Cambridge A Level Physics 9702 — 2012 Oct/Nov Paper 5 · Variant 2

1 As a bar magnet is dropped through a coil, an e.m.f. is induced in the coil. The maximum For e.m.f. E is induced as the magnet leaves the coil with speed v. Examiner’s Use It is suggested that E is directly proportional to v. Design a laboratory experiment to test the relationship between E and v. You should draw a diagram, on page 3, showing the arrangement of your equipment. In your account you should pay particular attention to (a) the procedure to be followed, (b) the measurements to be taken, (c) the control of variables, (d) the analysis of the data, (e) the safety precautions to be taken. 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Defining the Methods of Method of Safety Additional For problem data collection analysis considerations detail Examiner’s Use

Mark scheme: 1 Planning (15 marks) Defining the problem (3 marks) P v is the independent variable or vary v. [1] P E is the dependent variable or measure E. [1] P Keep the number of turns on the coil constant. [1] Methods of data collection (5 marks) M1 Labelled diagram showing magnet falling vertically through coil. [1] M2 Voltmeter or c.r.o. connected to the coil. Allow voltage sensor connected to datalogger. [1] M3 Method to change speed e.g. change height. [1] M4 Measurements to determine v. Use metre rule to measure distance magnet falls to the bottom of the coil or metre rule/ruler to measure length of coil or ruler to measure length of the magnet. [Allow timing instrument to measure the time of the fall from the start to the bottom of the coil.] [1] M5 Method of determining v corresponding to appropriate distance e.g. v = √2gh or v=2h/t (for height method) or v = L/t for length of magnet or coil and by stopwatch, timer or lightgate(s) connected to datalogger. [Allow v = gt for timing fall to bottom of coil.] [1] Method of analysis (2 marks) A Plot a graph of E against v. [Allow lg E against lg v] [1] A Relationship valid if straight line through origin. [1] [If lg-lg then straight line with gradient = (+)1 (ignore reference to y-intercept)] Safety considerations (1 mark) S Keep away from falling magnet/use sand tray/cushion to catch magnet. [1] Additional detail (4 marks) D1/2/3/4 Relevant points might include [4] Use coil with large number of turns/drop magnet from large heights/strong magnet 1 Detailed use of datalogger/storage oscilloscope to determine maximum E; allow video camera including slow motion play back 2 Use same magnet or magnet of same strength. 3 Use of short magnet so that v is (nearly) constant 4 Use short/thin coil so that v is (nearly) constant 5 Use a non-metallic vertical guide/tube 6 Method to support vertical coil or guide/tube 7 Repeat experiment for each v and average Do not allow vague computer methods. [Total: 15] GCE AS/A LEVEL – October/November 2012 9702 52

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2 Different light-emitting diodes (LEDs) can emit electromagnetic radiation of different For wavelengths. Examiner’s Use A student uses the circuit of Fig. 2.1 to investigate the minimum potential difference required to cause an LED to emit its characteristic wavelength. V Fig. 2.1 For different LEDs emitting radiation of wavelength λ, the minimum potential difference V is recorded. Question 2 continues on the next page. It is suggested that V and λ are related by the equation For Examiner’s hc Use = B + eV λ where c is the speed of light in a vacuum, e is the elementary charge, h is the Planck constant and B is a constant. (a) A graph is plotted of V on the y-axis against 1/λ on the x-axis. Determine expressions for the gradient and y-intercept in terms of B, c, e and h. gradient = ..................................................... y-intercept = ..................................................... [1] (b) Values of λ and V are given in Fig. 2.2. λ/ 10–9 m V / V 950 0.60 ± 0.05 875 0.70 ± 0.05 655 1.20 ± 0.05 560 1.55 ± 0.05 505 1.80 ± 0.05 430 2.25 ± 0.05 Fig. 2.2 Calculate and record values of (1 / λ) / 106 m–1 in Fig. 2.2. [2] (c) (i) Plot a graph of V / V against (1 / λ) / 106 m–1. Include error bars for V. [2] (ii) Draw the straight line of best fit and a worst acceptable straight line on your graph. Both lines should be clearly labelled. [2] (iii) Determine the gradient of the line of best fit. Include the uncertainty in your answer. gradient = ................................................ [2] 2.4 For Examiner’s Use V / V 2.2 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 (1 / h) / 106 m–1

Mark scheme: 2 Analysis, conclusions and evaluation (15 marks) Part Mark Expected Answer Additional Guidance (a) A1 Gradient = hc/e Note y-intercept must be negative y-intercept = – B/e (b) T1 1/λ / 106 m–1 Appropriate column heading T2 Must be values in table. 1.05 or 1.053 A mixture of 3 s.f. and 4 s.f. is allowed. 1.14 or 1.143 1.53 or 1.527 1.79 or 1.786 1.98 or 1.980 2.33 or 2.326 (c) (i) G1 Six points plotted correctly Must be within half a small square. Penalise ‘blobs’. Ecf allowed from table. U1 All error bars in V / V plotted Do not allow near misses correctly. (c) (ii) G2 Line of best fit If points are plotted correctly then lower end of line should pass between (1.12, 0.7) and (1.16,0.7) and upper end of line should pass between (2.32, 2.25) and (2.34, 2.25). Allow ecf from points plotted incorrectly – examiner judgement. G3 Worst acceptable straight line. Line should be clearly labelled or dashed. Steepest or shallowest possible line Should pass from top of top error bar to that passes through all the error bottom of bottom error bar or bottom of bars. top error bar to top of bottom error bar. Mark scored only if error bars are plotted. (c) (iii) C1 Gradient of best fit line The triangle used should be at least half the length of the drawn line. Check the read offs. Work to half a small square. Do not penalise POT. Should be about 1.3 × 10 –6. U2 Uncertainty in gradient Method of determining absolute uncertainty Difference in worst gradient and gradient. [± 0.08] (c) (iv) C2 y-intercept Must be negative Expect to see point substituted into y = mx + c FOX does not score. Do not penalise POT. Should be between –0.72 and –0.86 U3 Method of determining uncertainty in Difference in worst y-intercept and y-intercept y-intercept. [Should be about ± 0.14]. FOX does not score. Allow ecf from (c)(iv). GCE AS/A LEVEL – October/November 2012 9702 52 (d) (i) C3 h in the range 6.77 × 10–34 to Gradient must be used. Penalise 1 s.f. or >3 7.14 × 10–34 and given to 2 or 3 s.f. significant figures h = gradient × e/c = gradient × 5.33 × 10–28 Allow 6.8 × 10–34 to 7.1 × 10–34 to 2 s.f. (d) (ii) U4 Percentage uncertainty in h ∆m ∆h × 100 or × 100 m h [should be about 6%] (e) C4 B = –e × y-intercept Ignore ‘–’ signs. y-intercept must be used and J or C V or V C but allow ecf from FOX. Should be between 1.16 × 10–19 J and 1.37 × 10–19 J. If FOX 8.3 × 10–20 J U5 Absolute uncertainty in B Uncertainty = best B – worst B = ∆y-intercept × e [Total: 15] Uncertainties in Question 2 (c) (iii) Gradient [U2] Uncertainty = gradient of line of best fit – gradient of worst acceptable line Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) (iv) [U3] Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line Uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept) (d) (ii) [U4] ∆m Percentage uncertainty = × 100 m ∆h 1 (max h − min h ) Percentage uncertainty = × 100 = 2 × 100 h h (e) [U5] Absolute uncertainty = best B – worst B Absolute uncertainty = ∆y-intercept × e ∆c Absolute uncertainty = × B c

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