AQA A-level Physics · Topic 3.7
Practice questions covering fields and their consequences (a-level only) from this specification. Every question below is the real AQA question as it is printed on the paper, with its mark scheme.
6 questions on Fields and their consequences (A-level only), each with its mark scheme and worked method. Reveal the answer when you have had a go.
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Printable question pack — 3.7 Fields and their consequences (A-level only)All 180 questions across 18 specification points — 174 more than the 6 above — with the full mark scheme. Guest accessible · browser print.3.7 Fields and their consequences (A-level only) revision notes
30 questions shown as printed, 17 reference-only · newest series first
Question and mark-scheme extracts © AQA. Reproduced for educational use; the original paper is linked on every question. Contact hello@oscarsongtutoring.co.uk for removal.
AQA 7408 · June 2025 · Paper 2 · Q4 · 8 marks
8 marks
Examiner report
[4.3] Most students were able to realise that this was when the curved line was horizontal but significant numbers made mistakes reading the graph, either missing out the 108, writing 12.35 instead of 12.035 or failing to read to an appropriate number of significant figures for the range shown on the graph (e.g. 12.0).
Mimas, Enceladus and Titan are moons of Saturn.
AQA 7408 · June 2025 · Paper 2 · Q5 · 8 marks
8 marks
Examiner report
[5.2] Most students were able to determine the time constant from the discharge graph. However, a small number of students confused the time constant with the half-life or misread the scale on the time axis. A very small minority attempted to use the gradient or the area under the graph.
A circuit charges and discharges a capacitor using a two-position switch. In position A, an ideal battery and one resistor are in series with the capacitor, a second identical resistor and a microammeter. In position B, the battery-side resistor is disconnected and the capacitor discharges through the second resistor and meter.
AQA 7408 · June 2025 · Paper 2 · Q6 · 13 marks
13 marks
Examiner report
[6.4] Most students were able to determine the acceleration of the ion, although a minority of students mixed up the field and the potential.
Ions orbit Jupiter in a plasma torus. Consider gravitational, magnetic and electric fields acting separately on an ion in the torus.
AQA 7408 · June 2025 · Paper 3A · Q1 · 8 marks
8 marks
Examiner report
This question was about an investigation of how the frequency of a stationary wave on a string varied with tension. The main ideas tested were • identification of the control variables in required practical activity 1 (specification reference 3.7.5.5 and 8.3.2 PS2.4) • use of an oscilloscope as a voltmeter and as a clock (3.7.5.5) In part 01.1 what should have been a straightforward opening question proved tricky and only 17% obtained both marks. No credit was given for a careless annotation to Figure 1 to identify the length of string that should be controlled. In part 01.2 just under 50% correctly identified that m has a maximum range of 170 g. Part 01.3 was a bit mixed, although many could determine the peak voltage and divide by root 2. For the second mark we wanted a voltage between 13.0 and 13.2, so a 2-sf answer was not accepted. © 2025 AQA For part 01.4 nearly 70% correctly determined the period although a few thought that the time- base setting of 20 ms div−1 referred to minor divisions, thus getting about 530 ms. Here as elsewhere, some lost a mark, thinking they should round their 106 to 110 (mistakenly claiming this was 2 sf). A majority saw the intention of part 01.5 and made the expected quantitative comparison between the waveforms in Figure 2. The students who misinterpreted the time-base in the same way as above got about 375 ms and were not penalised. A few students took the novel approach of comparing the number of horizontal divisions for one cycle of each waveform. This was a valid idea and was not penalised.
This question is based on an investigation into how the frequency f of the first-harmonic stationary wave on a string varies with tension (required practical activity 1). A horizontal string runs from a vibration generator to a pulley, then vertically to a hanger carrying slotted masses of total mass m, so the tension is mg. A variable-frequency ac supply drives the vibration generator and an oscilloscope is connected across the supply. The frequency is varied until the first harmonic appears and the process is repeated for different m.
AQA 7408 · June 2024 · Paper 2 · Q2 · 9 marks
9 marks
Examiner report
[02.1] This question proved challenging for most students. Some realised that the current was given by the gradient × capacitance, although not all of these realised that as the initial current was asked for, the gradient must be taken at time 𝑡𝑡= 0. Those answers which did not specify the initial gradient either in words or by annotating the graph were only able to score one mark. The command word in the question was ‘explain’, so full marks required the answer to be justified. Thus, a full-mark answer had to identify Δ𝑉𝑉 Δ𝑡𝑡 as the gradient and also use 𝐼𝐼= Δ𝑄𝑄 Δ𝑡𝑡 and 𝐶𝐶= 𝑄𝑄 𝑉𝑉. Many students did not differentiate between a change in potential and the value of the potential.
An ideal battery, switch, resistor R and capacitor C are connected in series. A voltmeter can measure the potential difference across the resistor or capacitor. During charging, the capacitor pd rises from 0 V at t=0 to about 4.35 V at 5 s. During discharge, the resistor pd falls from about 0.62 V at 20 s to about 0.02 V at 45 s. The capacitor has capacitance 31.0 microfarad.
AQA 7408 · June 2024 · Paper 2 · Q3 · 7 marks
7 marks
Examiner report
[03.1] There were two approaches to this question. Students could calculate the flux cut as 𝐵𝐵𝐵𝐵cos 𝜃𝜃, the time for the rod to fall using 𝑠𝑠= 1 2 𝑔𝑔𝑡𝑡2, and then use Faraday’s law where the emf is given by the change in flux divided by the time taken. The most common mistakes with this route were: • missing out the cos 𝜃𝜃 term, by not realising that 𝐵𝐵 and 𝐴𝐴 had to be perpendicular • not realising that the area was given by the length of the rod × height the rod fell through. Making either mistake still allowed two marks to be scored. The most common approach was to use 𝜀𝜀= 𝐵𝐵𝐵𝐵𝐵𝐵. However, most who tried this approach also failed to realise that 𝐵𝐵 and 𝑣𝑣 should be perpendicular and therefore missed out the cos 𝜃𝜃 term. Students could calculate 𝑣𝑣 either by halving the speed with which it hit the ground or by dividing the distance by the time to fall. Attempts to use the equation for a rotating coil (𝜀𝜀= 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵sin 𝜔𝜔𝜔𝜔) were unlikely to score, although they could gain a mark for the time for the rod to fall.
The Earth's magnetic flux density is 4.9 x 10^-5 T and is directed downwards towards geographic north at 68 degrees to the horizontal. A conducting rod of length 2.0 m is initially horizontal and aligned east-west.
AQA 7408 · June 2024 · Paper 2 · Q5 · 8 marks
8 marks
Examiner report
[05.1] This question was very challenging, with most students gaining 1 or 0. The first mark was for calculating the angular speed of S1. Some students did gain this mark although they did not know how to progress from there. In order to make much progress with this question students had to realise that the centripetal acceleration was equal to the resultant gravitational field strength. They could then subtract the gravitational field strength of the Earth to determine the gravitational field strength of the moon. Some students attempted to determine an effective mass to allow this period and then use this with 𝑇𝑇2 = 4𝜋𝜋𝑟𝑟3 𝐺𝐺𝐺𝐺; this approach did not gain credit.
Satellite S1 is on the Earth-Moon line beyond the far side of the Moon so that it remains directly behind the Moon as viewed from Earth. It has the Moon's orbital period of 27.3 days and is 4.489 x 10^5 km from the Earth. The Earth's gravitational field strength at S1 is 1.98 x 10^-3 N kg^-1. The Moon's field strength at S1 is gM.
AQA 7408 · June 2024 · Paper 2 · Q6 · 12 marks
12 marks
Examiner report
[06.3] This question was more challenging and required some understanding of the context for full marks. Nearly 40% of students realised that 𝐸𝐸𝑘𝑘= Δ𝐸𝐸𝐸𝐸= 𝑒𝑒Δ𝑉𝑉 and scored one mark. However, determining the correct Δ𝑉𝑉 was more challenging: only around 14% students realised it was the largest change in potential between the two points and were able to score two marks.
The electric potential V varies periodically with horizontal position x from 0 to 1100 nm. The graph has maxima of about -0.2 V at x=300 nm and 800 nm and minima of about -3.4 V at x=50 nm, 550 nm and 1050 nm. Its steep, nearly straight sections cover approximately x=110-230 nm, 350-480 nm, 610-730 nm and 850-980 nm; on the first of these, for example, V rises from about -3.0 V at 110 nm to about -0.6 V at 230 nm. Electrons are confined near the potential maxima.
AQA 7408 · June 2024 · Paper 3A · Q3 · 15 marks
15 marks
Examiner report
03.1 is a type of question that students with limited practical experience struggle with. Once again, we have to point out that clamp stand rods may not be used as a vertical datum, but benches can be assumed to be horizontal. The question said a metre ruler must be involved in establishing that the rod was horizontal. Some used the ruler to measure between bench and rod at different points and check that the heights were the same. When this idea was combined with an appropriate use of a set-square to make the ruler vertical, usually explained by annotation to Figure 10, all three marks were awarded. Others placed the set square where the vertical ruler met the rod. We did not allow this because a set square is too short to accurately judge whether the rod is horizontal. Good sketches saved those who struggled to explain their methods, but poor sketches condemned others who showed gaps between their set squares and the bench. Most knew what was behind 03.2 but only the better ones were able to identify the direction of the force on the rod. Some good answers went through the action-reaction idea in great detail to make sure the examiner knew what they were thinking. Others just referred to ‘the force’ and assumed it was upwards because the balance reading had decreased when there was a current in the rod. We gave no credit for current ‘from positive to negative’. Those who did not mention a current direction could only get one mark. However, most could predict a suitable magnetic field direction having identified directions for their force and current although we wanted to see the left-hand rule mentioned in justification. The solution to 03.3 was, first, to produce a homogeneous expression involving k, flux density, current and mass; second, to eliminate flux density using force, current and length, then to eliminate force using mass, length and time. We allowed working in which quantities and units were mixed, but many failed to connect the parts of their solutions using equals signs, making the work difficult to assess. 17% reached the correct solution of s2 in a few lines and earned all three marks. Others slipped up at the last stage of re-arrangement to get s−2 which cost one mark. Some deduced a homogeneous expression with non- base units, usually k = kg T−1 A−1 which got one mark. We also saw much ‘kg = kTA + kg + kg’ which the student then tried to simplify by cancelling kg as if performing conventional algebra. Over 60% made some progress and use of F = BIL and N = kg m s−2 were often seen. The use of simultaneous equations to determine Y in 03.4 was widespread. Predictably, the use of the wrong scale to determine the M2 intercept was also common. The M1 and M2 intercepts were usually recorded to 2 decimal places as we required and 16% gave completely correct solutions for three marks. Those reading off M2 incorrectly as 135.35 could still get two marks. Some thought that reading off M1 and M2 intercepts at a particular non-zero current, usually 5.2 A, would allow them to find Y. The majority of these attempts failed because the student did not appreciate that B was different for each of the simultaneous equations so they could not cancel the kB terms. A very small number did recognise that problem and worked out both gradients to arrive at a completely correct value for Y. As in 03.2, the writing in 03.5 often betrayed a poor understanding of how the physics of the current balance worked. Several referred to flux cutting or alluded to induction in other ways. However, the general trend was to spot that B would be smaller so the gradient would be less steep, and the intercept stayed the same. Some saw that the gradient was kB so a smaller B meant a smaller gradient while others saw that the intercept was 2Z + Y and hence independent of B. However, very few rarely put the whole thing together, so we saw lots of two-mark answers but very few with all three. We expected students to know that the field between the magnets was uniform, but it did not end abruptly at the top edges of the yoke, yet some said that B at the rod was undetectable. They concluded that the gradient would be zero and the intercept was unchanged. We allowed one mark for this idea.
A horizontal current-carrying rod lies in a magnetic field and rests on a balance. A yoke holds removable magnets above and below the rod. The rod current is horizontal and the measured balance reading M changes linearly with current I. The model is M = kBI + nZ + Y, where B is magnetic flux density, n is the number of magnets, Z is the mass of one magnet and Y is the yoke mass.
AQA 7408 · June 2023 · Paper 2 · Q3 · 10 marks
10 marks
Examiner report
[03.3] It was most common for students to determine gS and gE individually rather than going straight to the ratio (as in the alternative on the mark scheme). The major problems for most students were the choice of correct distances to substitute into their working equations or failing to square the chosen distance in their calculations. Adding the Sun’s radius to the distance or just using the Sun’s radius in the calculation of gS was a common error.
A space probe of fixed mass moves radially away from Earth. A force-distance graph shows the gravitational force on it decreasing with height; the area from Earth's surface to height 8.0 x 10^6 m is shaded. At Earth's surface define g_E = 9.81 N kg^-1 and let g_S be the Sun's field there. Earth-Sun distance is 1.50 x 10^11 m.
AQA 7408 · June 2023 · Paper 2 · Q4 · 8 marks
8 marks
Examiner report
[04.2] Many students knew that the energy provided both to the electron and the argon ion was the same. Using this information to calculate the ratio of speeds then contained errors and so marks were missed by many. A significant minority (more than 10% of students) made no attempt at answering this question.
Figure 4 shows a spark detector used to detect alpha particles. Figure 4 The detector consists of a metal mesh placed 5.0 mm above a wire. A potential difference of 4000 V is applied between the mesh and the wire. Molecules in the air between the mesh and the wire are ionised by an alpha particle and a spark is produced. Figure 5 shows equipotentials between the mesh and the wire. Figure 5
AQA 7408 · June 2023 · Paper 3A · Q3 · 16 marks
16 marks
Examiner report
[03.2] Many recognised that cos 25 was involved in the calculation but to get credit we needed to see this used in a valid percentage-difference calculation. About a third of the students gave fully correct solutions.
A vertical circular coil of radius r carries constant direct current. Q is its centre and PR is the horizontal axis through Q. A sensor measures the horizontal flux density B_H along this axis; Earth's field is negligible. Later, two identical coaxial coils are separated by r and each carries 225 mA.
AQA 7408 · June 2022 · Paper 2 · Q3 · 12 marks
12 marks
Examiner report
[03.2] Almost half the students failed at the first stage by not using a tangent, but taking single values of V and r at r = 0.3 m. Those using a tangent approach began by drawing a tangent but then often failed to gain marks for a number of reasons. First, the triangle from which they calculated the gradient was too small. Also, at this stage powers of ten errors crept in because the potential scale was not recognised. This error was condoned in working out the gradient but was penalised in the final answer. Some students failed to give the final answer to better than one significant figure. Students who only scored one mark tended to get the final unit mark correct. Only a handful of students who went back to fundamental base units in this final part were successful.
An isolated solid conducting sphere of radius 0.10 m is initially uncharged. Electrons are transferred to it. Figure 3 shows V against r: V=-1.8 x 10^6 V from r=0 to 0.10 m, then rises smoothly through approximately (-0.9 x 10^6 V, 0.20 m), (-0.60 x 10^6 V, 0.30 m) and (-0.30 x 10^6 V, 0.60 m).
AQA 7408 · June 2022 · Paper 2 · Q4 · 12 marks
12 marks
Examiner report
[04.3] A very straightforward question in which most students were successful. A few failed by not realising that the equation, depending on how it was presented, had a square or square-root term. Others made numerical or re-arrangement errors.
Figure 4 shows gravitational field lines around stars G and H. There are more lines around G, and the zero-field point between the stars is closer to H. At X a line runs towards G; at Y a line curves towards H. Later, asteroid P is shown close to H with GP perpendicular to HP.
AQA 7408 · June 2022 · Paper 2 · Q5 · 11 marks
11 marks
Examiner report
[05.2] This is another question in which the marks were low because the students gave very generic answers rather than considering the specific situation. Only the best students appreciated that they had to describe a feature of the design and then explain how this improves the efficiency. Most students knew that the efficiency is reduced by eddy currents but made no reference to how the core in Figure 8 reduced these eddy currents. Those with the correct idea often failed to gain marks due to dealing in absolutes; they stated that eddy currents are ‘stopped’ rather than describing them as being ‘reduced’.
Figure 7 shows a transformer. Figure 7
AQA 7408 · November 2021 · Paper 2 · Q1 · 12 marks
12 marks
Examiner report
[01.3] About half of the students could calculate the initial stored energy from the standard equation but the majority failed to calculate the final stored energy correctly. Most did not realise that the capacitance increases by a factor of 6 and chose to increase either the charge or the energy by this factor.
An isolated 63 pF parallel-plate air capacitor holds charge 7.6 x 10^-10 C. Mica of dielectric constant 6.0 is inserted fully without discharge or plate-separation change. A separate variable capacitor has two overlapping semicircular plates.
AQA 7408 · November 2021 · Paper 2 · Q2 · 10 marks
10 marks
Examiner report
[02.5] A few able students justified the statement that the linear speed is smaller for higher orbits which they worked out either from the equation given in part 02.1 or from first principles. A majority thought the linear speed was greater, often justifying this by incorrectly taking the angular velocity to be constant.
A GPS satellite orbits Earth circularly at height h. Use Earth mass 5.97 x 10^24 kg, radius 6.37 x 10^6 m. For numerical parts h=2.02 x 10^7 m and satellite mass 1630 kg.
AQA 7408 · November 2021 · Paper 2 · Q4 · 8 marks
8 marks
Examiner report
[04.3] The misinterpretation of Figure 8 persisted here and very few students could determine the variation of magnetic flux density with time. Even fewer students realised that the question involved finding the maximum gradient of the graph within a distance range at 0.10 m x = .
A 200-turn search coil of area 3.5 x 10^-5 m^2 moves at 0.80 m s^-1 along an electromagnet axis. The B-x graph decreases with x; B=0.070+/-0.005 T at x=0.070 m. For x>=0.10 m its greatest magnitude gradient is at x=0.10 m, where a tangent crosses about (0 m,0.095 T) and (0.137 m,0 T).
AQA 7408 · November 2021 · Paper 2 · Q5 · 10 marks
10 marks
Examiner report
[05.1] Generally students did not explain the direction in which the force acts in a clear way. Only a few could identify the magnetic force as centripetal.
A cyclotron accelerates protons across its gap at peak alternating pd 10.0 kV. Uniform vertical magnetic field B bends each constant-speed path inside the dees. Proton exit energy is initially 14 MeV. For selection, cyclotrons are X: B=1.3 T,R=0.38 m; Y:1.1 T,0.50 m; Z:0.5 T,0.60 m.
AQA 7408 · November 2021 · Paper 3A · Q3 · 19 marks
19 marks
Examiner report
This question was set in the context of Required Practical Activity 9. In question 03.1 more students chose the distractor 13.5 than the correct value of 2.7, suggesting that many did not appreciate the distinction between ‘more sensitive’ and ‘less sensitive’. Question 03.2 is a question used in an earlier series. Some students improvised answers such as ‘the mirror was used to illuminate the scale to eliminate human error’. Only 10% could supply a completely correct answer. Question 03.3 was poorly answered in comparison with similar examples in previous papers, with many failing to halve the range before calculating the percentage uncertainty. In question 03.5 most found that the space provided was adequate for them to complete their answer; those who needed supplementary pages usually included information already provided in the question. Many could suggest something sensible about meter protection, checking that C was discharged being the most popular idea. Some recognised the advantage of timing for longer but suggested impractical ideas such as ‘measure the time for C to fully discharge’ or ‘add a resistor to the circuit’. Few could state how they would test for any relationship between range and time constant. Weaker students wrote that they would check whether the half life was 2.5 s. More than half were able to score one mark in question 03.6 by making at least one relevant observation about the values of V in Table 2, although finding an explanation to back up their observation was a harder proposition. In question 03.7 at least 30% knew how to find the voltmeter resistance (as per page 116 of the physics practical handbook). However, half of the students, despite often obtaining relevant data from Figure 8, then failed to make enough progress to score. In question 03.8 it was surprising to find that many, having correctly read relevant information from Figure 8, then attempted to calculate the current using 3 10 10 16 ln × V . Concluding Remarks The demands of this paper were no greater than in 2019. Besides demonstrating knowledge and understanding of the required practical activities, students must show that they can apply their core understanding of practical work in new and unfamiliar contexts. Some will not have seen an analogue meter and we would not expect any to have heard about the dead time of a radiation detector, but this would not have affected their ability to tackle the questions we set. Question 1 revisited Required Practical Activity 12 which featured in 2019 and this will have benefited those who looked at past papers and schemes. Specification coverage of theory leading up to Required Practical Activities 12 and 9 (see question 3) would normally take place later in the course, so both of these should have been fresh in students’ memory. However this assumes that students will have had uninterrupted study and this clearly will not be true in every case. The impression gained was that students felt more comfortable writing about capacitor discharge than about the inverse-square law. Where possible they should be given direct experience of using equipment relevant in each of the practical activities, including the safe manipulation of a sealed radioactive source. As might be expected, work at grade A was in very short supply, but a handful of capable students submitted answers and some of them found innovative variations on the expected approaches to question 03.7. At the other end of the spectrum we saw evidence of calculators set to read angles in radians (in question 02.1) and misguided attempts to use the Boltzmann constant (in question 01.3). Aside from questions 01.2, 01.3, 02.1 and 03.3, students at about the E boundary found marks difficult to obtain. Some of the work seen in questions 01.1 and 03.2 show that, while students know about parallax error, they are not always aware of techniques used to limit its effect. This is an area we will continue to test and the best way to acquire knowledge of such techniques is by direct practical experience. Work seen in question 01.4 suggests that some students wrongly assume that sealed sources emit γ radiation only through the open end of the source. Students should be made aware of the expectations behind the different command words used in the questions. When we use ‘Explain’ we expect a reasoned justification to be included and when, as in question 01.3, they are basing a decision based on a numerical outcome, the justification should be quantitative, eg a percentage-difference calculation. There is always scope in this paper to test the Maths Skills listed in section 6 of the specification. Question 01.6 exposed weak understanding about the use of logarithmic plots to test power law variations (MS3.11). Relevant opportunities for skills development are identified in the main body of the specification.
A dual-range analogue voltmeter has 3 V and 15 V ranges. On the 3 V scale its uncorrected zero is +0.3 V. A mirror runs along the scale. After zero correction, an unknown capacitor discharges through the meter. For the later 820 microF discharge, Figure 8 is a straight best-fit plot of ln(V / 1 V) against t passing approximately through (0 s,2.64) and (28 s,0.50).
AQA 7408 · November 2020 · Paper 2 · Q2 · 14 marks
14 marks
Examiner report
[02.2] A large number of students could not convert the equation into one showing proportionality and then substitute data correctly. Alternatively, they calculated k but then failed to rearrange the equation to make the radius the required subject. A few students failed to get started on the problem. However, a majority of the students found the question to be very straightforward and headed straight to the answer.
A moon of mass m moves in a circular orbit of radius r and period T around a planet of mass M, where m is much smaller than M and T^2=kr^3.
AQA 7408 · November 2020 · Paper 2 · Q3 · 11 marks
11 marks
Examiner report
[03.5] An energy approach to answering the question was the most successful and about one-sixth of the students gained full marks. Many other students were partially successful but these did not appreciate that the potential difference change for the ion was only half of the 6000 V. Those answers gaining no marks normally showed an inappropriate equation.
A positive lithium-ion beam travels left to right through crossed fields. The electric field is vertically downward and the magnetic field is into the page. Undeflected ions pass through an aperture into a second magnetic field.
AQA 7408 · November 2020 · Paper 2 · Q4 · 9 marks
9 marks
Examiner report
[04.2] A lack of experience or knowledge of the oscilloscope was shown in this straightforward question, with only about half of the students obtaining the correct answer.
An oscilloscope is connected across a resistor. A sinusoidal supply has peak voltage 15 V and appears with a peak displacement of 3 divisions.
AQA 7408 · June 2019 · Paper 2 · Q3 · 6 marks
6 marks
Examiner report
[3.2] More than half the students gave a quick answer and said why the field was not uniform. Many of the others did not really know what to say and answers were often very confused. Some who gave a clear statement thought incorrectly that in a uniform field the potential should not change at all.
A diagram gives equipotential lines outside the Moon; Moon radius is 1.74 x10^6 m and one line is V=-1.60 x10^6 J kg^-1 at r=3.06 x10^6 m.
AQA 7408 · June 2018 · Paper 2 · Q4 · 8 marks
8 marks
Examiner report
[04.5] Students found this more difficult than the previous question (42.8% completely correct). A minority did know the emf equation to use but they could not obtain a maximum value. They failed to replace the trigonometric factor with 1.
A horizontal solenoid is connected to an ammeter by two downward leads: terminal E is on the left lead and terminal F on the right lead. A bar magnet is to the right of the solenoid, with its north pole facing and approaching the solenoid's right-hand face. For current E-to-F through the bottom ammeter branch, current enters the solenoid at F and is anticlockwise when its right-hand face is viewed from the magnet. A separate one-turn horizontal square coil side32 mm moves wholly into uniform vertical B=0.38T, inducing2.9 x10^-4V, then rotates in field with peak5.1mV.
AQA 7408 · June 2018 · Paper 3A · Q1 · 19 marks
19 marks
Examiner report
[01.4] The first row of Table 1 attracted the most correct responses and thereafter things got patchier. Examiners needed to see correct responses in both row 3 and row 4 to earn the third marking point; this proved beyond all but the most able, with only 11.4% obtaining full credit.
A practical uses a metre ruler pivoted at1.0 cm and supported by a prism on a zeroed balance at30.0 cm; with the horizontal ruler alone the balance reads214.66 g. In the side-view diagram, terminal P is at the left end and Q at the right end of a wire along the ruler. The supply makes conventional current flow P-to-Q, rightwards across a5.0 cm horizontal magnetic-field region at the85-90 cm marks. The straight M-against-I graph has gradient -0.28 g A^-1.
AQA 7408 · June 2017 · Paper 2 · Q2 · 5 marks
5 marks
Examiner report
[02.4] A majority of students tackled this question with little trouble. The most common error was in failing to convert the mass from g to kg.
A horizontal wire DE, length 50.0 mm, lies perpendicular to a uniform field between magnets standing on an electronic balance. The balance reading increases by 0.620 g when current 3.43 A is switched on.
AQA 7408 · June 2017 · Paper 2 · Q3 · 6 marks
6 marks
Examiner report
[03.2] More than half the students could present the starting equation relating the force due to the magnetic field, to the centripetal force. Many continued to successfully find an expression for the period of motion. However, only some of these referred back to the question to give the time to travel round one semi-circle. Many of the other students got into difficulties if their initial expression for the centripetal force did not involve velocity. From this point on, they often took numerous false trails and the work presented looked very disorganised. Another route taken by some was to reduce their first equation to R = mv / BQ and simply say that this radius does not involve time and is constant. This not only re-writes the question, but misses the point that v is different on each transit.
A proton of mass m and charge Q enters a uniform magnetic field B at right angles and follows a semicircle of radius r before leaving the field.
AQA 7408 · June 2017 · Paper 2 · Q6 · 9 marks
9 marks
Examiner report
[06.2] Many students appreciated that the gravitational field was stronger at K but then did not give a good reason for it. They often referred to the earth not being flat, or stated that K is up a mountain or down a hole. Other students did state that there was a mass at K or it was an area where the Earth’s density is greater, but they did not refer to the field being stronger. Also a few students referred to gravitational force when they were really referring to gravitational field.
Diagrams compare gravitational and electric fields. In the gravitational diagram, field lines converge near point K beneath point J. In the electric diagram, lines leave a larger object L and terminate on a smaller object M; point N lies between them.
AQA 7408 · June 2017 · Paper 3A · Q2 · 14 marks
14 marks
Examiner report
This question addressed the ideas behind assessed practical activity 9. Students who had seen the second set of specimen questions should have been well prepared for the parts relating to the use of the oscilloscope. It was clear from the descriptive parts of this question, however, that many students had not had direct experience of using an oscilloscope, as shown by the lack of much, if any, use of correct terminology to describe the controls. Even if some students thought the waveform shown in Figure 8 showed an ac signal, the instruction to determine the peak to peak voltage should have made question 02.1 straightforward. However, a disappointing number of students found a variety of ways to get this wrong and were similarly unsuccessful with question 02.2. Those students who obtained 6.3 V and 250 Hz were well-placed to score in question 02.3 and a popular and successful approach was to use RC t e V V ∆ − = 0 ; a transparent, viable attempt along these lines could earn a mark, even if the final answer fell outside the required range. Others read off the time interval for V to fall to 37% when the capacitor was discharging, but poor communication often made it harder to give any credit, if the final result obtained by this method was unsuccessful. Those using the idea that the capacitor would completely discharge in five time constants were almost invariably unsuccessful due to difficulty in accurately pinpointing the time for 5RC from Figure 10. In question 02.4, most students knew what would happen to the waveform displayed in the oscilloscope when the time-base was adjusted, but often struggled to give a convincing explanation. Stating that the ‘waveform would not fit’ left examiners looking for further evidence to decide whether the student was referring to vertical or horizontal direction. Some students referred to the ‘wavelength increasing’, or equivocated by suggested that the waveform ‘might not fit’. While better answers suggested that either the charging or discharging parts (but not both) would now be displayed, very few students took a quantitative approach, e.g. that the discharge curve would now be 2.5 times wider. Here and elsewhere in the paper, some students explained that uncertainty would be reduced, but without supporting reasoning this gained no credit. Question 02.5 discriminated well, with better students stating that the resistance, and hence the time constant, is halved, and spotted the opportunity to give the quantitative detail suggested by the information in the question. The sketches on Figure 11 showed that some students thought that adding the extra resistor would change the amplitude or the period of the waveform; relatively few students supplied the sketch examiners were looking for. In the sketch for question 02.6, examiners wanted to see a complete half cycle while the signal generator output was V, and a complete half cycle when the output was zero. Students were generally more successful with the latter, but often failed to consider how Figures 12a and 12b could be used in combination to deduce the other part of the waveform. In question 02.7, “reduce Y-gain”, or “increase volt per division” could score but an equivocal “change the Y-gain” could not. Once again, extra credit was available for quantitative detail such as “set the gain to 2 V per division”. Some students clearly failed to appreciate that Figure 12c showed a graph and was not the way the waveform would appear on the oscilloscope; these students stated that the waveform was already fully visible, so no changes were necessary.
A signal generator produces a square wave. The oscilloscope settings are 1 V division^-1 and 0.5 ms division^-1. The trace spans 6.3 vertical divisions peak-to-peak and one period spans 8 horizontal divisions. With a series R-C circuit and the oscilloscope across C, each transition is exponential between +3.15 V and -3.15 V; after a step, the trace crosses 0 V about 0.175 ms later.
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