Outline what is meant by electric field strength.

An electron is placed at X and released from rest. Draw, on the diagram, the direction of the force acting on the electron due to the field.

The electron is replaced by a proton which is also released from rest at X. Compare, without calculation, the motion of the electron with the motion of the proton after release. You may assume that no frictional forces act on the electron or the proton.

force per unit charge

acting on a small/test positive charge

horizontally to the left

Arrow does not need to touch X

proton moves to the right/they move in opposite directions

force on each is initially the same

proton accelerates less than electron initially «because mass is greater»

field is stronger on right than left «as lines closer»

proton acceleration increases «as it is moving into stronger field»

OR

electron acceleration decreases «as it is moving into weaker field»

Allow ECF from (b)

Accept converse argument for electron

Show that the energy of photons from the UV lamp is about 10 eV.

Calculate, in J, the maximum kinetic energy of the emitted electrons.

Suggest, with reference to conservation of energy, how the variable voltage source can be used to stop all emitted electrons from reaching the collecting plate.

The variable voltage can be adjusted so that no electrons reach the collecting plate. Write down the minimum value of the voltage for which no electrons reach the collecting plate.

On the diagram, draw and label the equipotential lines at –0.4 V and –0.8 V.

An electron is emitted from the photoelectric surface with kinetic energy 2.1 eV. Calculate the speed of the electron at the collecting plate.

E₁ = –13.6 «eV» E₂ = – $\frac{13.6}{4}$ = –3.4 «eV»

energy of photon is difference E₂ – E₁ = 10.2 «≈ 10 eV»

Must see at least 10.2 eV.

[2 marks]

10 – 5.1 = 4.9 «eV»

4.9 × 1.6 × 10^–19 = 7.8 × 10^–19 «J»

Allow 5.1 if 10.2 is used to give 8.2×10⁻¹⁹ «J».

EPE produced by battery

exceeds maximum KE of electrons / electrons don’t have enough KE

For first mark, accept explanation in terms of electric potential energy difference of electrons between surface and plate.

[2 marks]

4.9 «V»

Allow 5.1 if 10.2 is used in (b)(i).

Ignore sign on answer.

[1 mark]

two equally spaced vertical lines (judge by eye) at approximately 1/3 and 2/3

labelled correctly

[2 marks]

kinetic energy at collecting plate = 0.9 «eV»

speed = «$\sqrt{\frac{2\times 0.9\times 1.6\times {10}^{-19}}{9.11\times {10}^{-31}}}$» = 5.6 × 10⁵ «ms^–1»

Allow ECF from MP1

[2 marks]

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» ✔

Identify, on the diagram, the direction of the electric field between the plates.

The following data are available.

Determine the strength of the magnetic field B.

The velocity of the electrons is now increased. Explain the effect that this will have on the path of the electron beam.

direction indicated downwards, perpendicular to plates

Arrows must be between plates but allow edge effects if shown. Only one arrow is required.

$E=\frac{V}{d}=55000$ «Vm^–1»

B = «$\frac{55000}{5\times {10}^5}=$» 0.11 «T»

ECF applies from MP1 to MP2 due to math error.

Award [2] for a bald correct answer.

ALTERNATIVE 1

magnetic force increases
OR
magnetic force becomes greater than electric force

electron beam deflects “downwards” / towards S
OR
path of beam is downwards

ALTERNATIVE 2

when v increases, the B required to maintain horizontal path decreases
«but B is constant» so path of beam is downwards

Do not apply an ecf from (a).

Award [1 max] if answer states that magnetic force decreases and therefore path is upwards.

Ignore any statement about shape of path

Do not allow “path deviates in direction of magnetic force” without qualification.

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”.

Outline why component X is considered non-ohmic.

Determine the resistance of the variable resistor.

Calculate the power dissipated in the circuit.

State the range of current that the ammeter can measure as the slider S of the potential divider is moved from Q to P.

Slider S of the potential divider is positioned so that the ammeter reads $20 \text{mA}$. Explain, without further calculation, any difference in the power transferred by the potential divider arrangement over the arrangement in (b).

current is not «directly» proportional to the potential difference
OR
resistance of X is not constant
OR
resistance of X changes «with current/voltage» ✓

ALTERNATIVE 1

voltage across X$=2.3 «\mathrm{V}»$ ✓

voltage across R$«=4.0-2.3»=1.7 «\mathrm{V}»$ ✓

resistance of variable resistor $«=\frac{1.7}{0.020}»=85 «\mathrm{\Omega }»$ ✓

ALTERNATIVE 2

overall resistance $«=\frac{4.0}{0.020}»=200 «\mathrm{\Omega }»$ ✓

resistance of X $«=\frac{2.3}{0.020}»=115 «\mathrm{\Omega }»$ ✓

resistance of variable resistor $«=200-115»=85 «\mathrm{\Omega }»$ ✓

power $«=4.0\times 0.020»=0.080 «\mathrm{W}»$ ✓

from $0$ to $60 \mathrm{mA}$ ✓

ALTERNATIVE 1

current from the cell is greater «than $20 \text{mA}$» ✓

because some of the current must flow through section SQ of the potentiometer ✓

overall power greater «than in part (b)» ✓

ALTERNATIVE 2

total/overall resistance decreases ✓

because SQ and X are in parallel ✓

overall power greater «than in part (b)» ✓

Allow the reverse argument.

Most answers that didn't score simply referred to the shape of the graph without any explanation as to what this meant to the relationship between the variables.

This question produced a mixture of answers from the 2 alternatives given in the markscheme. As a minimum, many candidates were able to score a mark for the overall resistance of the circuit.

A straightforward calculation question that most candidates answered correctly.

Surprisingly a significant number of candidates had difficulty with this. Answers of 20 mA and 4 V were often seen.

HL only. This question challenged candidate's ability to describe clearly the changes in an electrical circuit. It revealed many misconceptions about the nature of electrical current and potential difference, of those who did have a grasp of what was going on the explanations often missed the second point in each of the markscheme alternatives as detail was missed about where the current was flowing or what was in parallel with what.

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

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

Bohr modified the Rutherford model by introducing the condition mvr = n$\frac{h}{2\pi }$. Outline the reason for this modification.

Show that the speed v of an electron in the hydrogen atom is related to the radius r of the orbit by the expression

$$v=\sqrt{\frac{k{e}^2}{m_{\text{e}}r}}$$

where k is the Coulomb constant.

Using the answer in (b) and (c)(i), deduce that the radius r of the electron’s orbit in the ground state of hydrogen is given by the following expression.

$$r=\frac{h^2}{4{\pi }^{2}k{m}_{\text{e}}{e}^2}$$

Calculate the electron’s orbital radius in (c)(ii).

Explain what may be deduced about the energy of the electron in the β^– decay.

Suggest why the β^– decay is followed by the emission of a gamma ray photon.

Calculate the wavelength of the gamma ray photon in (d)(ii).

the electrons accelerate and so radiate energy

they would therefore spiral into the nucleus/atoms would be unstable

electrons have discrete/only certain energy levels

the only orbits where electrons do not radiate are those that satisfy the Bohr condition «mvr = n$\frac{h}{2\pi }$»

[3 marks]

$\frac{m_{\text{e}}{v}^2}{r}=\frac{k{e}^2}{r^2}$

OR

KE = $\frac{1}{2}$PE hence $\frac{1}{2}$m_ev² = $\frac{1}{2}\frac{k{e}^2}{r}$

«solving for v to get answer»

Answer given – look for correct working

[1 mark]

combining v = $\sqrt{\frac{k{e}^2}{m_{\text{e}}r}}$ with m_evr = $\frac{h}{2\pi }$ using correct substitution

«eg ${m_e}^2\frac{k{e}^2}{m_{\text{e}}r}{r}^2=\frac{h^2}{4{\pi }^2}$»

correct algebraic manipulation to gain the answer

Answer given – look for correct working

Do not allow a bald statement of the answer for MP2. Some further working eg cancellation of m or r must be shown

[2 marks]

« r = $\frac{{(6.63\times {10}^{-34})}^2}{4{\pi }^2\times 8.99\times {10}^9\times 9.11\times {10}^{-31}\times {(1.6\times {10}^{-19})}^2}$»

r = 5.3 × 10^–11 «m»

[1 mark]

the energy released is 3.54 – 0.48 = 3.06 «MeV»

this is shared by the electron and the antineutrino

so the electron’s energy varies from 0 to 3.06 «MeV»

[3 marks]

the palladium nucleus emits the photon when it decays into the ground state «from the excited state»

[1 mark]

Photon energy

E = 0.48 × 10⁶ × 1.6 × 10^–19 = «7.68 × 10^–14 J»

λ = «$\frac{hc}{E}=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{7.68\times {10}^{-14}}$ =» 2.6 × 10^–12 «m»

Award [2] for a bald correct answer

Allow ECF from incorrect energy

[2 marks]

Calculate the resistance of the conductor.

Calculate the drift speed v of the electrons in the conductor in cm s^–1.

Determine the electric field strength E.

Show that $\frac{v}{E}=\frac{1}{ne\rho }$.

1.7 × 10^–8 × $\frac{0.10}{{(0.02\times {10}^{-2})}^2}$

0.043 «Ω»

[2 marks]

v «= $\frac{I}{neA}$» = $\frac{2}{8.5\times {10}^{22}\times 1.60\times {10}^{-19}\times {0.02}^2}$

0.37 «cms^–1»

[2 marks]

V = RI = 0.086 «V»

«$\frac{V}{d}=\frac{0.086}{0.10}=$» 0.86 «V m^–1»

Allow ECF from 4(a).

Allow ECF from MP1.

[2 marks]

ALTERNATIVE 1

clear use of Ohm’s Law (V = IR)

clear use of R = $\frac{\rho L}{A}$

combining with I = nAve and V = EL to reach result.

ALTERNATIVE 2

attempts to substitute values into equation.

correctly calculates LHS as 4.3 × 10⁹.

correctly calculates RHS as 4.3 × 10⁹.

For ALTERNATIVE 1 look for:

V = IR

R = $\frac{\rho L}{A}$

V = EL

I = nAve

V = I$\frac{\rho L}{A}$

EL = I$\frac{\rho L}{A}$

E = I$\frac{\rho }{A}$

E = nAve$\frac{\rho }{A}$ = nveρ

$\frac{v}{E}=\frac{1}{ne\rho }$

[3 marks]

State what is meant by the emf of a cell.

Show that the resistance of the wire AC is 28 Ω.

Determine E.

Cell X is replaced by a second cell of identical emf E but with internal resistance 2.0 Ω. Comment on the length of AC for which the current in the second cell is zero.

the work done per unit charge

in moving charge from one terminal of a cell to the other / all the way round the circuit

Award [1] for “energy per unit charge provided by the cell”/“power per unit current”

Award [1] for “potential difference across the terminals of the cell when no current is flowing” 

Do not accept “potential difference across terminals of cell”

[2 marks]

the resistance is proportional to length / see 0.35 AND 1«.00»

so it equals 0.35 × 80

«= 28 Ω»

[2 marks]

current leaving 12 V cell is $\frac{12}{80}$ = 0.15 «A»

OR

E = $\frac{12}{80}$ × 28

E = «0.15 × 28 =» 4.2 «V»

Award [2] for a bald correct answer

Allow a 1sf answer of 4 if it comes from a calculation.

Do not allow a bald answer of 4 «V»

Allow ECF from incorrect current

[2 marks]

since the current in the cell is still zero there is no potential drop across the internal resistance

and so the length would be the same

OWTTE

[2 marks]

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.

Determine the initial acceleration of the spacecraft.

(i) Estimate the maximum speed of the spacecraft.

(ii) Outline why the answer to (i) is an estimate.

Outline why scientists sometimes use estimates in making calculations.

Outline why the ions are likely to spread out.

Explain what effect, if any, this spreading of the ions has on the acceleration of the spacecraft.

change in momentum each second = 6.6 × 10⁻⁶ × 5.2 × 10⁴ «= 3.4 × 10⁻¹kg m s⁻¹» ✔

acceleration = «$\frac{3.4\times {10}^{-1}}{740}$ =» 4.6 × 10⁻⁴ «m s⁻²» ✔

(i) ALTERNATIVE 1:

(considering the acceleration of the spacecraft)

time for acceleration = $\frac{30}{6.6\times {10}^{-6}}$ = «4.6 × 10⁶» «s» ✔

max speed = «answer to (a) × 4.6 × 10⁶ =» 2.1 × 10³ «m s⁻¹» ✔

ALTERNATIVE 2:

(considering the conservation of momentum)

(momentum of 30 kg of fuel ions = change of momentum of spacecraft)

30 × 5.2 × 10⁴ = 710 × max speed ✔

max speed = 2.2 × 10³ «m s⁻¹» ✔

(ii) as fuel is consumed total mass changes/decreases so acceleration changes/increases
OR
external forces (such as gravitational) can act on the spacecraft so acceleration isn’t constant ✔

problem may be too complicated for exact treatment ✔

to make equations/calculations simpler ✔

when precision of the calculations is not important ✔

some quantities in the problem may not be known exactly ✔

ions have same (sign of) charge ✔

ions repel each other ✔

the forces between the ions do not affect the force on the spacecraft. ✔

there is no effect on the acceleration of the spacecraft. ✔

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.

Explain why the electric potential decreases from A to B.

Draw, on the axes, the variation of electric potential $V$ with distance $r$ from the centre of the sphere.

Calculate the electric potential difference between points A and B.

Determine the charge $Q$ of the sphere.

The concept of potential is also used in the context of gravitational fields. Suggest why scientists developed a common terminology to describe different types of fields.

ALTERNATIVE 1
work done on moving a positive test charge in any outward direction is negative ✓
potential difference is proportional to this work «so $V$ decreases from A to B» ✓

ALTERNATIVE 2
potential gradient is directed opposite to the field so inwards ✓
the gradient indicates the direction of increase of $V$ «hence $V$ increases towards the centre/decreases from A to B» ✓

ALTERNATIVE 3
$V=\frac{kQ}{R}$ so as $r$ increases $V$ decreases ✓
$V$ is positive as $Q$ is positive ✓

ALTERNATIVE 4
the work done per unit charge in bringing a positive charge from infinity ✓
to point B is less than point A ✓

curve decreasing asymptotically for $r>R$ ✓

non $-$ zero constant between $0$ and $R$ ✓

$«\frac{W}{q}=\frac{1.7\times {10}^{-16}}{1.60\times {10}^{-19}}=»1.1\times {10}^3 «\mathrm{V}»$ ✓

$8.99\times {10}^9\times Q\times \left(\frac{1}{5.0\times {10}^{-2}}-\frac{1}{1.0\times {10}^{-1}}\right)=1.1\times {10}^3$ ✓

$Q=1.2\times {10}^{-8} «\mathrm{C}»$ ✓

to highlight similarities between «different» fields ✓

The majority who answered in terms of potential gained one mark. Often the answers were in terms of work done rather than work done per unit charge or missed the fact that the potential is positive.

This was well answered.

Most didn't realise that the key to the answer is the definition of potential or potential difference and tried to answer using one of the formulae in the data booklet, but incorrectly.

Even though many were able to choose the appropriate formula from the data booklet they were often hampered in their use of the formula by incorrect techniques when using fractions.

This was generally well answered with only a small number of answers suggesting greater international cooperation.

Identify the laws of conservation that are represented by Kirchhoff’s circuit laws.

State the emf of the cell.

Deduce the internal resistance of the cell.

Calculate the reading on the voltmeter.

Comment on the implications of your answer to (c)(i) for cell B.

Outline why electricity is a secondary energy source.

Some fuel sources are renewable. Outline what is meant by renewable.

A fully charged cell of emf 6.0 V delivers a constant current of 5.0 A for a time of 0.25 hour until it is completely discharged.

The cell is then re-charged by a rectangular solar panel of dimensions 0.40 m × 0.15 m at a place where the maximum intensity of sunlight is 380 W m⁻².

The overall efficiency of the re-charging process is 18 %.

Calculate the minimum time required to re-charge the cell fully.

Outline why research into solar cell technology is important to society.

« conservation of » charge ✓

« conservation of » energy ✓

Allow [1] max if they explicitly refer to Kirchhoff’ laws linking them to the conservation laws incorrectly.

12 V ✓

I = 2.0 A OR 12 = I (r + 4) OR 4 = Ir OR 8 = 4I ✓

«Correct working to get » r = 2.0 «Ω» ✓

Allow any valid method.

Allow ECF from (b)(i)

Loop equation showing EITHER correct voltages, i.e., 10 – 4 on one side or both emf’s positive on different sides of the equation OR correct resistances, i.e. I (1 + 2) ✓

10−4 = I (1 + 2) OR I = 2.0 «A» seen ✓

V = 8.0 «V» ✓

Allow any valid method

Charge is being driven through the 4.0 V cell OR it is being (re-)charged ✓

is generated from primary/other sources ✓

«a fuel » that can be replenished/replaced within a reasonable time span

OR

«a fuel» that can be replaced faster than the rate at which it is consumed

OR

renewables are limitless/never run out

OR

«a fuel» produced from renewable sources

OR

gives an example of a renewable (biofuel, hydrogen, wood, wind, solar, tidal, hydro etc..) ✓

OWTTE

ALTERNATIVE 1

«energy output of the panel =» Vlt OR 6 x 5 x 0.25 x 3600 OR 27000 «J» ✓

«available power =» 380 x 0.4 x 0.15 x 0.18 OR 4.1 «W» ✓

$t=$ «$\frac{27000}{4.1}$=» 6600 «s» ✓

ALTERNATIVE 2

«energy needed from Sun =» $\frac{vlt}{eff}$ OR $\frac{6\times 5\times 0.25\times 3600}{0.18}$ OR 150000 «J» ✓

« incident power=» 380 x 0.4 x 0.15 OR 22.8 «W» ✓

$t=$ «$\frac{150000}{22.8}$=» 6600 «s» ✓

Allow ECF for MP3

Accept final answer in minutes (110) or hours (1.8).

coherent reason ✓

e.g., to improve efficiency, is non-polluting, is renewable, does not produce greenhouse gases, reduce use of fossil fuels

Do not allow economic reasons

a) Most just stated Kirchhoff's laws rather than the underlying laws of conservation of energy and charge, basically describing the equations from the data booklet. When it came to guesses, energy and momentum were often the two, although even a baryon and lepton number conservation was found. It cannot be emphasised enough the importance of correctly identifying the command verb used to introduce the question. In this case, identify, with the specific reference to conservation laws, seem to have been explicit tips not picked up by some candidates.

bi) This was probably the easiest question on the paper and almost everybody got it right. 12V. Some calculations were seen, though, that contradict the command verb used. State a value somehow implies that the value is right in front to be read or interpreted suitably.

bii) In the end a lot of the answers here were correct. Some obtained 2 ohms and were able to provide an explanation that worked. A very few negative answers were found, suggesting that some candidates work mechanically without properly reflecting in the nature of the value obtained.

ci) A lot of candidates figured out they had to do some sort of loop here but most had large currents in the voltmeter. Currents of 2 A and 10 A simultaneously were common. Some very good and concise work was seen though, leading to correct steps to show a reading of 8V.

cii) This question was cancelled due to an internal reference error. The paper total was adjusted in grade award. This is corrected for publication and future teaching use.

di) The vast majority of candidates could explain why electricity was a secondary energy source.

dii) An ideal answer was that renewable fuels can be replenished faster than they are consumed. However, many imaginative alternatives were accepted.

ei) This question was often very difficult to mark. Working was often scattered all over the answer box. Full marks were not that common, most candidates achieved partial marks. The commonest problem was determining the energy required to charge the battery. It was also common to see a final calculation involving a power divided by a power to calculate the time.

eii) Almost everybody could give a valid reason why research into solar cells was important. Most answers stated that solar is renewable. There were very few that didn't get a mark due to discussing economic reasons.

Label with arrows on the diagram the magnetic force F on the proton.

Label with arrows on the diagram the velocity vector v of the proton.

For this proton, determine, in m, the radius of the circular path. Give your answer to an appropriate number of significant figures.

For this proton, calculate, in s, the time for one full revolution.

F towards centre ✔

v tangent to circle and in the direction shown in the diagram ✔

«$qvB=\frac{m{v}^2}{R}\Rightarrow »R=\frac{mv}{qB}/\frac{1.673\times {10}^{-27}\times 2.16\times {10}^6}{1.60\times {10}^{-19}\times 0.042}$  ✔

R = 0.538«m»  ✔

R = 0.54«m»   ✔

$T=\frac{2\pi R}{v}/\frac{2\pi \times 0.54}{2.16\times {10}^6}$  ✔

$T=1.6\times {10}^{-6}$«s»   ✔

Examiners were requested to be lenient here and as a result most candidates scored both marks. Had we insisted on e.g. straight lines drawn with a ruler or a force arrow passing exactly through the centre of the circle very few marks would have been scored. For those who didn’t know which way the arrows were supposed to be the common guesses were to the left and up the page. Some candidates neglected to label the arrows.

Examiners were requested to be lenient here and as a result most candidates scored both marks. Had we insisted on e.g. straight lines drawn with a ruler or a force arrow passing exactly through the centre of the circle very few marks would have been scored. For those who didn’t know which way the arrows were supposed to be the common guesses were to the left and up the page. Some candidates neglected to label the arrows.

This was generally well answered although usually to 3 sf. Common mistakes were to substitute 0.042 for F and 1 for q. Also some candidates tried to answer in terms of electric fields.

This was well answered with many candidates scoring ECF from the previous part.

The charge per unit area on the surface of the wall is σ. It can be shown that the electric field strength E due to the charge on the wall is given by the equation

$E=\frac{\sigma }{2{\epsilon }_0}$.

Demonstrate that the units of the quantities in this equation are consistent.

The thread makes an angle of 30° with the vertical wall. The ball has a mass of 0.025 kg.

Determine the horizontal force that acts on the ball.

The charge on the ball is 1.2 × 10⁻⁶C. Determine σ.

The thread breaks. Explain the initial subsequent motion of the ball.

Calculate the charge on Q. State your answer to an appropriate number of significant figures.

Outline, without calculation, whether or not the electric potential at P is zero.

identifies units of $\sigma$ as ${\text{C\hspace{0.05em}m}}^{-2}$ ✓

$\frac{C}{m^2}\times \frac{N{m}^2}{C^2}$ seen and reduced to ${\text{N\hspace{0.05em}C}}^{-1}$ ✓

Accept any analysis (eg dimensional) that yields answer correctly

horizontal force $F$ on ball $=T \sin 30$ ✓

$T=\frac{mg}{\cos 30}$ ✓

$F «=mg \tan 30 = 0.025\times 9.8 \times \tan 30» = 0.14 «\text{N}»$ ✓

Allow g = 10 N kg⁻¹

Award [3] marks for a bald correct answer.

Award [1max] for an answer of zero, interpreting that the horizontal force refers to the horizontal component of the net force.

$E=\frac{0.14}{1.2\times {10}^{-6}}«=1.2\times {10}^5»$ ✓

$\sigma =«\frac{2\times 8.85\times {10}^{-12}\times 0.14}{1.2\times {10}^{-6}}»=2.1\times {10}^{-6} «{\text{C\hspace{0.05em}m}}^{-2}»$ ✓

Allow ECF from the calculated F in (b)(i)

Award [2] for a bald correct answer.

horizontal/repulsive force and vertical force/pull of gravity act on the ball ✓

so ball has constant acceleration/constant net force ✓

motion is in a straight line ✓

at 30° to vertical away from wall/along original line of thread ✓

$\frac{Q}{0.{22}^2}=\frac{1.2\times {10}^{-6}}{0.{18}^2}$ ✓

$«+»1.8\times {10}^{-6} «\text{C}»$✓

2sf ✓

Do not award MP2 if charge is negative

Any answer given to 2 sig figs scores MP3

work must be done to move a «positive» charge from infinity to P «as both charges are positive»
OR
reference to both potentials positive and added
OR
identifies field as gradient of potential and with zero value ✓

therefore, point P is at a positive / non-zero potential ✓

Award [0] for bald answer that P has non-zero potential

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.

Identify a property of electrons demonstrated by this experiment.

Show that the energy E of each electron in the beam is about 7 × 10⁻¹¹J.

The de Broglie wavelength for an electron is given by $\frac{hc}{E}$. Show that the diameter of an oxygen-16 nucleus is about 4 fm.

Estimate, using the result in (a)(iii), the volume of a tin-118 $\left({}_{50}{}^{118}\text{Sn}\right)$ nucleus. State your answer to an appropriate number of significant figures.

wave properties ✓

Accept reference to diffraction or interference.

440 x 10⁶ x 1.6 x 10^-19  OR  7.0 × 10^-11 «J» ✓

$\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{7\times {10}^{-11}}$  OR  $\frac{1.24\times {10}^{-6}}{440\times {10}^6}$  OR  2.8 × 10^-15«m» seen ✓

read off graph as 46° ✓

«Use of $D=\frac{\lambda }{\sin \theta }$=» 3.9 × 10^-15 m ✓

Accept an angle between 45 and 47 degrees.

Allow ECF from MP2

ALTERNATIVE 1

use of $R\propto A^{\frac{1}{3}}$   OR  $V\propto A$ ✓

volume of $\text{Sn}=\frac{4}{3}\pi \left(\frac{A_{Sn}}{A_O}\right)r_O^{ 3}$ or equivalent working ✓

2.3 to 2.5 × 10^-43«m³»✓

answer to 1 or 2sf ✓

ALTERNATIVE 2

use of $R=R_{\text{o}}\times A^{\frac{1}{3}}$ ✓

volume of $\text{Sn}=\frac{4}{3}\pi R^3$  OR  5.9 x 10^-15 seen ✓

8.5 × 10^-43 «m³»✓

answer to 1 or 2sf ✓

Although the question expects candidates to work from the oxygen radius found, allow ALT 2 working from the Fermi radius.

MP4 is for any answer stated to 1 or 2 significant figures.

ai) Well answered.

aii) Well answered.

aiii) This was generally well done but quite a few attempted the small angle approximation. Probably worth a mention in the report.

b) Most gained credit from the first alternative solution, trying to use the data as the question intended. There were the inevitable slips and calculator mistakes. Most got the fourth mark.

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]

Calculate the power dissipated in the cable.

power = «35² × 0.028» = 34 «W»

Allow 35 – 36 W if unrounded figures for R or I are used.
Allow ECF from (a)(i) and (a)(ii).

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

Show that the charge on the surface of the sphere is +18 μC.

Describe, in terms of electron flow, how the smaller sphere becomes charged.

Predict the charge on each sphere.

$Q=«\frac{VR}{k}=»\frac{3.4\times {10}^5\times 0.48}{8.99\times {10}^9}$

OR

$Q=18.2$ «μC» ✓

electrons leave the small sphere «making it positively charged» ✓

$k\frac{q_1}{48}=k\frac{q_2}{24}\Rightarrow q_1=2q_2$ ✓

$q_1+q_2=18$ ✓

so $q_1=12$ «μC», $q_2=6.0$ «μC» ✓

Award [3] marks for a bald correct answer.

Calculate the electric potential at O.

Sketch, on the axes, the variation of the electric potential V with distance between X and Y.

Identify the direction of the resultant force acting on Z as it oscillates.

Deduce whether the motion of Z is simple harmonic.

use of $\frac{kQ}{r}$ ✓

$\frac{\left(8.99\times {10}^9\right)\left(4\times {10}^{-6}\right)}{0.15}$ OR 240 «kV» for one charge calculated ✓

480 «kV» for both ✓

MP1 can be seen or implied from calculation.

Allow ECF from MP2 for MP3.

symmetric curve around 0 with potential always positive, “bowl shape up” and curve not touching the horizontal axis. ✓

clear asymptotes at X and Y ✓

force is towards O ✓

always ✓

ALTERNATIVE 1

motion is not SHM ✓

«because SHM requires force proportional to r and» this force depends on $\frac{1}{r^2}$ ✓

ALTERNATIVE 2

motion is not SHM ✓

energy-distance «graph must be parabolic for SHM and this» graph is not parabolic ✓

This question was generally well approached. Two common errors were either starting with the wrong equation (electric potential energy or Coulomb's law) or subtracting the potentials rather than adding them.

Very few candidates drew a graph that was awarded two marks. Many had a generally correct shape, but common errors were drawing the graph touching the x-axis at O and drawing a general parabola with no clear asymptotes.

Many candidates were able to identify the direction of the force on the particle at position Z, but a common error was to miss that the question was about the direction as the particle was oscillating. Examiners were looking for a clear understanding that the force was always directed toward the equilibrium position, and not just at the moment shown in the diagram.

This was a challenging question for candidates. Most simply assumed that because the charge was oscillating that this meant the motion was simple harmonic. Some did recognize that it was not, and most of those candidates correctly identified that the relationship between force and displacement was an inverse square.