Write down the maximum magnitude of the rate of change of flux linked with the coil.

State the fundamental SI unit for your answer to (a)(i).

Explain why the graph becomes negative.

Part of the graph is above the t-axis and part is below. Outline why the areas between the t-axis and the curve for these two parts are likely to be the same.

Predict the changes to the graph when the magnet is dropped from a lower height above the coil.

«−» 5.0 «mV»  OR  5.0 × 10⁻³ «V» ✓

Accept 5.1

kg m²A⁻¹s⁻³ ✓

ALTERNATIVE 1

Flux linkage is represented by magnetic field lines through the coil ✓

when magnet has passed through the coil / is moving away ✓

flux «linkage» is decreasing ✓

suitable comment that it is the opposite when above ✓

when the magnet goes through the midpoint the induced emf is zero ✓

ALTERNATIVE 2

reference to / states Lenz’s law ✓

when magnet has passed through the coil / is moving away ✓

«coil attracts outgoing S pole so» induced field is downwards ✓

before «coil repels incoming N pole so» induced field is upwards
OR
induced field has reversed ✓

when the magnet goes through the midpoint the induced emf is zero ✓

OWTTE

area represents the total change in flux «linkage» ✓

the change in flux is the same going in and out ✓

«when magnet is approaching» flux increases to a maximum ✓

«when magnet is receding» flux decreases to zero ✓

«so areas must be the same»

magnet moves slower ✓

overall time «for interaction» will be longer ✓

peaks will be smaller ✓

areas will be the same as before ✓

Allow a graphical interpretation for MP2 as “graph more spread out”

ai
Well answered, with wrong answers stating 8 for the difference or 3 without realising that the sign does not matter.

aii
Very few candidates managed to get the correct fundamental SI unit for V. All kinds of errors were observed, from power errors to the use of C as a fundamental unit instead of A.

bi) Most scored best by marking using an alternative method introduced to the markscheme in standardisation. There were some confused and vague comments. Clear, concise answers were rare.

bii) It was common to see conservation of energy invoked here with suggestions that energy was the area under the graph. Many candidates described the shapes to explain why the areas were the same rather than talking about the physics e.g. one peak is short and fat and the other is tall and thin so they balance out.

c) A surprising number didn't pick up on the fact that the magnet would be moving slower. As a result, they discussed everything happening sooner, i.e. the interaction with the magnet and the coil, and that led onto things happening quicker so peaks being bigger.

Explain the role of the diode in the circuit when the switch is at position A.

Show that the maximum energy stored by the capacitor is about 160 J.

Calculate the maximum charge Q₀ stored in the capacitor.

Identify, using the label + on the diagram, the polarity of the capacitor.

Describe what happens to the energy stored in the capacitor when the switch is moved to position B.

Show that the charge remaining in the capacitor after a time equal to one time constant $\tau$ of the circuit will be 0.37 Q₀.

The graph shows the variation with time of the charge in the capacitor as it is being discharged through the heart.

Determine the electrical resistance of the closed circuit with the switch in position B.

In practice, two electrodes connect the heart to the circuit. These electrodes introduce an additional capacitance.

Explain the effect of the electrode capacitance on the discharge time.

to charge a capacitor current must be direct ✓

diode will only allow current to flow in one direction

OR

the diode provides half wave rectification ✓

$V_{\text{s}}=15\times 220=«3300 \text{V»}$ ✓

$E=\frac{1}{2}C{V}^2=\frac{1}{2}\times 30\times {10}^{-6}\times {3300}^2$

OR

163 «J» ✓

Allow use of 220 V as an RMS value to calculate V_s = 467 V and E = 327 J for full marks if appropriate work is provided.
Answer must be to 3 or more sf or working shown for MP2

$Q_0=0.0989\approx 0.1 «\text{C}»$ ✓

Allow ECF from (b)(i) (Q = 30 μF x V)

labels + on the lower side of the capacitor ✓

the energy stored in the capacitor is delivered to the resistor/heart ✓

$\text{use of }Q=Q_0{e}^{-\frac{t}{\tau }} \text{to show that} 0.37=\frac{1}{e}$ ✓

ALTERNATIVE 1

reads from the graph $\tau =1.6 \text{ms}$ ✓

so $R=\frac{0.0016}{30\times {10}^{-6}}=53 «\mathrm{\Omega }»$✓

ALTERNATIVE 2

reads a correct value from the graph for $\frac{Q}{Q_0}$ and $t$ ✓

so $R=\frac{t}{\ln \left({\displaystyle \frac{Q}{Q_0}}\right)\left(3\times {10}^{-5}\right)}$ ✓

«the capacitors are in parallel hence» capacitances are added / more charge is stored
OR
C_eq is larger
OR
electrode capacitor charges and discharges ✓

«therefore» discharge takes longer/increases ✓

Electrical power output is produced by several alternating current (ac) generators which use transformers to deliver energy to the national electricity grid.

The following data are available. Root mean square (rms) values are given.

(i) Calculate the current output by the transformer to the grid. Give your answer to an appropriate number of significant figures.

(ii) Electrical energy is often delivered across large distances at 330 kV. Identify the main advantage of using this very high potential difference.

In an alternating current (ac) generator, a square coil ABCD rotates in a magnetic field.

The ends of the coil are connected to slip rings and brushes. The plane of the coil is shown at the instant when it is parallel to the magnetic field. Only one coil is shown for clarity.

The following data are available.

(i) Explain, with reference to the diagram, how the rotation of the generator produces an electromotive force (emf ) between the brushes.

(ii) Calculate, for the position in the diagram, the magnitude of the instantaneous emf generated by a single wire between A and B of the coil.

(iii) Hence, calculate the total instantaneous peak emf between the brushes.

i
$I=0.96\times \left(\frac{25\times {10}^3\times 3.9\times {10}^3}{330\times {10}^3}\right)$
Award [2] for a bald correct answer to 2 sf.
Award [1 max] for correct sf if efficiency used in denominator leading to 310 A or if efficiency ignored (e=1) leading to 300 A (from 295 A but 295 would lose both marks).

=280 «A»
Must show two significant figures to gain MP2.

ii
higher V means lower I «for same power»

thermal energy loss depends on I or is ∝I² or is I²R so thermal energy loss will be less
Accept “heat” or “heat energy” or “Joule heating” for “thermal energy”.
Reference to energy/power dissipation is not enough.

i

«long» sides of coil AB/CD cut lines of flux
OR
flux «linkage» in coil is changed  

«Faradays law:» induced emf depends on rate of change of flux linked
OR
rate at which lines are cut

“Induced” is required
Allow OWTTE or defined symbols if “induced emf” is given.
Accept “induced” if mentioned at any stage in the context of emf or accept the term “motional emf”.
Award [2 max] if there is no mention of “induced emf”.

emfs acting in sides AB/CD add / act in same direction around coil

process produces an alternating/sinusoidal emf

ii

Blv = 0.34×8.5×10^–2×2 = 0.058 «V»  

Accept 0.06V.

iii

160×(c)(ii) = 9.2 or 9.3 or 9.6 «V»  

Allow ECF from (c)(ii)
If 80 turns used in cii, give full credit for cii x 2 here.

Satellite X orbits 6600 km from the centre of the Earth.

Mass of the Earth = 6.0 x 10²⁴ kg

Show that the orbital speed of satellite X is about 8 km s^–1.

the orbital times for X and Y are different.

satellite Y requires a propulsion system.

The cable between the satellites cuts the magnetic field lines of the Earth at right angles.

Explain why satellite X becomes positively charged.

Satellite X must release ions into the space between the satellites. Explain why the current in the cable will become zero unless there is a method for transferring charge from X to Y.

The magnetic field strength of the Earth is 31 μT at the orbital radius of the satellites. The cable is 15 km in length. Calculate the emf induced in the cable.

Estimate the value of k in the following expression.

T = $2\pi \sqrt{\frac{m}{k}}$

Give an appropriate unit for your answer. Ignore the mass of the cable and any oscillation of satellite X.

Describe the energy changes in the satellite Y-cable system during one cycle of the oscillation.

«$v=\sqrt{\frac{G{M}_E}{r}}$» = $\sqrt{\frac{6.67\times {10}^{-11}\times 6.0\times {10}^{24}}{6600\times {10}^3}}$

7800 «m s^–1»

Full substitution required

Must see 2+ significant figures.

Y has smaller orbit/orbital speed is greater so time period is less

Allow answer from appropriate equation

Allow converse argument for X

to stop Y from getting ahead

to remain stationary with respect to X

otherwise will add tension to cable/damage satellite/pull X out of its orbit

cable is a conductor and contains electrons

electrons/charges experience a force when moving in a magnetic field

use of a suitable hand rule to show that satellite Y becomes negative «so X becomes positive»

Alternative 2

cable is a conductor

so current will flow by induction flow when it moves through a B field

use of a suitable hand rule to show current to right so «X becomes positive»

Marks should be awarded from either one alternative or the other.

Do not allow discussion of positive charges moving towards X

electrons would build up at satellite Y/positive charge at X

preventing further charge flow

by electrostatic repulsion

unless a complete circuit exists

«ε = Blv =» 31 x 10^–6 x 7990 x 15000

3600 «V»

Allow 3700 «V» from v = 8000 m s^–1.

use of k = «$\frac{4{\pi }^{2}m}{T^2}=$» $\frac{4\times {\pi }^2\times 350}{{5.2}^2}$

510

N m^–1 or kg s^–2

Allow MP1 and MP2 for a bald correct answer

Allow 500

Allow N/m etc.

E_p in the cable/system transfers to E_k of Y

and back again twice in each cycle

Exclusive use of gravitational potential energy negates MP1

Explain, by reference to Faraday’s law of induction, how an electromotive force (emf) is induced in the coil.

The average power output of the generator is $8.5\times {10}^5 \mathrm{W}$. Calculate the root mean square (rms) value of the generator output current.

The voltage output from the generator is stepped up before transmission to the consumer. Estimate the factor by which voltage has to be stepped up in order to reduce power loss in the transmission line by a factor of $2.5\times {10}^2$.

The frequency of the generator is doubled with no other changes being made. Draw, on the axes, the variation with time of the voltage output of the generator.

there is a magnetic flux «linkage» in the coil / coil cuts magnetic field ✓

this flux «linkage» changes as the angle varies/coil rotates ✓

«Faraday’s law» connects induced emf with rate of change of flux «linkage» with time ✓

Do not award MP2 or 3 for answers that don’t discuss flux.

$V_{\mathrm{rms}}=\frac{25\times {10}^3}{\sqrt{2}}«=17.7\times {10}^3 \mathrm{V}»$ ✓

$I_{\mathrm{rms}}=\frac{8.5\times {10}^5}{17.7\times {10}^3}=48 «\mathrm{A}»$ ✓

«power loss proportional to $I^2$ hence the step-up factor is $\sqrt{2.5\times {10}^2}» 16$ ✓

peak emf doubles ✓

$T$ halves ✓

Must show at least 1 cycle.

This question was well answered with the majority discussing changes in flux rather than wires cutting field lines, which was good to see.

Generally well answered.

This was well answered by many, but some candidates left the answer as a surd. The most common guess here involved the use of root 2.

Well answered, with the majority of candidates scoring at least 1 mark.

Each rod is to have a resistance no greater than 0.10 Ω. Calculate, in m, the minimum radius of each rod. Give your answer to an appropriate number of significant figures.

Calculate the maximum number of lamps that can be connected between the rods. Neglect the resistance of the rods.

One advantage of this system is that if one lamp fails then the other lamps in the circuit remain lit. Outline one other electrical advantage of this system compared to one in which the lamps are connected in series.

Outline how eddy currents reduce transformer efficiency.

Determine the peak current in the primary coil when operating with the maximum number of lamps.

ALTERNATIVE 1:

$r=\sqrt{\frac{\rho l}{\pi \text{R}}}$ OR $\sqrt{\frac{7.2\times {10}^{-7}\times 12.5}{\pi \times 0.1}}$ ✔

r = 5.352 × 10⁻³ ✔

5.4 × 10⁻³ «m» ✔

For MP2 accept any SF

For MP3 accept only 2 SF

For MP3 accept ANY answer given to 2 SF

ALTERNATIVE 2:

$A=\frac{7.2\times {10}^{-7}\times 12.5}{0.1}$ ✔

r = 5.352 × 10⁻³ ✔

5.4 × 10⁻³ «m» ✔

For MP2 accept any SF

For MP3 accept only 2 SF

For MP3 accept ANY answer given to 2 SF

current in lamp = $\frac{5}{24}$ «= 0.21» «A»

OR

n = 24 × $\frac{8}{5}$ ✔

so «38.4 and therefore» 38 lamps ✔

Do not award ECF from MP1

when adding more lamps in parallel the brightness stays the same ✔

when adding more lamps in parallel the pd across each remains the same/at the operating value/24 V ✔

when adding more lamps in parallel the current through each remains the same ✔

lamps can be controlled independently ✔

the pd across each bulb is larger in parallel ✔

the current in each bulb is greater in parallel ✔

lamps will be brighter in parallel than in series ✔

In parallel the pd across the lamps will be the operating value/24 V ✔

Accept converse arguments for adding lamps in series:

when adding more lamps in series the brightness decreases

when adding more lamps in series the pd decreases

when adding more lamps in series the current decreases

lamps can’t be controlled independently

the pd across each bulb is smaller in series

the current in each bulb is smaller in series

in series the pd across the lamps will less than the operating value/24 V

Do not accept statements that only compare the overall resistance of the combination of bulbs.

«as flux linkage change occurs in core, induced emfs appear so» current is induced ✔

induced currents give rise to resistive forces ✔

eddy currents cause thermal energy losses «in conducting core» ✔

power dissipated by eddy currents is drawn from the primary coil/reduces power delivered to the secondary ✔

power = 190 OR 192 «W» ✔

required power $=190\times \frac{100}{95}$ «200 or 202 W» ✔

so $\frac{200}{240}=0.83$ OR 0.84 «A rms» ✔

peak current = «$0.83\times \sqrt{2}$ OR $0.84\times \sqrt{2}$» = 1.2/1.3 «A» ✔

The battery has an emf of 7.5 V. Determine the charge that flows through the motor when the mass is raised.

The motor can transfer one-third of the electrical energy stored in the capacitor into gravitational potential energy of the mass. Determine the maximum height through which a mass of 45 g can be raised.

An additional identical capacitor is connected in series with the first capacitor and the charging and discharging processes are repeated. Comment on the effect this change has on the height and time taken to raise the 45 g mass.

charge stored on capacitor = 12 × 10⁻³ × 7.5 = 0.09 «C» ✔

energy stored in capacitor «$\frac{1}{2}$CV² or $\frac{1}{2}$QV =» $\frac{1}{2}$ × 12 × 10⁻³ × 7.5² «= 0.338 J» ✔

height = «$\frac{1}{3}\times \frac{0.338}{9.81\times 4.5\times {10}^{-2}}=$» 0.25/0.26 «m»

Allow use of g = 10 m s⁻² which gives 0.25 «m»

C halved ✔

so energy stored is halved/reduced so rises «less than» half height ✔

discharge time/raise time less as RC halved/reduced ✔

Allow 6 mF

The primary coil has 3300 turns. Calculate the number of turns on the secondary coil.

Determine the total resistance of the lamps when they are working normally.

Calculate the current in the primary of the transformer assuming that it is ideal.

Flux leakage is one reason why a transformer may not be ideal. Explain the effect of flux leakage on the transformer.

A pendulum with a metal bob comes to rest after 200 swings. The same pendulum, released from the same position, now swings at 90° to the direction of a strong magnetic field and comes to rest after 20 swings.

Explain why the pendulum comes to rest after a smaller number of swings.

$«\frac{15}{110}\times 3300=» 450 «\text{turns}»$ ✓

ALTERNATIVE 1

calculates total current $\mathrm{=}\frac{35}{15}\times 8«=18.7\text{\hspace{0.05em}A}»$✓

resistance $=«\frac{15}{18.7}=» 0.80 «\text{Ω}»$ ✓

ALTERNATIVE 2

calculates total power $= 35\times 8 \text{« = 280\hspace{0.05em}W»}$ ✓

resistance $=«\frac{{15}^2}{280}=» 0.80 «\text{Ω}»$✓

ALTERNATIVE 3

calculates individual resistance $=\frac{{15}^2}{35}«=6.43 \text{Ω}»$✓

resistance $=«\frac{6.43}{8}»=0.80 «\text{Ω}»$✓

total power required = 280 «W»
OR
uses factor $\frac{3300}{450}$
OR
total current = 18.7 « A» ✓

current = 2.5 OR 2.6 «A» ✓

Award [2] marks for a bald correct answer.

Allow ECF from (a)(ii).

the secondary coil does not enclose all flux «lines from core» ✓

induced emf in secondary
OR
power transferred to the secondary
OR
efficiency is less than expected ✓

Award [0] for references to eddy currents/heating of the core as the reason.

Award MP2 if no reason stated.

bob cuts mag field lines
OR
there is a change in flux linkage ✓

induced emf across bob ✓

leading to eddy/induced current in bob ✓

eddy/induced current produces a magnetic field that opposes «direction of» motion ✓

force due to the induced magnetic field decelerates bob ✓

damping of pendulum increases/there is additional «magnetic» damping ✓

MP4 and MP5 can be expressed in terms of energy transfer from kinetic energy of bob to electrical/thermal energy in bob

Calculate the radius of each wire.

Calculate the peak current in the cable.

Determine the power dissipated in the cable per unit length.

Calculate the root mean square (rms) current in each cable.

The two cables in part (c) are suspended a constant distance apart. Explain how the magnetic forces acting between the cables vary during the course of one cycle of the alternating current (ac).

Suggest the advantage of using a step-up transformer in this way.

The use of alternating current (ac) in a transformer gives rise to energy losses. State how eddy current loss is minimized in the transformer.

area = $\frac{1.7\times {10}^{-3}\times 35\times {10}^3}{64}$ «= 9.3 x 10^–6 m²»

radius = «$\sqrt{\frac{9.3\times {10}^{-6}}{\pi }}=$» 0.00172 m

I_peak «$=\frac{P_{peak}}{V_{peak}}$» = 730 « A »

resistance of cable identified as «$\frac{64}{32}=$» 2 Ω

$\frac{\text{a power}}{35000}$ seen in solution

plausible answer calculated using $\frac{\text{2}{\text{I}}^{\text{2}}}{35000}$ «plausible if in range 10 W m^–1 to 150 W m^–1 when quoted answers in (b)(ii) used» 31 «W m^–1»

Allow [3] for a solution where the resistance per unit metre is calculated using resistivity and answer to (a) (resistance per unit length of cable = 5.7 x 10^–5 m )

Award [2 max] if 64 Ω used for resistance (answer x32).

An approach from $\frac{V^2}{R}$ or VI using 150 kV is incorrect (award [0]), however allow this approach if the pd across the cable has been calculated (pd dropped across cable is 1.47 kV).

«$\frac{\text{response to (b)(ii)}}{2\sqrt{2}}$» = 260 «A»

wires/cable attract whenever current is in same direction

charge flow/current direction in both wires is always same «but reverses every half cycle»

force varies from 0 to maximum

force is a maximum twice in each cycle

Award [1 max] if response suggests that there is repulsion between cables at any stage in cycle.

higher voltage gives lower current

«energy losses depend on current» hence thermal/heating/power losses reduced

laminated core

Do not allow “wires are laminated”.

Show that the charge stored on C₁ is about 0.04 mC.

Calculate the energy transferred from capacitor C₁.

Explain why the energy gained by capacitor C₂ differs from your answer in (b)(i).

The switch is closed at time t = 0. Explain how the voltmeter reading varies after the switch is closed.

Determine the average emf induced across coil Y in the first 3.0 ms.

$Q=$«$CV=1.5\times 28\times {10}^{-6}$» = 0.042 «mC» ✓

Award MP for full replacement or correct answer to at least 2 significant figures.

$E_{\text{i}}=\frac{1}{2}\times \left(28\times {10}^{-6}\right){\left(1.5\right)}^2=3.15\times {10}^{-5}$ «J»  ✓

total capacitance = 50 «μF»

OR

pd = «$\frac{42\times {10}^{-6}}{50\times {10}^{-6}}=$» 0.84 «V»

OR

charge on C₁ after switch moved to B = 0.0235 «mC» ✓

$E_{\text{f}}=\frac{1}{2}\times \left(28\times {10}^{-6}\right){\left(0.84\right)}^2=9.9\times {10}^{-6}$ «J» ✓

energy lost $=3.2\times {10}^{-5}-9.9\times {10}^{-6}=22$ «μJ» ✓

energy transferred to electromagnetic radiation «to environment»

OR

energy is transferred as thermal energy / heat «to circuit components» ✓

initial deflection by voltmeter falling to zero reading ✓

emf is induced «only» while the field / flux is changing ✓

attempted use of $\epsilon =\frac{∆\Phi }{∆t}$ OR $NA\frac{∆B}{∆t}$ ✓

$\epsilon =\frac{5\times 7.5\times {10}^{-3}\times 6.4\times {10}^{-4}}{3\times {10}^{-3}}$ ✓

8.0 «mV» ✓

This was a "show that" question, and it was very well done by most candidates.

This question was challenging for many candidates. A large number successfully calculated the initial energy of C1, but then seemed confused about the next steps. Few candidates successfully calculated the energy in C1 after the switch was closed. There was an ECF opportunity for candidates who recognized that the final answer was the difference between these two values.

This question used an "explain" command term, so examiners were looking for more than a generic "energy was lost". Candidates needed to specify a form of energy that was lost (such as thermal energy) for the mark. A very common incorrect response was simply stating that the difference was due to the capacitors having a different capacitance.

This question was well answered by some candidates who recognized that it was an electromagnetic induction question and understood that eventually the current in coil X would hit a steady state and the voltmeter reading would return to zero. Common issues were candidates thinking that the potential would fluctuate in a manner similar to an alternating current, candidates discussing this more as a transformer, and candidates who missed that there were two separate coils and wrote responses suggesting that a simple circuit had been formed and the voltmeter would read the potential of the cell.

This question was well approached with most candidates recognizing that this was a Faraday's law question. Many made an attempt to use the correct equation, but common errors were choosing incorrect values from the graph and incorrectly converting the given area. Examiners were generous with ECF for candidates who clearly showed work leading to an incorrect result.

Calculate the total length of aluminium foil that the student will require.

The plastic film begins to conduct when the electric field strength in it exceeds 1.5 MN C^–1. Calculate the maximum charge that can be stored on the capacitor.

The resistor R in the circuit has a resistance of 1.2 kΩ. Calculate the time taken for the charge on the capacitor to fall to 50 % of its fully charged value.

The ammeter is replaced by a coil. Explain why there will be an induced emf in the coil while the capacitor is discharging.

Suggest one change to the discharge circuit, apart from changes to the coil, that will increase the maximum induced emf in the coil.

length = $\frac{d\times C}{\text{width}\times \epsilon }$ ✔

= 0.33 «m» ✔

so 0.66/0.67 «m» «as two lengths required» ✔

1.5 × 10⁶ × 55 × 10^-6 = 83 «V» ✔

q «= CV»= 5.6 × 10^-6 «C»✔

$0.5={\text{e}}^{-\frac{t}{RC}}={\text{e}}^{-\frac{t}{1200\times 6.8\times {10}^{-8}}}$

t = «−» 1200 × 6.8 × 10⁻⁸ × ln0.5 ✔

5.7 × 10⁻⁵ «s» ✔

OR

use of t $\frac{1}{2}$ = RC × ln2 ✔

1200 × 6.8 × 10⁻⁸ × 0.693 ✔

5.7 × 10⁻⁵ «s» ✔

mention of Faraday’s law ✔

indicating that changing current in discharge circuit leads to change in flux in coil/change in magnetic field «and induced emf» ✔

decrease/reduce ✔

resistance (R) OR capacitance (C) ✔

Many candidates were able to use the proper equation to calculate the length of one piece of aluminum foil for the first two marks, but very few doubled the length for the final mark.

This question was challenging for many candidates. While some candidates were able to use proper equations for capacitors to determine the charge some of the candidates attempted to use electrostatic equations for the electric field around a point charge to solve this problem.

This question was also challenging for many candidates, with not an insignificant number leaving it blank. The candidates who did attempt it generally set up a correct equation, but ran into some simple calculation and power of ten errors. Some candidates attempted to solve the equation using basic circuit equations, which did not receive any marks.

This is an explain question, so there was an expectation for a fairly detailed response. Many candidates missed the fact that the discharging capacitor is causing the current in the coil to change in time, and that this is what is inducing the emf in the coil. Many simply stated that the current created a magnetic field with not complete explanation of induction.

Candidates who recognized that something about the discharge circuit (not the charging circuit) needed to be changed generally suggested that something had to change with the resistance or capacitance. It should be noted that even though this was the last question on the exam, it was attempted at a higher rate than many of the other questions on the exam.

Show that the capacitance of this arrangement is C = 6.6 × 10^–7 F.

Calculate in V, the potential difference between the thundercloud and the Earth’s surface.

Calculate in J, the energy stored in the system.

Show that about –11 C of charge is delivered to the Earth’s surface.

Calculate, in A, the average current during the discharge.

State one assumption that needs to be made so that the Earth-thundercloud system may be modelled by a parallel plate capacitor.

C = «ε$\frac{A}{d}$ =» 8.8 × 10^–12 × $\frac{1.2\times {10}^8}{1600}$

«C = 6.60 × 10^–7 F»

[1 mark]

V = «$\frac{Q}{C}$ =» $\frac{25}{6.6\times {10}^{-7}}$

V = 3.8 × 10⁷ «V»

Award [2] for a bald correct answer

[2 marks]

ALTERNATIVE 1

E = «$\frac{1}{2}$QV =» $\frac{1}{2}$ × 25 × 3.8 × 10⁷

E = 4.7 × 10⁸ «J»

ALTERNATIVE 2

E = «$\frac{1}{2}$CV² =» $\frac{1}{2}$ × 6.60 × 10^–7 × (3.8 × 10⁷)²

E = 4.7 × 10⁸ «J» / 4.8 × 10⁸ «J» if rounded value of V used

Award [2] for a bald correct answer

Allow ECF from (b)(i)

[2 marks]

Q = «$Q_0{e}^{-\frac{t}{\tau }}$ =» 25 × $e^{-\frac{18}{32}}$

Q = 14.2 «C»

charge delivered = Q = 25 – 14.2 = 10.8 «C»

«≈ –11 C»

Final answer must be given to at least 3 significant figures

[3 marks]

I «= $\frac{\mathrm{\Delta }Q}{\mathrm{\Delta }t}=\frac{11}{18\times {10}^{-3}}$» ≈ 610 «A»

Accept an answer in the range 597 − 611 «A»

[1 mark]

the base of the thundercloud must be parallel to the Earth surface

OR

the base of the thundercloud must be flat

OR

the base of the cloud must be very long «compared with the distance from the surface»

[1 mark]

The capacitance of the capacitor is 22 mF. Calculate the energy stored in the capacitor when it is fully charged.

The resistance of the wire is 8.0 Ω. Determine the time taken for the capacitor to discharge through the resistance wire. Assume that the capacitor is completely discharged when the potential difference across it has fallen to 0.24 V.

The mass of the resistance wire is 0.61 g and its observed temperature rise is 28 K. Estimate the specific heat capacity of the wire. Include an appropriate unit for your answer.

Suggest one other energy loss in the experiment and the effect it will have on the value for the specific heat capacity of the wire.

«$\frac{1}{2}C{V}^2=\frac{1}{2}\times 0.22\times {24}^2$» = «J»

$\frac{1}{100}=e^{-\frac{t}{8.0\times 0.022}}$

$\ln 0.01=-\frac{t}{8.0\times 0.022}$

0.81 «s»

c = $\frac{Q}{m\times \mathrm{\Delta }T}$

OR

$\frac{6.3}{0.00061\times 28}$

370 J kg^–1 K^–1

Allow ECF from 3(a) for energy transferred.

Correct answer only to include correct unit that matches answer power of ten.

Allow use of g and kJ in unit but must match numerical answer, eg: 0.37 J kg^–1 K^–1 receives [1]

ALTERNATIVE 1

some thermal energy will be transferred to surroundings/along connecting wires/to
thermometer

estimate «of specific heat capacity by student» will be larger «than accepted value»

ALTERNATIVE 2

not all energy transferred as capacitor did not fully discharge

so estimate «of specific heat capacity by student» will be larger «than accepted value»

While the magnet is moving towards the ring, state why the magnetic flux in the ring is increasing.

While the magnet is moving towards the ring, sketch, using an arrow on Diagram 2, the direction of the induced current in the ring.

While the magnet is moving towards the ring, deduce the direction of the magnetic force on the magnet.

the magnetic field at the position of the ring is increasing «because the magnet gets closer to the ring» ✔

the current must be counterclockwise «in diagram 2» ✔

eg:

since the induced magnetic field is upwards

OR

by Lenz law the change «of magnetic field/flux» must be opposed

OR

by conservation of energy the movement of the magnet must be opposed ✔

therefore the force is repulsive/upwards ✔

This was well-answered.

Answers here were reasonably evenly split between clockwise and anti-clockwise, with the odd few arrows pointing left or right.

The majority of candidates recognised that the magnetic force would be upwards and the most common way of explaining this was via Lenz’s law. Students needed to get across that the force is opposing a change or a motion.

Calculate the distance between the plates.

The capacitor is connected to a 16 V cell as shown.

Calculate the magnitude and the sign of the charge on plate A when the capacitor is fully charged.

The capacitor is fully charged and the space between the plates is then filled with a dielectric of permittivity ε = 3.0ε₀.

Explain whether the magnitude of the charge on plate A increases, decreases or stays constant.

In a different circuit, a transformer is connected to an alternating current (ac) supply.

The transformer has 100 turns in the primary coil and 1200 turns in the secondary coil. The peak value of the voltage of the ac supply is 220 V. Determine the root mean square (rms) value of the output voltage.

Describe the use of transformers in electrical power distribution.

d = «$\frac{8.85\times {10}^{-12}\times {0.025}^2}{4.3\times {10}^{-12}}$ =» 1.3 × 10^–3 «m»

[1 mark]

6.9 × 10^–11 «C»

negative charge/sign

[2 marks]

charge increases

because capacitance increases AND pd remains the same.

[2 marks]

ALTERNATIVE 1

ε_s = $\frac{1200}{100}$ × 220

= 2640 «V»

V_rms = $\frac{2640}{\sqrt{2}}$ = 1870 «V»

ALTERNATIVE 2

(Primary) V_rms = $\frac{220}{\sqrt{2}}$ = 156 «V»

(Secondary) V_rms = $\frac{156\times 1200}{100}$

V_rms = 1870 «V»

Allow ECF from MP1 and MP2.

Award [2] max for 12.96 V (reversing N_p and N_s).

[3 marks]

step-up transformers increase voltage/step-down transformers decrease voltage

(step-up transformers increase voltage) from plants to transmission lines / (step-down transformers decrease voltage) from transmission lines to final utilizers

this decreases current (in transmission lines)

to minimize energy/power losses in transmission

[3 marks]

The switch S is initially open. Calculate the total power dissipated in the circuit.

The switch is now closed. State, without calculation, why the current in the cell will increase.

The switch is now closed. $\text{Deduce the ratio }\frac{\text{power dissipated in Y with S open}}{\text{power dissipated in Y with S closed}}$.

The cell is used to charge a parallel-plate capacitor in a vacuum. The fully charged capacitor is then connected to an ideal voltmeter.

The capacitance of the capacitor is 6.0 μF and the reading of the voltmeter is 12 V.

Calculate the energy stored in the capacitor.

Calculate the change in the energy stored in the capacitor.

Suggest, in terms of conservation of energy, the cause for the above change.

total resistance of circuit is 8.0 «Ω» ✔

$P=\frac{{12}^2}{8.0}=18$«W» ✔

«a resistor is now connected in parallel» reducing the total resistance

OR

current through YZ unchanged and additional current flows through X ✔

evidence in calculation or statement that pd across Y/current in Y is the same as before ✔

so ratio is 1 ✔

$E=«\frac{1}{2}C{V}^2=\frac{1}{2}\times 6\times {10}^{-6}\times {12}^2=»4.3\times {10}^{-4}$«$\text{J}$» ✔

ALTERNATIVE 1

capacitance doubles and voltage halves ✔

since $E=\frac{1}{2}C{V}^2$ energy halves   ✔

so change is «–»2.2×10^–4 «J»  ✔

ALTERNATIVE 2

$E=\frac{1}{2}C{V}^2\text{ and }Q=CV\text{ so }E=\frac{Q^2}{2C}$  ✔ 

capacitance doubles and charge unchanged so energy halves ✔

so change is «−»2.2 × 10⁻⁴ «J» ✔

it is the work done when inserting the dielectric into the capacitor ✔

Most candidates scored both marks. ECF was awarded for those who didn’t calculate the new resistance correctly. Candidates showing clearly that they were attempting to calculate the new total resistance helped examiners to award ECF marks.

Most recognised that this decreased the total resistance of the circuit. Answers scoring via the second alternative were rare as the statements were often far too vague.

Very few gained any credit for this at both levels. Most performed complicated calculations involving the total circuit and using 12V – they had not realised that the question refers to Y only.

Most answered this correctly.

By far the most common answer involved doubling the capacitance without considering the change in p.d. Almost all candidates who did this calculated a change in energy that scored 1 mark.

Very few scored on this question.

Show that the magnitude of the resultant electric field at P is 3 MN C⁻¹

State the direction of the resultant electric field at P.

Explain why q will perform simple harmonic oscillations when it is released.

Calculate the period of oscillations of q.

At t = 0, the switch is connected to X. On the axes, draw a sketch graph to show the variation with time of the voltage V_R across R.

The switch is then connected to Y and C discharges through the 20 MΩ resistor. The voltage V_c drops to 50 % of its initial value in 5.0 s. Determine the capacitance of C.

«electric field at P from one charge is $\frac{kQ}{r^2}=$» $\frac{8.99\times {10}^9\times 44\times {10}^{-6}}{0.{48}^2}$

OR

$1.7168\times {10}^6$ «NC⁻¹» ✓

« net field is » $2\times 1.7168\times {10}^6\times \cos 30°=2.97\times {10}^6$ «NC⁻¹» ✓

directed vertically up «on plane of the page» ✓

Allow an arrow pointing up on the diagram.

force «on q» is proportional to the displacement ✓

and opposite to the displacement / directed towards equilibrium ✓

$«a=\frac{F}{m}=»{\omega }^{2}x=\frac{115x}{0.25}$ ✓

$T=«\frac{2\mathrm{\pi }}{\omega }=» 0.29 «\text{s}»$ ✓

Award [2] marks for a bald correct answer.

Allow ECF for MP2.

decreasing from 12 ✓

correct shape as shown ✓

Do not penalize if the graph does not touch the t axis.

$\frac{1}{2}=e^{-\frac{5.0}{20\times {10}^6 C}}$ ✓

$C=3.6\times {10}^{-7}$ «F» ✓

Award [2] for a bald correct answer.

State Faraday’s law of induction.

Explain, using Faraday’s law of induction, how the transformer steps down the voltage.

The input voltage is 240 V. Calculate the output voltage.

Outline how energy losses are reduced in the core of a practical transformer.

Step-up transformers are used in power stations to increase the voltage at which the electricity is transmitted. Explain why this is done.

the size of the induced emf
is proportional/equal to the rate of change of flux linkage

The word ‘induced’ is required here.
Allow correctly defined symbols from a correct equation. ‘Induced’ is required for MP1.

varying voltage/current in primary coil produces a varying magnetic field

this produces a change in flux linkage / change in magnetic field in the secondary coil

a «varying» emf is induced/produced/generated in the secondary coil

voltage is stepped down as there are more turns on the primary than the secondary

Comparison of number of turns is required for MP4.

output voltage $=\frac{90\times 240}{1800}$

= 12 «V»

laminated core reduces eddy currents

less thermal energy is transferred to the surroundings

for a certain power to be transmitted, large V means low I

less thermal energy loss as P = I²R / joule heating

Show that the speed of the loop is 20 cm s⁻¹.

Sketch, on the axes, a graph to show the variation with time of the magnetic flux linkage $\Phi$ in the loop.

Sketch, on the axes, a graph to show the variation with time of the magnitude of the emf induced in the loop.

There are 85 turns of wire in the loop. Calculate the maximum induced emf in the loop.

The resistance of the loop is 2.4 Ω. Calculate the magnitude of the magnetic force on the loop as it enters the region of magnetic field.

Show that the energy dissipated in the loop from t = 0 to t = 3.5 s is 0.13 J.

The mass of the wire is 18 g. The specific heat capacity of copper is 385 J kg⁻¹ K⁻¹. Estimate the increase in temperature of the wire.

$\frac{70}{3.5}$ ✓

shape as above ✓

shape as above ✓

Vertical lines not necessary to score.

Allow ECF from (b)(i).

ALTERNATIVE 1

maximum flux at «$5.0\times 5.0\times {10}^{-4}\times 85\times 0.94$» $=0.19975\approx 0.20$«Wb» ✓

emf = «$\frac{0.20}{0.25}=$» $0.80$«V» ✓

ALTERNATIVE 2

emf induced in one turn = BvL = $0.94\times 0.20\times 0.05=0.0094$«V» ✓

emf $=85\times 0.0094=0.80$«V» ✓

Award [2] marks for a bald correct answer.

Allow ECF from MP1.

$I=«\frac{V}{R}=»\frac{0.8}{2.4}$  OR  $0.33$ «A» ✓

$F=«NBIL=85\times 0.94\times 0.33\times 0.05=»=1.3$ «N» ✓

Allow ECF from (c)(i).

Award [2] marks for a bald correct answer.

Energy is being dissipated for 0.50 s ✓

$E=Fvt=1.3\times 0.20\times 0.50=$«$0.13$ J»

OR

$E=Vlt=0.80\times 0.33\times 0.50=$«$0.13$ J» ✓

Allow ECF from (b) and (c).

Watch for candidates who do not justify somehow the use of 0.5 s and just divide by 2 their answer.

$∆T=\frac{0.13}{0.018\times 385}$ ✓

$∆T=1.9\times {10}^{-2}$ «K» ✓

Allow [2] marks for a bald correct answer.

Award [1] for a POT error in MP1.

Calculate, in J, the energy stored in X with the switch S open.

Calculate the final charge on X and the final charge on Y.

Calculate the final total energy, in J, stored in X and Y.

Suggest why the answers to (a) and (b)(ii) are different.

$E=\frac{1}{2}\frac{Q^2}{C}$ OR $V=\frac{Q}{C}$✔

$E=«\frac{1}{2}\frac{\left(48\times {10}^{-6}\right)}{18\times {10}^{-6}}=» 6.4\times {10}^{-5} «\mathrm{J}»$✔

ALTERNATIVE 1
$Q_{\mathrm{X}}+Q_{\mathrm{Y}}=48$ ✔

$\frac{Q_{\mathrm{X}}}{18}=\frac{Q_{\mathrm{Y}}}{12}$ ✔

solving to get $Q_{\mathrm{X}}=29 «\mu \mathrm{C}» Q_{\mathrm{Y}}=19 «\mu \mathrm{C}»$ ✔

ALTERNATIVE 2

$48=18 \mathrm{V}+12 \mathrm{V}\Rightarrow V=1.6 «\mathrm{V}»$✔

$Q_{\mathrm{X}}=«1.6\times 18=» 29 «Q_{\mathrm{X}}=1.6\times 18=29 «\mu \mathrm{C}»$ ✔

$Q_{\mathrm{Y}}=«1.6\times 12=» 19 «\mu \mathrm{C}»$ ✔

NOTE: Award [3] for bald correct answer

ALTERNATIVE 1

$E_{\mathrm{T}}=\frac{1}{2}\frac{{\left(29\times {10}^{-6}\right)}^2}{18\times {10}^{-6}}+\frac{1}{2}\frac{{\left(19\times {10}^{-6}\right)}^2}{12\times {10}^{-6}}$✔

$=3.8\times {10}^{-5}«\mathrm{J}»$✔

ALTERNATIVE 2

$E_{\mathrm{T}}=\frac{1}{2}\times 18\times {10}^{-6}\times 1.6^2+\frac{1}{2}\times 12\times {10}^{-6}\times 1.6^2$ ✔

$=3.8\times {10}^{-5}«\mathrm{J}»$✔

NOTE: Allow ECF from (b)(i)
Award [2] for bald correct answer
Award [1] max as ECF to a calculation using only one charge

charge moves/current flows «in the circuit» ✔
thermal losses «in the resistor and connecting wires» ✔

NOTE: Accept heat losses for MP2