE OF radius r. The component of the magnetic fi eld parallel to the path B// is B which is of constant magnitude along the path and is always in the same direction as the line element, l. s !PPLYING THE EQUATION 6(B// 'l) = μ0 I, 6('l)(l) = 2πr and the current enclosed by the closed path is I. Therefore, B//(2πr) = μ0 I B = μ0 I 2Sr Example 4 Two parallel long straight rigid wires X and Y carry the same current of 20 A in opposite directions as shown in the diagram. The wires are 10 mm apart. Find the magnitude and direction of the magnetic fl ux density (a) at the point P distance 5.0 mm from wire X, (b) at the point Q which is equidistant from wire X and wire Y. Solution: (a) Figure (a) shows the view from the top of the magnetic fi elds at the point P due to the current in wire X (BX) and the current in wire Y (BY). BX = μ0 I ——2Sr1 (r1 = PX = 5.0 mm) BY = μ0 I ——2Sr2 (r2 = PY = 15.0 mm = 3r1 ) = μ0 I ——— 2S(3r1 ) 10 mm 20 A P Q Wire X Wire Y 20 A BY BX X P Y Figure (a)
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 161 Since BX > BY , the resultant magnetic fl ux density at P is BP = BX – BY = μ0 I ——2Sr2 – μ0 I ——— 2S(3r1 ) = — 2 3 μ0 I ——2Sr1 = —2 3 (4S u 10–7)(20) ——————— 2S u (5.0 u 10–3) = 5.33 u 10–4 T (b) At the point Q the magnetic fi elds due to the current in wires X and Y are in the same direction and of the same magnitude. BX = BY = μ0 I ——2Sr1 (r1 = 5.0 mm) Hence the resultant magnetic fl ux density at Q is BQ = BX + BY = μ0 I ——2Sr1 + μ0 I ——2Sr1 = (4S u 10–7)(20) ——————– S(5.0 u 10–3) = 1.60 u 10–3 T Example 5 A magnet AB is freely pivoted at its centre of gravity on a non-conducting rod CD, supported midway between two long straight conducting wires EF and GH as shown below. R S H E A S N D F S E N W C G B The wires are parallel, in the same horizontal plane, and lie in the north-south direction. With the switch S open, the magnetic fi eld of the Earth causes the magnet to dip as shown. The switch is then closed, and the variable resistor R is adjusted so that the magnetic fi eld due to the current I causes the magnet to be horizontal. This happens when the fi eld due to the current in the wires neutralises the vertical componen of the Earth’s magnetic fi eld. (a) Calculate the vertical component of the Earth’s magnetic fl ux density if I = 3.0 A, and EH = FG = 0.50 m. (b) Discuss what would happen to the pivoted magnet if the current I were increased to a very large value (i) in the same direction as described above, (ii) in a direction opposite to that shown. BX + BY X Y Q Figure (b)
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 162 Solution: (a) Since the currents in the wires EF and HG are in opposite directions, the magnetic fl ux density midway between the wire due to the currents in both wires is in the same direction and of magnitude B = μ0 I —–– 2Sr + μ0 I —–– 2Sr (r = 1 ––2 EH) = μ0 I —– Sr when the magnet is horizontal. Vertical component of = resultant fl ux density midway Earth’s magnetic fi eld between EF and HG = μ0 I —– Sr = (4S u 10–7)(3.0) ———————– S(0.25) = 4.8 u 10–6 T (b) (i) The direction of the resultant magnetic fi eld due to the currents in the wires EF and HG is vertically upwards. When the current is increased, the resultant magnetic fi eld due to the current becomes greater than the vertical component of the Earth’s magnetic fi eld. Hence the N-pole (end B) of the magnet tilts upwards until the magnet becomes vertical. (ii) When the direction of the current is reversed, the resultant magnetic fl ux density midway between the wires EF and HG is vertically downwards. Hence, when the current is increased, the magnet would fi nally be vertical with the N-pole (end B) at the bottom. Circular Coil 1. Figure 16.12 shows the magnetic fi eld due to a current in a plane circular coil. The direction of the magnetic fi eld can be determined using the right-hand grip rule. Figure 16.12 2. The magnitude of the magnetic fl ux density at the centre of the coil is B = μ0 NI ——– 2r where N = number of turns r = radius of the coil
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 163 Example 6 Two circular coils P and Q are concentric and placed of a fl at surface. Coil P has 10 turns of radius 4.0 cm and carries a current of 1.0 A. Coil Q has 20 turns of radius 12.0 cm and the magnitude and direction of the current it carries are adjusted so that the resultant magnetic fi eld at the common centre is zero. What is the direction and magnitude of the current in coil Q? Solution: Since the resultant magnetic fi eld at the common centre is zero, the direction of the fi eld due to current in the coils P and Q must be of the same magnitude but in opposite directions. The direction of the current I in coil Q must be opposite to that in P. Using B = μ0 NI ——– 2r , BQ = BP μ0 (20)I ———– 2(0.12) = μ0 (10)(1.0) ————–– 2(0.04) I = 1.5 A Long Solenoid 1. When a current fl ows in a long solenoid, the magnetic fi eld produced is identical to that of a bar magnet as shown in Figure 16.13. 2. The pole at each end of the solenoid can be determined by observing the direction of the current at each of the solenoid. For the end A in Figure 16.13, the direction of the current is anticlockwise. Hence the pole at A is an N-pole. 3. Whereas an S-pole is obtained at the end B of the solenoid since the direction of the current is clockwise. 4. The magnetic fi eld at the centre of the coil is uniform and of fl ux density, B = μ0 nI, where n = number of turns per unit length of solenoid. 5. At the centre of each end of the solenoid, the fl ux density, B = — 1 2 μ0 nI, which is half that at the centre of the solenoid. P OQ B′ B′ B′ B′ P O Q I B′ B′ + B′ = 2B′ (a) (b) B′ Figure 16.14 P Q Magnetic flux density B′ B = 2B′ (c) n1 = 10 I1 = 1.0 A n2 = 20 I P Q Figure 16.13 Current A B Solenoid
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 164 6. Figure 16.14(a) shows that a long solenoid may be considered to be made up of two identical solenoids joined in series. Suppose that the magnetic fl ux density at each end of the two solenoids is B’. At the centre O, the directions of the magnetic fi elds due to current in the two solenoids are the same. Hence, the resultant fl ux density at the centre O of the long solenoid, B = B’ + B’ = 2B’ or B’ = —1 2 B Flux density at the ends = —1 2 (fl ux density at the centre of the solenoid) 7. Figure 16.14(c) shows the variation of the magnetic fl ux density along the axis of the solenoid from the end P to the end Q. 8. If the solenoid is wound on a soft iron rod, then the magnetic fl ux density at the centre of the solenoid is given by B = μr μ0 nI where μr is the relative permeability of soft iron. 9. The value of μr for soft iron is 370. This implies that the strength of the magnetic fi eld increases multifold with the soft iron as core of the solenoid. 10. There are alloys of iron, nickel and cobalt whose value of μr exceeds 10 000. On the other hand the value of μr for copper is 0.99999. Example 7 (a) Calculate the value of the magnetic fl ux density B and the magnetic fl ux Φ in an air-core solenoid of diameter 2.50 cm and 2 000 turns m–1 length when the current in the solenoid is 0.10 A. (b) How does the value of B and Φ change when a material of relative permeability 5 000 is used as the core of the solenoid? Solution: (a) Using B = μ0 nI = (4S u 10–7)(2 000)(0.10) = 2.51 u 10–4 T Magnetic fl ux, Φ = fl ux density area normal to B = B Sd2 —–4 = (2.51 u 10–4) S u (2.50 u 10–2)2 ———————— 4 = 1.23 u 10–7 Wb (b) When μr = 5 000, B’ = μr B = (5 000) (2.51 u 10–4) = 1.26 T Φ’ = μrΦ = (5 000) (1.23 u 10–7) = 6.15 u 10–4 Wb
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 165 Quick Check 4 1. An overhead power cable carries an alternating current of peak value 200 A. At what distance would the peak magnetic fl ux density due to the current be 10 T? A 2.8 m C 5.7 m B 4.0 m D 8.0 m 2. Two long parallel, long, straight, thin wires carry equal currents I in the same direction as shown below. P I I Q O R S Which of the following graphs shows correctly the variation of the magnetic fl ux density due to the two wires from P to S? A C B P Q O R S P Q O B R S B D P Q O B R S P Q O B R S 3. The fi gure shows a plane OXY with axes OX and OY at right-angles. Y P Q O X Which of the following currents in a straight conductor will produce a magnetic fi eld at O in the direction OX? A At P into the plane of the fi gure. B At P out of the plane of the fi gure. C At Q into the plane of the fi gure. D At Q out of the plane of the fi gure. 4. Four parallel current-carrying conductors, pass vertically through the four corners of a square WXYZ. In two of the conductors, the current is fl owing into the page, and in the other two out of the page. W X O B Y Z In which directions must the current fl ow to produce a resultant magnetic fi eld B in the direction shown at O, the centre of the square? Into the page Out of the page A W and X Y and Z B W and Y X and Z C X and Z W and Y D Y and Z W and X 5. Two fl at circular coils, X and Y, each with 200 turns are arranged coaxially as shown below. X Y The radius of X is 0.050 m and carries a current of 2.0 A. Y has a radius of 0.10 m, and carries a current of 4.0 A in the same direction. What is the magnitude of the resultant magnetic fl ux density at the centre of the coils? A 0 C 4 000μ0 B 2 000μ0 D 8 000μ0
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 166 6. On the axis of a long, uniform, currentcarrying solenoid the ratio flux density of the centre ———————–———— flux density at an end is A 1 : 1 C 1 : 2 B 2 : 1 D 1 : 4 7. A long, air-cored solenoid is of length l, has N turns, cross-sectional area A and carries current I. What is the magnetic flux through the cross section in this region? A μ0 NIA C μ0 NIA —–—– l B μ0 NI —–—A D μ0 NI —–—lA 8. A plotting compass is placed near one end of a solenoid. When there is no current in the solenoid, the compass needle points north as shown. X Y N W E S When a current flows in the solenoid from X to Y, the magnetic field produced is of the same magnitude as the magnitude of the Earth’s magnetic field. The compass then points to A North-east. C North-west. B East. D West. 9. (a) Sketch the pattern of the magnetic field around a long straight wire carrying a current. Show clearly the direction of the current and the magnetic field. (b) Show the change to the field pattern if a uniform magnetic field acts at right angles to the wire. 10. The figure shows a flat circular coil carrying a current I. B C A D I I (a) Sketch the magnetic field pattern in the plane ABCD. Show the direction of the field. (b) Explain why there is no force on a straight conductor carrying a current placed along the axis of the coil. 11. (a) The magnetic flux density at the end of a long solenoid is B. What is the flux density at the centre of the solenoid? Explain your answer. (b) State the factors that determine the magnitude of the magnetic flux density at the centre of a long, current-carrying solenoid. State the dependence of the magnetic flux density on each factor. 12. (a) An infinite long straight wire carries a current I. Use Ampere’s law to derive the expression for the magnetic flux density B at distance r from the wire. (b) Four long vertical wires are arranged parallel to each other and pass through holes at the vertices of a square of sides 0.10 m drawn on a piece of cardboard which is horizontal as shown in the diagram. Currents I1 = 15 A, I2 = 10 A, I3 = 25 A and I4 = 30 A flow in the wires in the directions shown in the diagram below. O 0.10 m 0.10 m I 1 I 2 I 4 I 3 (i) Copy the diagram and mark the directions of the magnetic field due to each of the currents at the centre O of the square. Hence deduce the direction of the resultant magnetic field at O. (ii) Determine the magnitude of the resultant magnetic field at O.
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 167 16.5 Force between Two Current-Carrying Conductors Students should be able to: GHULYHDQGXVHWKHIRUPXODF = μ0 I1 I2 l 2Sd IRUWKHIRUFHEHWZHHQWZRSDUDOOHOFXUUHQWFDUU\LQJFRQGXFWRUV Learning Outcome 1. Figure 16.15 shows two long, thin, straight, parallel wires P and Q each carrying current I1 and I2 respectively in the same direction in a vacuum. The wires are separated by a distance d. Figure 16.15 P d Q B1 F2 B1 F2 I2 I1 2. At a distance d from the wire P, the magnetic fl ux density due to the wire P is B1 = μ0 I ——1 2Sd Since the wire Q that carries current I2 is at a distance d from wire P, the force on wire Q is F2 = B1 (I2 l) in the direction shown. Force on wire Q is F2 = μ0 I ——1 2Sd I2 l = μ0 I1 I2 l ——–– 2Sd ............................. { 3. Similarly, the fl ux density at a distance d from wire Q due to the current I2 is B2 = μ0 I ——2 2Sd The force on wire P is F1 = B2 (I1 l) in the direction as shown in Figure 16.16. Force on wire P, F1 = μ0 I ——2 2Sd I2 l = μ0 I1 I2 l ——–– 2Sd ..............................| P d Q B2 B2 F1 F1 I1 I2 Figure 16.16 2014/P2/Q10, 2018/P2/Q16
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 168 4. Comparing equations { and |, the forces on wires P and Q of the same magnitude but in opposite directions. Hence the forces F1 and F2 form an action-reaction pair consistent with Newton’s third law of motion. 5. From F1 = F2 = μ0 I1 I2 l ——–– 2Sd , Force per unit length, F —1 l = F —2 l = μ0 I1 I ——–2 2Sd This expression is used in the defi nition of the ampere (1 A), one of the base SI units. 6. From F —1 l = F —2 l = μ0 I1 I ——–2 2Sd , when I1 = I2 = 1 A, d = 1 m then F —1 l = F —2 l = (4S u 10–7) u 1 u 1 ——————––— 2S u 1 = 2 u 10–7 N m–1 This gives the defi nition of the ampere (1 A). The ampere (1 A) is that constant current which if maintained in two straight parallel thin wires, of infi nite length and placed one metre apart in vacuum, would produce on each wire a force of 2 u 10–7 N m–1. 7. The force between the wires is attraction if the directions of the currents are the same (Figure 16.17(a)), and repulsion when the currents are in opposite directions (Figure 16.17(b)). Figure 16.17 (a) Attraction (b) Repulsion P Q B2 B1 F1 F2 I1 I2 B2 B1 F1 F2 I 1 I 2 Example 8 Two long, parallel vertical wires separated by a distance of 0.3 m, are placed east-west of one another. The current in the westerly wire is 20 A and the other is 30 A. Both currents fl ow upwards. If the horizontal component of the Earth’s magnetic fl ux density is 2.0 u 10–5 T, calculate the resultant force on one metre length of each wire. Solution: The magnetic fl ux density at a distance r = 0.30 m to the east of the wire carrying a current I1 = 20 A is B1 = μo I1 2πd = (4S u 10–7)(20) ——————— 2S u 0.30 = 1.33 u 10–5 T to the north 0.3 m B2 B1 BE F BE I1 = 20 A I2 = 30 A
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 169 The resultant magnetic fi eld 0.30 m from the wire, B = BE + B1 = (2.0 u 10–5) + (1.33 u 10–5) = 3.33 u 10–5 T Force per metre length on the wire carrying current I 2 = 30 A is — F l = (BI2 l) ——– l = (3.33 u 10–5) (30)N m–1 = 1.0 u 10–3 N m–1 towards the west Magnetic fl ux density at a distance r = 0.30 m to the west of the wire carrying current I2 = 30 A is B2 = μo I2 2πd = (4Su 10–7)(30) ——————— 2Su 0.30 = 2.0 u 10–5 T towards the south Since the horizontal component of the Earth’s magnet is towards the north, the resultant magnetic fl ux density at a distance r = 0.30 m to the west of the wire carrying current I2 = 30 A is B’ = BE – B2 = (2.0 u 10–5) – (2.0 u 10–5) T = 0 Hence force on a metre length of the westerly wire is zero. Example 9 Two straight, parallel horizontal conductors each of length 20.0 cm are arranged one on top of the other in air. The conductors are connected in series to a battery so that the currents in the conductors are in opposite directions. The lower conductor is fi xed, the upper conductor is free to move vertically along guides so that it is always parallel to the lower conductor. If the mass of the upper conductor is 1.20 g, estimate the value of the current in the conductors so that the conductors are separated at a fi xed distance of 0.75 cm. (Neglect the effect of the Earth’s magnetic fi eld). Solution: The magnetic fl ux density at a distance r = 0.75 cm above the lower conductor is B = μo I 2πd The force on the upper conductor carrying current I, F = BIl = μo I2 l 2πd Since the upper conductor is stationary, F = μo I2 l 2πd = mg (mg is weight of conductor) I 2 = (1.2 u 10–3)(9.81)(2Su 0.75 u 10–2) ———————————————– (4Su 10–7 ) u (20 u 10–2) I = 47 A F mg 0.75 cm 20 cm I I
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 170 Example 10 Derive a formula for the approximate force between two identical fl at coils, each having a big radius R and carrying current I. The coils are coaxial and separated by a small distance r. There are N turns in each coil. As an approximation, you may assume that the conductors are straight and parallel. What determines the direction of the force? Identify the symbols you use and give the unit for each quantity. Solution: Assuming that the conductors are straight, length of each conductors, l = N(2πR) N = number of turns The magnetic fl ux density at a distance r from a straight conductor carrying current I is B = μo I 2πr Magnetic fl ux density at a distance r from the coil having N turns, B = N μ0 I —–– 2Sr Hence the force between the conductors, F = BIl = μo I2 l 2πr (l = 2RN) = μo I2 RN r where μ0 = permeability of free space = 4π u 10–7 H m–1 I = current in A R = radius of each coil in m N = number of turns r = distance between the coils in m The direction of the force is determined by the directions of the currents relative to each other. (a) If the currents I are in the same direction, the force F is attraction (b) If the currents I are in opposite directions, the force F is repulsion. F I I F F I I Quick Check 5 1. Two long and parallel straight conductors are separated by a distance, d. When current fl ows in the two conductors, the force, F between the conductors is directly proportional to A d C — 1 d2 B — 1 d D d2 2. Two long parallel conductors are separated by 1 cm and free to move. A current of 20 A fl ows in the same direction in both conductors. The conductors A remain stationary. B move away from each other. C move towards each other. D rotate in the clockwise direction.
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 171 3. The currents in two long parallel wires X and Y are 3 A and 5 A respectively. The force per unit length on the wire X is 5 10–5 N m–1 towards the right as shown in the diagram. Force X Y The force per unit length on the wire Y is A 3 u 10–5 N m–1 to the right B 3 u 10–5 N m–1 to the left C 5 u 10–5 N m–1 to the right D 5 u 10–5 N m–1 to the left 4. Two parallel conductors carrying alternating sinusoidal current of the same frequency but with a phase difference of radians. Which of the following graphs best shows the variation of the force F between the conductors with time t? A C F 0 t + – F 0 t B D F 0 t + – F 0 t 5. A small coil lies inside a large coil. The two coils are horizontal, concentric and carry current in the same direction. View from above The small coil will experience A no resultant force. B an upward force along the axis. C a downward force along the axis. D a torque about the vertical axis. 6. Three long, parallel, straight wires X, Y, and Z are placed in the same plane in vacuum as shown in the diagram. 1 A 2 A 1 A 10 cm 10 cm X Y Z If the force per unit length between two long, parallel, straight wires placed 10 cm apart, each carrying a current of 1 A, is F, what is the resultant force per unit length acting on the wire X? A 0.5F B 1F C 1.5F D 2.5F 7. A long straight wire XY lies in the same PLANE AS A SQUARE LOOP OF WIRE 0123 WHICH IS FREE TO MOVE 4HE SIDES 03 AND 12 ARE initially parallel to XY. The wire and loop carry steady currents as shown in the diagram. X Y S R P Q What is the effect on the loop? A It moves towards the wire XY. B It moves away from the wire XY. C It rotates about an axis parallel to XY. D It contracts in size.
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 172 8. The diagram shows three long, straight PARALLEL WIRES 0 1 AND 2 NORMAL TO THE PLANE OF THE PAPER 7IRES 0 AND 2 CARRY CURRENTS directed into the plane of the paper, and wire Q carries a current directed out of the paper. All three currents are of the same magnitude. Which arrow shows correctly the direction of the resultant force on wire P? C A B D R Q P 9. Given that the force between two long parallel conductors of length l, a distance a apart, carrying currents, I 1 and I2 , is F = μ0 I1 I2 l –——– 2Sa , use the defi nition of the ampere to show that μ0, the permeability of free space, is 4π u 10–7 J s2 C–2 m–1. 10. Two wires X and Y, carrying currents I1 and I2 out of the plane of the paper, are separated by a distance r as shown in the diagram. X r I1 I2 Y (a) Write an expression for the magnetic fl ux density at Y due to the current I1 . Show in the fi gure the direction of the magnetic fi eld. (b) What is the force per unit length of wire which I1 causes on wire Y ? Show the direction of the force. 16.6 Determination of the Ratio e —m Students should be able to: GHVFULEHWKHPRWLRQRIDFKDUJHGSDUWLFOHLQWKHSUHVHQFHRIERWKPDJQHWLFDQGHOHFWULFÀHOGV IRUvB and ESHUSHQGLFXODUWRHDFK RWKHU H[SODLQWKHSULQFLSOHVRIWKHGHWHUPLQDWLRQRIWKHUDWLR e mIRUHOHFWURQVLQ7KRPVRQҋVH[SHULPHQW TXDQWLWDWLYHWUHDWPHQWLVUHTXLUHG Learning Outcomes Determination of the Specifi c Charge, q m of a Particle 1. The apparatus used to determine the specifi c charge, q —m of a charged particle is known as a mass spectrometer. 2. Figure 16.18 shows a form of the mass spectrometer which uses the principle of the force on a charge moving in a magnetic fi eld. Positive ion To vacuum pump Uniform magnetic field B out of the plane of the paper Collector Detector Anode Electron beam Filament Cathode V A S2 S1 S3 D r P Q Figure 16.18
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 173 3. The region A is filled with a gas or vapour at a low pressure. 4. Electrons are emitted from the filament when current flows in it. The electrons are accelerated to the anode by the high potential difference between the cathode and anode. 5. Collisions between fast electrons and the gas in A ionises the gas. Positive ions and electrons are produced. 6. The positive ions are attracted to the electrode Q which is at a negative potential relative to the anode. 7. These positive ions are further accelerated by the strong electric field between the electrodes Q and P. A beam of fast moving positive ions emerges from the slit S1 into a uniform magnetic field. 8. If V = potential difference between the electrodes Q and P, the velocity v of the positive ions that emerge from S1 is given by —1 2 mv 2 = qV v 2 = 2qV —–– m ............................. { where q = charge on ion m = mass of ion 9. Since the magnetic field of flux density B acts out of the plane of the page, the ions are subjected to the force F = Bqv. 10. The subsequent path of the ions in the magnetic field is a semicircle of radius r given by mv 2 —–– m = Bqv qB = mv —– r v 2 = 2qV —–– m = —m r 2qV —–– m 1 –– 2 q2 B2 = m2 —– r 2 . 2qV —–– m q —m = —— 2V B2 r 2 11. The accelerating potential difference, V between the electrodes Q and P is adjusted until the detector shows a high reading. This implies that a large number of positive ions are collected by the collector. Knowing the values of V, B and r, the specific charge q —m of these ions can be determined. 12. Usually the gas sample in the spectrometer has a number of isotopes. Ions of each of these isotopes have the same charge, q, but different mass, m. 13. When the value of the accelerating potential difference is changed, the collector reading would be large for values of V corresponding to the masses of the other isotopes.
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 174 Example 11 The specifi c charge —e me of an electron can be determined using the apparatus shown in the diagram. Electrons are emitted from a cathode C and are accelerated by a potential difference V between C and a narrow slit S1 . A magnetic fi eld of fl ux density B causes the electrons to pass through slits S2 and S3 following a semicircular path of radius R with centre at O. A detector D records the electron beam current, which is adjusted to its maximum value by altering V. (a) Deduce an expression for the speed of the electron at S1 in terms of V, e and me . (b) Deduce (i) the direction of the force on the electron at P, (ii) the direction of the magnetic fi eld. (c) Derive an expression for —e me in terms of V, B and R. (d) Explain why the apparatus needs to be evacuated. (e) If the detector shows a reading of 8 u 10–14 A, how many electrons per second passes the slit S3 ? (f) Describe how you would modify the apparatus to determine the specifi c charge of proton. Solution: (a) If v = speed of electron, —1 2 me v2 = eV v = —–– 2eV me (b) (i) The direction of the force on the electron at P is in the direction PO. (ii) Using the right-hand rule (or Fleming’s left-hand rule) the direction of the magnetic fi eld is into the plane of the page. (c) Using F = Bev = me v2 —–– R , Be = —– me R —–– 2eV me 1 –– 2 B2 e2 = me 2 —–– R2 —–– 2eV me —e me = —–– 2V B2 R2 (d) The apparatus is evacuated so that the electrons do not collide with air molecules and the speed v of the electrons remains constant. (e) Current, I = —– dQ dt = ne Number of electrons per second, n = —I e = (8 u10–14 ————–––) (1.6 u10–19) = 5.0 u 105 s–1 F s #ATHODE REPLACED BY A SOURCE OF PROTONS s !CCELERATING POTENTIAL DIFFERENCE V is reversed, potential of S1 negative. s $IRECTION OF MAGNETIC l ELDS REVERSED AND ITS MAGNITUDE IS DECREASED D P R O C S1 S3 S2 Detector
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 175 Mass Spectrometer 2013/P2/Q11 1. The mass spectrometer is shown in Figure 16.19. Figure 16.19 – + Photographic plate Magnetic field into the plane of the paper S1 S2 S3 P1 FM P2 FE 2. Positive ions from a source enter a velocity selector through two narrow slits S1 and S2 . 3. The velocity selector consists of a region where a uniform magnetic field B and a uniform electric field E which are mutually perpendicular act. 4. The uniform electric field E between two parallel plates P1 and P2 produces a force FE = qE q = charge of ion 5. The uniform magnetic field B produces a force FM = qvB in the opposite direction to FE where v = velocity of ion. 6. Ions go through the velocity selector undeflected if FE = FM qE = qvB Velocity selected, v = — E B 7. Ions of different isotopes which have the same charge and velocity v = —E B enter the uniform magnetic field through S3 . 8. The force FM = qvB which is due to the magnetic field B alone, causes the ions to move in a semicircular path of radius r given by FM = qvB = mv2 —––r r = —–– mv qB v = — E B = —–– mE qB2 r ∞m since charge q, electric field E and magnetic field B are constants.
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 176 9. Ions of different isotopes move along different semicircles and hit a photographic plate at different points. By measuring the distances of the marks on the photographic plates from the slit S3 , the diameters and hence the radii r of the circular paths of the ions of the various isotopes can be measured. The mass m of the ion can be calculated using m = qrB2 —–– E . Example 12 In a mass spectrometer, the electric fi eld of the velocity selector is 1.19 u 105 V m–1 and the magnetic fl ux density is 0.53 T. The radii of the paths of the singlely charged oxygen ions are 7.08 cm, 7.53 cm, and 7.97 cm. If the charge on each oxygen ion is 1.60 u 10–19 C, calculate (a) the mass of each of the oxygen ions, (b) the relative atomic mass of each of the ions. Solution: (a) Common velocity of ions emerging from the velocity selecton, v = E ––B . In the uniform magnetic fi eld, qvB = mv 2 —– r Mass of ion, m = qBr —– v = qB2 r —–– E m1 = (1.60 u 10–19)(0.53)2 (7.08 u 10–2) (1.19 u 105 ) = 2.67 u 10–26 kg m2 = (1.60 u 10–19)(0.53)2 (7.53 u 10–2) (1.19 u 105 ) = 2.84 u 10–26 kg m3 = (1.60 u 10–19)(0.53)2 (7.97 u 10–2) (1.19 u 105 ) = 3.00 u 10–26 kg (b) The relative atomic mass of the fi rst ion = m1 u = 2.67 u 10–26 1.66 u 10–27 = 16.1 the second ion = 17.1 the third ion = 18.1
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 177 Thomson’s Experiment for Determination of e m for Electron 1. J. J. Thomson used a cathode-ray tube to determine the specific charge e m, which is the charge per unit mass of the electron. The modern form of the cathode-ray tube is as shown in Figure 16.20. Fluorescent screen Light spot Cathode Anode FE FM v E B v Figure 16.20 2. Electrons are emitted from a hot cathode and accelerated by the potential difference between the cathode and the anode. The anode is at a positive potential relative to the cathode. 3. The electrons emerge from a hole in the anode and travel with a constant velocity v. The electrons then enter a region of cross electric and magnetic fields. 4. The electric field E is produced by the potential difference between two parallel horizontal plates. The magnetic field B is produced by two current-carrying coils outside the cathode ray-tube (not shown in Figure 16.20) The E and B fields are perpendicular to each other. 5. An electron experiences an electric force, FE towards the positive plate, and a magnetic force, FM in the opposite direction. FE = eE and FM = Bev 6. The magnitudes of the electric field E and the magnetic field B are adjusted such that FM = FE so that the electron passes straight through the cross field and strikes the centre of the fluorescent screen. A spot of light is produced on the screen. 7. When FM = FE, Bev = eE Velocity of electron, v = E B Hence, the velocity v of the electron can be determined from the magnitude of E and B. 8. Keeping the magnitude of the electric field E, the magnetic field B is turned off. The force FE on the electron deflects the electron upwards. Its path in the electric field is a parabola. After emerging from the electric field its path is a straight line as shown in Figure 16.21. A light spot is formed on the screen at P. 2009/P2/Q8, 2017/P2/Q11
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 178 x y s L P C O L FE v — 2 Figure 16.21 9. In the electric field E, there is no force on the electron in the horizontal direction. Hence the horizontal component of velocity of the electron is constant at v. If the length of the plates is L, then time taken by the electron to move through the electric field, E is t = L v 10. The force FE = eE produces an acceleration a in the vertical direction. Acceleration, a = F m = eE m (m is the mass of the electron.) 11. When the electron emerges from the electric field E in between the plates, it is displaced through a distance s given by s = ut + 1 2 at2 = 0 + 1 2 1 eE m 21L v 2 2 .............................. { 12. In Figure 16.21, C is the centre of the space between the plates, and CP is a straight line. s y = L/2 x s = Ly 2x .............................. | 13. Equating { and |, s = Ly 2x = 1 2 1 eE m 21L v 2 2 Specific charge of electron, e m = yv2 xEL 1v = E B2 = yE xB2 L which is in terms of measurable quantities y, E, x, B and L.
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 179 Quick Check 6 1. In a mass spectrometer, an ion of mass m and charge Q is accelerated by a potential difference V. The ion then enters a uniform magnetic fi eld of fl ux density, B. Its subsequent path is circular of radius, R. Which of the following is equal to —Q m ? A —–– 2V BR2 C —–– V B2 R2 B —– V BR D —–– 2V B2 R2 2. In Thomson’s experiment a beam of electrons passes through a region P which consists of an electric fi eld E and a magnetic fi eld B as shown in the diagram. Electron beam Region P E B + – The function of the region P is A to accelerate the electrons B to focus the electron beam C to balance the weight of the electrons D to select electrons of a certain velocity 3. Ions of charge q from a source S move through a velocity selector of a mass spectrometer where a uniform electric field E and a mutually perpendicular magnetic fi eld B act. + – S Slit S1 B E Slit S2 Ions which emerge through the slit S2 have a velocity of A —B E C qE —–B B —E B D B —– qE 4. In a mass spectrometer an ion of mass m and charge q enters at right angles into a uniform magnetic fi eld. The subsequent path of the ion is A a circle whose radius is directly proportional to —m q . B a circle whose radius is directly proportional to q —m . C a helix whose radius is directly proportional to q —m . D a straight line perpendicular to its initial motion. 5. In the mass spectrometer shown in the diagram, ions which follow the paths P1 and P2 have the same + – S1 S2 S3 P1 P2 A mass. C speed. B specifi c charge. D acceleration. 6. The diagram below shows the path of a charged particle which moves in a plane at right-angles to a uniform magnetic fi eld. Uniform magnetic field Which of the following explanation about the shape of the path is correct? A The magnetic fl ux density is decreasing. B The energy of the charged particle is decreasing. C The charge of the particle is decreasing. D The mass of the particle is increasing.
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 180 7. A beam of electrons travelling with a speed v passes undeflected through a region where an electric field E and a magnetic field B act. What is the relationship between the two fields in terms of magnitude and direction? 8. In a mass spectrometer, negatively charged particles enter a region in which the uniform magnetic flux density is 0.087 T and the uniform electric field strength is 3.04 u 106 V m–1, as shown in the diagram below. The particles are not deflected. – Path of charged particles + S N Draw a sketch showing the direction of the force which each field exerts on the particles and find the velocity of the particles. 9. The diagram below shows a cathode-ray tube that contains helium gas at low pressure. Electrons from a hot cathode emerge from a hole at the anode. The subsequent path of the electrons can be observed because of the helium gas in the tube. Anode Path of electrons Cathode (a) The potential difference between the cathode and anode in 5.0 kV. Calculate the speed of electrons that emerge from the anode. (b) A uniform magnetic field acts perpendicularly to the plane of the page. Explain why the path of the electrons is circular. Calculate the radius of the circular path, if the magnetic flux density is 2.0 u 10–3 T. (c) Name the process that occurs in the gas which causes the path of the electrons to be visible. 10. (a) The diagram below shows a proton moving with a velocity v entering a uniform magnetic field B at right-angles. + Proton v Uniform magnetic field B Mark on the above figure, the direction of the electric field E which is required so that the proton moves through the combined fields undeflected. (b) Derive an expression for the magnitude of E in terms of v and B. (c) An ion of deuterium 2 1H which is singularly charged enters the combined field in (b) above with a velocity v. Describe and explain the path of the deuterium ion. 11. (a) Explain what is meant by the ‘statement neon has isotopes of nucleon numbers 20, 21 and 22’. (b) Describe briefly how you would obtain a beam of positively charged neon ions. (c) In the diagram shown, a beam containing ions of 20 10 Na, 21 10 Na and 22 10 Na which are singlely charged passes through narrow slits A and B, and enters a vacuum C.
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 181 P1 P2 Photographic plate A B C S A uniform magnetic fi eld is applied in C, and the electric fi eld between the plates P1 and P2 is 1.50 u 105 V m–1. (i) Show on a labelled diagram the polarity of the plates P1 and P2 , and the direction of the magnetic fi eld in order for the ions to follow the path shown. (ii) If the magnetic fl ux density is 0.60 T, fi nd the velocity of the ions emerging from the slit S. (iii) Ions of one of the isotopes of neon produce a trace on the photographic plate at a distance 0.190 m from S. Find the mass number of this isotope. (iv) Calculate the distance of the trace on the photographic plate produced by each of the ions of the other two isotopes of neon. 16.7 Hall Effect Students should be able to: H[SODLQ+DOOHIIHFWDQGGHULYHDQH[SUHVVLRQIRU+DOOYROWDJHV+ VWDWHWKHDSSOLFDWLRQVRI+DOOHIIHFW Learning Outcomes 1. Another effect of an electric fi eld crossing a magnetic fi eld is in Hall effect. 2. Figure 16.22(a) shows a piece of p-type semiconductor whose majority charge carriers are positive holes carrying a current in a uniform magnetic fi eld. (a) (b) + + E FM FM v d v Q P Q P produced produced + + + Magnetic field Magnetic field + + – ––––– –––––– ++++++ I II I E – Figure 16.22 3. The force FM on the positive charge carriers due to the magnetic fi eld defl ects them to the side P of the semiconductor. This produces a potential difference and an electric fi eld across PQ. 4. As more positive holes are defl ected to the side P, the transverse potential difference and the electric fi eld across PQ increase. Figure 16.23 FM = qvB FM = qE + v 2011/P1/Q30, 2012/P1/Q31, 2013/P2/Q20, 2018/P2/Q11
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 182 5. Besides the force FM due to the magnetic fi eld, a force FE due to the electric fi eld acts on the charge carriers. When FM = FE, the charge carriers are no longer defl ected and a steady potential difference VH known as Hall’s voltage is produced across PQ. 6. The electric fi eld, E = V —–H – d When FM = FE, qvB = q V —–H – d Hall’s voltage, VH = Bvd Exam Tips Direction of Hall’s voltage: perpendicular to the GLUHFWLRQRIWKHFXUUHQWñRZ Using I = nAvq, v = —––I nAq = —–——–– I n(d u w)q w = thickness d = width Hence, VH = B —–I –— ndwq d VH = —– BI–– nwq 7. The Hall’s voltage is high for semiconductors compared to metals because n, the number of charge carriers per unit volume for semiconductors is very much smaller than for metals. 8. Figure 16.23(b) shows a piece n-type of semiconductor carrying the current I in the same magnetic fi eld. The free electrons are defl ected to the side P. Hence, the direction of the Hall’s voltage VH is opposite to that for a piece of p-type semiconductor. 9. Hence, the sign + or –, for the majority charge carrier in a piece of semiconductor can be determined from the direction of the Hall’s voltage. Use of Hall Effect–Comparing Flux Density of Magnetic Fields 1. From the equation: Hall’s voltage, VH = Bvd The Hall’s voltage VH is directly proportional to the magnetic fl ux density B. 2. Hence, the fl ux densities of two magnetic fi elds can be compared by measuring the Hall’s voltages produced in a piece of semiconductor placed fi rst in one fi eld and then in the other. Figure 16.24 I I Q Semiconductor Magnetic field B P INFO /HSS,MMLJ[
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 183 3. The advantages of this method include s IT IS SUITABLE FOR MEASURING THE m UX DENSITY AT ISOLATED POINTS BECAUSE OF THE SMALL SIZE OF THE semiconductor piece. s IT IS SENSITIVE ABLE TO DETECT SMALL CHANGES IN m UX DENSITY s IT IS RESPONSIVE ABLE TO MEASURE FAST CHANGING MAGNETIC l ELD Example 13 (a) A slice of indium antimonide is 2.5 mm thick and carries a current of 150 mA. A magnetic fi eld of fl ux density 0.50 T, if correctly applied, produces a maximum Hall voltage of 8.75 mV between the edges of the slice. Calculate the number of free charge carriers per unit volume, assuming that they have a charge of – 1.6 u 10–19 C. (b) What can you conclude from the observation that the Hall voltage in different conductors can be positive, negative or zero? Solution: (a) If d = width of indium antimonide slice VH = Hall’s voltage B = 0.5 T I w = 2.5 mm I = 150 mA VH = 8.75 mV d Electric fi eld, E = V —–H – d Force due to magnetic fi eld = force due to electric fi eld FM = FE Bev = eE From I = nAve, v = —–I – nAe Be —–I – nAe = e V —–H – d Number of free charge carriers per unit volume, n = ——– BId VHAe = ————–– BId VH(d u w)e = ——– BI VHwe = 0.50 u (150 u 10–3) ———————————————– (8.75 u 10–3)(2.5 u 10–3)(1.6 u 10–19) = 2.14 u 1022 m–3 (b) If Hall’s voltage is positive, the majority charge carriers are positively charged. If Hall’s voltage is negative, the majority charge carriers are negatively charged. If Hall’s voltage is zero, then there are equal numbers of negative charged carriers and positive charged carriers.
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 184 Quick Check 7 1. The diagram shows a piece of semiconductor in the form of a square of sides ωcarrying a current I in a uniform magnetic fi eld of fl ux density B. R B S P Q I W The current is due to free electrons of charge e moving with a velocity v. (a) Explain for the set up of a Hall’s voltage BETWEEN EDGES 01 AND 23 OF THE semiconductor. (b) Write the equation for the force on a free electron in the semiconductor due to Hall’s voltage VH , and show that VH = Bvω. (c) Describe and explain the effect on the Hall’s voltage if (i) the current is increased, (ii) the piece of semiconductor is replaced by a piece of copper of the same size and carrying the same current. 2. Explain what is meant by Hall effect, and outline an experiment to demonstrate it. Using a simple free electron model, show that the Hall coeffi cient RH, i.e. —————————————– the transverse electric fi eld fl ux density u current density , is given by RH = —– 1 ne , where n is the number of free electrons per unit volume and e is the charge on the electron. What information may be provided by the magnitude and the sign of the Hall coeffi cient? 3. A thin piece of metal carrying a current is placed in a magnetic fi eld which is normal to its surface. (a) Show on a sketch, the directions of the current, motion of free electrons, magnetic fi eld and the polarities of the Hall e.m.f. (b) Explain why the Hall e.m.f. is greater for semiconductors compared to metals. (c) State a practical application of Hall’s effect. 4. (a) State the conditions for a charge in a magnetic fi eld to experience a force F. Write an expression for F and identify all the symbols in the expression. (b) The diagram shows a metal block of sides a, b, and c parallel to the x, y and z axes respectively. I b a c B x O z y The current I in the block is parallel to the positive Oz direction and a uniform magnetic fi eld of fl ux density B acts in the positive Oy direction. Describe how a Hall potential difference is set up, and derive an expression for this potential difference in terms of e, the charge on an electron, and n the number of free electrons per unit volume. (c) The block is then moved mechanical with a uniform acceleration along the positive Oz direction. The direction and magnitude of B and I remain unchanged. (i) Sketch a graph to show how the Hall’s voltage VH changes with time. (ii) Explain why the Hall’s voltage dissapears for a certain value of the velocity and fi nd this value of the velocity, if I = 12 A, a = 1.5 mm, b = 8.0 mm, c = 2.0 mm and n = 1.0 u 1029 m–3 (d) What deduction can be made about the charge carriers of a conductor which does not show any Hall effect?
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 185 Important Formulae 1. Force on a charge q in a magnetic fi eld B F = q(v u B) Magnitude of force, F = qvB sin θ When θ = 90o , F = qvB = mv2 ——r 2. In velocity selector of Spectrometer, FE = FM qE = qvB 3. Hall voltage, VH = Bvd where d is the width of the sample. 4. Force on a current-carrying conductor F = I(l u B) Magnitude of force, F = BIl sin θ 5. Magnetic fi eld produced by a current I in a straight wire: B = μo I 2πd , circular coil: B = μo NI 2r , solenoid: B = μ0 nI 6. Force between two parallel wires carrying currents I1 and I2 and separated by a distance d, —F l = μ0 I1 I ——–2 2Sd STPM PRACTICE 16 1. Blood fl ow in an artery constitutes a fl ow of electrically charged fluid. By applying a magnetic field across the artery and measuring the potential difference produced across it, the velocity of blood fl ow can be determined. This procedure is based on which of the following? A Hall’s effect B Electromagnetic induction C Electrostatic induction D Photoelectric effect 2. A conductor of length 0.12 m, cross-sectional area of 1.8 cm2 and carries a current I is placed in a magnetic fi eld of fl ux density B = 1.5 T. The force F on the conductor is 3.6 N as shown in the diagram. B F I If the number density of free electrons in the conductor is 3.4 u1028 m s-3, what is the value of the drift velocity of the free electrons in the conductor? A 3.2 u10-6 m s-1 B 2.3 u10-5 m s-1 C 4.1 u10-5 m s-1 D 1.1 u10-4 m s-1 3. Hall’s effect is used to measure A magnetic fl ux density. B specifi c charge of electrons. C mass of different isotopes. D dielectric constant of a material. 4. A conductor of length L and weight W is free to move between vertical guides inside a uniform magnetic field B as shown in the diagram. When a current I fl ows in the conductor, the conductor is able to fl oat under the action of the magnetic force. B I
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 186 Which is the correct expression for the current I? A I =WLB C I = WB L B I = WL B D I = W BL 5. Two particles P and Q enter a uniform electric field B with the same velocity v. The semicircular paths of P and Q are as shown in the diagram. The radii of the paths are r and 2r. The magnitude of the charge on P and Q is the same. B P Q v Which statement about the particles P and Q is correct? A P and Q are charged positive, mass of P is twice the mass of Q. B P and Q are charged positive, mass of P is half the mass of Q. C P and Q are charged negative, mass of P is twice the mass of Q. D P and Q are charged negative, mass of P is half the mass of Q. 6. A piece of n-type semiconductor of dimensions l u b u w is placed in a uniform magnetic field B. The direction of B is into the page. A current I flows in the semiconductor in the direction as shown in the diagram. The number of free electrons per unit volume of the semiconductor is n. B w b l I I What is the magnitude of the Hall voltage VH, and which of the surfaces is at a negative potential? Magnitude of Surface at a Hall voltage negative potential A VH = BI new Upper surface B VH = BI new Lower surface C VH = BI nel Front surface D VH = BI nel Back surface 7. A steady current flows in a long straight wire. Which graph correctly shows the variation of the magnetic flux density B with the perpendicular distance r from the wire? A C B 0 r B 0 r B D B 0 r B 0 r 8. A wire PQ hangs horizontally from strings attached to its ends carries a current I in the direction shown in the diagram. P Q I x y z In which direction must a magnetic field be applied so that the tension in the strings is reduced? A Positive- x C Positive- z B Negative-x D Negative-z
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 187 9. Positive ions move straight through a velocity selector that consists of mutual perpendicular magnetic field B and electric field E. If B = 0.40 T and E = 80.0 kV m–1, what is the velocity of the ions? A 5.0 u 10–6 m s–1 B 2.0 u 102 m s–1 C 3.2 u 104 m s–1 D 2.0 u 105 m s–1 10. A slice of semiconductor is placed in a uniform magnetic field B as shown in the diagram. I I B Semiconductor A Hall voltage is produced when a current I flows through the semiconductor. The value of Hall voltage is independent of A the number of charge carriers per unit volume. B the current density in the semiconductor. C the length of the slice of semiconductor. D the flux density of the magnetic field. 11. A conductor carrying a steady current from X to Y lies between the poles of two similar horse shoe magnets as shown in the diagram. Current X Y N S N S The wire experiences A a force in the direction XY. B a force vertically upwards. C a torque tending to rotate the conductor. D a force at right angle to XY and the plane of the paper. 12. Two long, straight, parallel wires X and Y are separated by a distance d. The force between the wires is F when the current in both the wires is I. If the current in both the wires is increased to 2I, what is the separation between the wires so that the force between the wires remains as F? A —d 4 C 2d B —d 2 D 4d 13. Two long, straight, parallel wires carry currents of 10 A and 20 A. Which diagram shows the directions and relative magnitudes of F1 and F2 of the force per unit length on each of the wires? A C F1 F2 F1 = F2 10 A 20 A B D F1 F2 F2 = 2F1 10 A 20 A 10 A 20 A F2 = 2F1 F1 F2 14. Three long, vertical, parallel wires pass through the corners of an equilateral triangle 012 4HEY CARRY EQUAL CURRENTS INTO OR OUT of the paper in the directions shown in the figure. P Q R Which of the following shows the forces F1 and F2 ON THE WIRE 2 DUE TO THE WIRES 0 AND Q respectively? F1 F2 F1 = F2 10 A 20 A
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 188 A C F1 F2 P Q R F1 F2 P Q R B D F1 F2 P R Q F1 F2 P Q R 15. Two long, straight, parallel wires separated by a distance r carry equal currents I but in opposite directions. The magnetic flux density midway between the wire is A μ0 I ——2Sr C 2μ0 I ——Sr B μ0 I ——Sr D μ0 I ——4Sr 16. A horizontal wire which lies in the N-S direction is connected to a battery and a switch S. A plotting compass P is placed above the wire and another compass Q below the wire as shown in the diagram. Q S P S N W E The switch S is then closed. The deflections of the compass needles are correctly stated in Compass P Compass Q A towards the east towards the west B towards the west towards the east C towards the east towards the east D towards the west towards the west 17. A piece of aluminium of thickness 0.05 mm carries a current of 10.0 A in a magnetic field B of flux density 1.5 T as shown in the diagram. The Hall voltage produced is 12.0 V. d = 0.05 mm I = 10.0 A B = 1.5 T What is the free electron density of aluminium? A 7.1 u 1024 m–3 B 1.6 u 1027 m–3 C 7.1 u 1028 m–3 D 1.6 u 1029 m–3 18. Two protons are injected with velocities v and 2v respectively at right angles to a uniform magnetic field. The protons then move in circular paths of radii r1 and r2 , taking time T1 and T2 respectively to complete a revolution. Which of the following pairs of relations is correct? A r1 = r2 , T2 = T1 B r1 = r2 , T2 = 0.5T1 C r2 = 2r1 , T2 = T1 D r2 = 2r1 , T2 = 2T1 19. An ion of charge 1.6 u 10–19 C and mass 13 u, enters at right angles to a uniform magnetic field of flux density 0.50 T with a velocity 1.2 u 105 m s–1. What is the diameter of its circular path in the magnetic field? A 3.0 mm C 25 mm B 4.0 mm D 65 mm 20. The diagram shows the paths of ions of different isotopes of an element in a mass spectrometer. 30 cm 2 cm 4 cm The ratio of the masses of the three isotopes is A 15 : 16 : 18 C 31 : 32 : 36 B 9 : 10 : 12 D 33 : 34 : 36
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 189 21. A proton and an D-particle are injected with the same velocity at right angles to a uniform magnetic field. What is the ratio of the radius of the circular path of the proton to that of the D-particle? A —1 4 C 2 B — 1 2 D 4 22. The figure shows the apparatus used to determine the specific charge e/m of an electron. Electrons enter the region of cross electric field E and magnetic field B with a velocity v, and leave the combined field undeflected. P Electron Q B O Screen v (a) Which of the parallel plates P or Q should be at a positive potential? (b) Derive an expression for the velocity v of the electrons. (c) State what happens to an electron whose velocity is (i) less than, and (ii) greater than the velocity v in (b) above. 23. (a) Write the expression for the magnetic flux Φ through an area A. (b) Determine the magnetic flux density (i) at a point 15.0 cm from a straight wire carrying a current of 8.0 A. (ii) at the centre of a plane circular coil of 20 turns, each of radius 5.0 cm and carries a current of 8.0 A. (c) Explain what is Hall effect. A piece of n-type semiconductor of dimensions a u b u c carries a current I in a uniform magnetic field B as shown in the digaram. The Hall voltage produced is VH. a I B c b (i) State the direction of the Hall voltage VH. (ii) Derive the expression for the number of charge carriers per unit volume of the semiconductor. (iii) If B = 1.20 T, the dimensions of the piece of semiconductor is 2.00 mm u 4.00 mm u 8.00 mm and VH = 25.0 mV when I = 0.050 A, calculate the number of charge carriers per unit volume of the semiconductor. 24. A rectangular coil WXYZ of dimensions 20.0 cm 10.0 cm having 50 turns is in a uniform magnetic field of 0.25 T. The plane of the coil is parallel to the magnetic field. A current of 1.50 A flows in the direction WXYZ. W X Z Y B (a) Show on the diagram, the direction of forces on the sides XY and WZ. (b) What is the magnitude of the forces? (c) What is the effect of the forces? 25. (a) The diagram shows the screen of a cathode-ray tube when viewed from the front. E B
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 190 The beam of electrons travelling towards the screen in the tube passes through the combined electric field E, and magnetic field B acting in the directions shown. The electric field E and the magnetic field B separately deflects the spot of light on the screen through the same distance from the centre of the screen. Show on a figure, the position of the spot of light on the screen when both the fields act. (b) A particle of mass 200 mg and charge 2.0 u 10–8 C moves in a uniform horizontal magnetic field at constant velocity of 5.0 u 104 m s–1. Determine the minimum magnetic flux density of the magnetic field in order that the particle moves at a constant height. 26. (a) Explain how a proton is able to move with constant velocity in a uniform magnetic field. (b) A proton moves in a circular path with a speed of 1.50 u 105 m s–1 inside a solenoid that has 25 turns per cm, and carries a current of 5.00 A. Find (i) the magnitude of the magnetic field inside the solenoid. (ii) the radius of the circular path Sketch a diagram to show the circular path relative to the direction of the magnetic field and velocity of the proton. 27.(a) (i) Define magnetic flux density in terms of the force on a currentcarrying conductor. (ii) Hence derive an expression for the force on an electron moving in a magnetic field of flux density B. (b) The diagram shows a wire of length 0.57 m placed at right angles to a uniform magnetic field of flux density 1.8 u 10–3 T. The density of the wire is 7.8 u 103 kg m–3 and its resistivity is 8.8 u 10–8 m. Electromagnetic force Magnetic field (i) Calculate the potential difference required between the ends of the wire to produce an electromagnetic force on the wire that equals its weight. Mark the direction of current in the wire. (ii) The horizontal component of the Earth’s magnetic field is 1.8 u 10–5 T. Explain why in practice, current-carrying wires are not seen to lift off the ground. 28. (a) (i) Explain what is meant by Hall effect. (ii) Experiments to demonstrate Hall effect normally use a semiconductor specimen instead of a metal. Explain why. (b) B a c b I I A steady current I passes a piece of metal placed in a uniform magnetic field of flux density B as shown in the diagram. Show that the magnitude of the Hall potential difference is given by VH = —– BI nbe where n = the number of free electrons per unit volume in the metal. Show the polarity of the Hall potential difference. (c) A long metal bar, 40 mm wide, is pulled along a horizontal table at a uniform velocity of 6.0 mm s–1 as shown in the figure. B Y 40 mm 6 mm s–1 X V Voltmeter
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 191 Stationary brushes X and Y make electrical contact with the sides of the bar. When a uniform magnetic fi eld of fl ux density B is directed downwards normal to the table, the voltmeter gives a reading of 50 μV. (i) Explain how this potential difference is set up. (ii) Find the flux density B of the magnetic fi eld. 29. (a) (i) Write the equation for the force F acting on an electron of charge e travelling with velocity v in, and at right angles to, a uniform magnetic fi eld of fl ux density B. (ii) Describe and explain the path of the electron in the magnetic fi eld. (b) The diagram shows part of a mass spectrometer. The whole arrangement is in a vacuum. H P S –3 000 V Beam of negative ions Uniform electric field Uniform magnetic field Metal container Metal plate Negative ions of mass 2.84 u 10–26 kg and charge –1.60 u 10–19 C are produced at S, which is at a potential of –3 000 V. The ions are accelerated in a narrow beam towards H, a hole in a hollow metal container which is earthed. Inside the container, the negative ions enter a region where an electric fi eld of strength E, and a magnetic fi eld of fl ux density B = 0.83 T act. The negative ions continue to move in these fi elds in a straight line with constant velocity. (i) Calculate the velocity of the ions when they reach H. (ii) Explain, using a sketch, why it is possible for the ions not be defl ected in the fi elds. Mark the direction of the electric fi eld, and calculate the electric fi eld strength E. (iii) If the magnetic fi eld is then removed, what happens to the path of the negative ions? 30. (a) Defi ne the magnetic fl ux density in terms of the force on a conductor carrying a current. (b) A light but rigid wire frame XYZ is pivoted about a vertical axis XZ so that it can rotate freely in a uniform horizontal magnetic fi eld. Initially the plane of the frame is parallel to the magnetic fi eld as shown in the diagram below. Y X Z Uniform magnetic field Axis of rotation (i) Describe the motion of the wire frame XYZ when a short pulse of current passes through it from X to Z. (ii) Sketch a graph to show how the kinetic energy of the frame varies with the angle of rotation θ for 0° θ 180°. Explain the shape of the graph. (c) The diagram shows a square coil of wire ABCD of sides 20.0 cm long and a long straight wire XY on a smooth horizontal table. Y X D C A B 20 A 5.0 cm 20 cm 10 A 10 A 20 cm XY and AD are parallel and at a distance of 5.0 cm apart. A current of 20 A is in wire XY, and 10 A in the coil ABCD. By considering the electromagnetic forces on the sides of the coil, calculate the resultant force on the coil ABCD.
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 192 1 1. D $OLQHRIIRUFHLQDPDJQHWLFÀHOGVKRZVWKH SDWKWKDWDIUHHPDJQHWLFSROHZRXOGPRYHLQ WKHÀHOG E 'LUHFWLRQVKRZQRQWKHOLQHRIIRUFHLQGLFDWHV WKHGLUHFWLRQRIPRWLRQRIDIUHH1SROHLQWKH ÀHOG 2. 0DJQHWLFOLQHVRIIRUFHDUHFORVHUWRJHWKHULQD VWURQJPDJQHWLFÀHOG 3. D 2 1. C: VLVSHUSHQGLFXODUWRB 2. B 3. D 4. D 5. B 6. u–3 T x x x x x x x x x x x x x x x x 3 1. D fl1 E F fl1 2. D $ E PV–2 R F = BIl mg 37° 3. D 5HIHUSDJH (b) F = BevVLQθ (c) I = W BdFRVθ FXUUHQWIURP<WR; 4 1. B 2. D 3. B 4. D 5. D 6. B 7. C 8. A 9. D 5HIHUSDJHffi (b) N S 10. &RQGXFWRUSDUDOOHOWRB θ F = BIlVLQθ 11. (a) 2B E 8VHB = μ0 nI B ∞ I B ∞ nQXPEHURIWXUQVP² 12. D 5HIHUWRSDJH (b) (i) B = μ0 I ——– 2πr vI O 0.10 m 0.10 m I 1 I 2 I 4 I 3 B3 B1 B4 B2 B (B + B3 ) v (I + I 3 $ (B2 + B4 ) v (I 2 + I 4 $ +HQFHUHVXOWDQWBLVLQWKHGLUHFWLRQVKRZQ (ii) (B + B3 ) = μ0 2π(0.10/2) u T (B2 + B4 ) = μ0 2π(0.10/2) u T 5HVXOWDQWB = u) T flu T 5 1. B 2. C 3. D 4. D 5. A 6. C 7. A 8. C 9. 6XEVWLWXWHI = I 2 $ffl Pa P 10. (a) B = μ0 I ۛr (b) F — l = μ0 I I2 ۛr 6 1. D 2. 'ffl(OHFWURQVZLWKDVHOHFWHGYHORFLW\SDVVHV VWUDLJKWWKURXJKWKHUHJLRQ3 3. B 4. A 5. C 6. B 7. v = E — B FM = FE eE = Bev B v E ANSWERS
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 193 8. FM v FE v ffiu PV² 9. (a) v u PV² — 2 mv2 = eV E )RUFHDOZD\VQRUPDOWRYHORFLW\v r P F ,RQLVDWLRQ 10. (a) E v (b) E = vB FE = FM F 6WUDLJKWOLQHXQGHÁHFWHG 11. D 6DPHQXPEHURIS+ EXWGLͿHUHQW QXPEHUVRIQHXWURQV LQQXFOHXV E 8VHDGLVFKDUJHWXEHFRQWDLQLQJQHRQJDV F L 3 QHJDWLYH32 SRVLWLYH BLQWRWKHSODQHRIWKHÀJXUH (ii) v u PV² v = E B (iii) 22 LY d P d m d flP 7 1. D 5HIHUSDJHfl (b) F = e 1—– VH ω 2 (c) (i) VHLQFUHDVHVI = nAve (ii) VHGHFUHDVHVn larger for copper 2. RH large nVPDOO RHSRVLWLYH PDMRULW\FKDUJHFDUULHUVSRVLWLYHO\ FKDUJHG RHQHJDWLYH PDMRULW\FKDUJHFDUULHUVQHJDWLYHO\ FKDUJHG 3. D 5HIHUSDJHfl)LJXUH E (b) VH ∞ —n , nIRUVHPLFRQGXFWRUVPDOOHU (c) Determination of B 4. (a) From F = qvBVLQθ FǂLIvǂθǂ E 'HULYHVH = BI —– neb (c) (i) VH Bva 0 t (ii) VH ZKHQYHORFLW\RIEORFNHTXDOWRGULIW YHORFLW\RIIUHHHOHFWURQV v u²PV² I = nAve G (TXDOQXPEHUVRISRVLWLYHDQGQHJDWLYHFKDUJH FDUULHUV STPM PRACTICE 16 1. A 2. C: F = BIlVLQo DQGI = nAve v = 2F BnAel ðPV 3. A 4. D: F = BIL = W, I = W —– BL 5. &ffl 8VH)OHPLQJ·VOHIWKDQGUXOHF = qvB = mv2 —– r , m r 6. $ffl 8VH)OHPLQJOHIWKDQGUXOHWRREWDLQWKH GLUHFWLRQRIGHÁHFWLRQRIWKHHOHFWURQV Bev = e1—– VH l 2, I = n(lω)ve VH = BI —– neω 7. C: B = μ I 2Sr 8. $ffl$SSO\F = I(l B) 9. D: v = — E B = 2.0 u 105 m s–1 10. C 11. C 12. D 13. A 14. B 15. C 16. $ffl8VH ULJKWKDQG JULS UXOH WR GHWHUPLQH WKH GLUHFWLRQ RI WKH PDJQHWLF ILHOG SURGXFHG E\ FXUUHQWLQWKHZLUH 17. D: VH = Bvb (I = n(bd)ve) = B1——– I nbde 2b n = ——– BI deVH = 1.6 u 1029 m–3 18. C 19. D 20. A 21. B 22. D 3ODWH3ffl3RVLWLYH (b) Bev = eE, v = E — B (c) (i) FM < FEHOHFWURQGHÁHFWHGXSZDUGV (ii) FM > FEHOHFWURQGHÁHFWHGGRZQZDUGV 23. (a) Φ = BAFRVθZKHUHθLVWKHDQJOHEHWZHHQ BDQGWKHQRUPDOWRWKHDUHDA (b) (i) B = μ I 2Sd = (4S u² fl ———–——— 2S T u² T (ii) B = μ NI 2r = (4S u² fl —————–——— T u–3 T
16 Physics Term 2 STPM Chapter 16 Magnetic Fields 194 F +DOOHͿHFWLVDSURGXFWLRQRIDWUDQVYHUVHHPI LQDFRQGXFWRUFDUU\LQJFXUUHQWLQDPDJQHWLF ÀHOG (i) VH EHWZHHQ VXUIDFHV VHSDUDWHG E\ D GLVWDQFH F ZLWK WKH IURQW VXUIDFH DW QHJDWLYHSRWHQWLDO (ii) FE = FM, I = nAve = n(ab)ve e1—— VH b 2 = Bev = Be1——– I nabe 2 n = 1——– BI VHae 2 (iii) n = 1——– BI VHae 2 = ———————————–– u²ffi) m–3 u m–3 24. (a) W X Z Y B F F (b) F = BIl = 1 F (ͿHFWffl7RURWDWHWKHFRLODERXWWKHYHUWLFDOD[LV 25. (a) (b) mg FM = qvB v B 'LUHFWLRQRIPDJQHWLFÀHOGBLVDVVKRZQLQ WKHÀJXUH FM = qvB = mg B = mg ——qv = u–3 ffifl ————————— u²fl u4 ) T ffi7 26. D 9HORFLW\SDUDOOHOWRPDJQHWLFÀHOG Magnetic force FM (b) (i) B = μ nI P7 (ii) mv2 ——r = qvB r = ——mv qB = u² u ) ——————————– u²ffi m ffiffiFP 27. D L 5HIHUSDJH LL 5HIHUSDJH E L 9 BIl = mg &XUUHQWÁRZVIURPOHIWWRULJKW LL )RUFHGXHWRPDJQHWLFÀHOGF = BIl << mg 28. (a) L 5HIHUSDJHfl LL )RU VHPLFRQGXF WRU +DOO·V YRO WDJH ODUJHU EHFDXVH VH∞ —n , n VPDOOHU IRU VHPLFRQGXFWRU (b) VH – + I I (c) L 5HIHUSDJHfl (ii) B fl7 Bev = e VH d 29. (a) (i) F = Bev LL &LUFXODU FDOZD\VSHUSHQGLFXODUWRv E L flu PV² —2 mv2 = qV (ii) FM = FEEXWLQRSSRVLWHGLUHFWLRQ u V m² v = E —B GRZQZDUGV LLL 'HÁHFWHGXSZDUGV²SDWKfflSDUDEROD 30. D 5HIHUSDJH E L 7RUTXHRQIUDPH )UDPH VWDUWV WR URWDWH FORFNZLVH DQG WKHQVORZVGRZQGXHWRHOHFWURPDJQHWLF LQGXFWLRQ (ii) Kinetic energy 0 180° θ F flu–41