Year 11 Physics: Electricity and Magnetism Practice Exam

This original tutorgum practice paper covers the supplied Year 11 Physics Electricity and Magnetism content for the 2025 syllabus edition. It tests calculations, circuit analysis, experimental uncertainty, field models, and explanations of physical processes. All questions are original, and the worked solutions and marking guide appear after the complete paper.

01Exam instructions

Paper conditions

• Suggested working time: 120 minutes.
• Reading time: 5 minutes.
• Total marks: 100.
• Permitted equipment: Scientific calculator, ruler, pencil, eraser, and graph paper.
• Working: Show calculations, include units, and give answers to an appropriate number of significant figures.
• Assumptions: Treat connecting wires and meters as ideal unless stated otherwise. Treat air as a vacuum for electrostatic calculations. Ignore gravity where instructed.

This is an original tutorgum practice paper. It is not an official or endorsed NESA examination, and it is not a prediction of future exam questions. The time, marks, and permitted equipment are suggested conditions for this practice paper, not claims about official examination rules.

Assessed scope

The paper assesses the supplied Year 11 Electricity and Magnetism content across:

• Electric current, charge flow, AC and DC, and circuit measurement.
• Series, parallel, and combination circuits; Ohm’s law; Kirchhoff’s laws; resistance; power; and electrical energy.
• Measurement uncertainty and conductor geometry.
• Electrostatic charging, Coulomb’s law, electric fields, potential difference, work, and electrical potential energy.
• Uniform electric fields, charged-particle motion, static electricity, and lightning.
• Magnetic field models, permanent magnets, current-carrying wires, solenoids, magnetic materials, and electromagnets.

This is a topic revision paper, not a full-course Year 11 or Year 12 Physics examination.

Exclusions: Electromagnetic induction in coils, transformers, generators, capacitors, semiconductor devices, magnetic forces on moving charges, and detailed AC impedance or phase analysis are not assessed. Electrostatic charging by induction is included.

Useful constants

Use these values where required:

\[ k = 8.99\times10^9\ \mathrm{N\,m^2\,C^{-2}} \]

\[ \mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}} \]

Also use \(1\,\mathrm{\mu C}=10^{-6}\,\mathrm C\), \(1\,\mathrm{nC}=10^{-9}\,\mathrm C\), and \(1\,\mathrm{min}=60\,\mathrm s\).

02Questions

Question 1 (8 marks)

(a) A steady current transfers \(360\,\mathrm C\) of charge through a wire in \(2.0\,\mathrm{min}\).

Calculate the current. (2 marks)

(b) Explain how electric current is produced in a metal wire. Distinguish the direction of electron drift from the direction of conventional current. (2 marks)

(c) The following current-time graph contains two traces, A and B.

• A
• B

Identify which trace represents steady direct current (DC) and which represents alternating current (AC). Justify both choices using the direction and time variation of current. (2 marks)

(d) Describe how an ammeter and a voltmeter should be connected to measure the current through a resistor and the potential difference across it. State the ideal resistance of each meter. (2 marks)

Question 2 (12 marks)

An ideal \(12\,\mathrm V\) DC supply is connected to a \(2.0\,\Omega\) resistor in series with a parallel combination of a \(6.0\,\Omega\) resistor and a \(3.0\,\Omega\) resistor.

Assume all resistors are ohmic at their operating temperatures.

(a) Calculate the equivalent resistance of the parallel section and the total resistance of the circuit. (3 marks)

(b) Calculate the total current supplied by the battery and the potential difference across the \(2.0\,\Omega\) resistor. (2 marks)

(c) Calculate the current through each resistor in the parallel section. Use your results to verify Kirchhoff’s current law at the junction. (3 marks)

(d) Calculate the power dissipated by the \(3.0\,\Omega\) resistor and the electrical energy it transfers in \(2.0\,\mathrm{min}\). (2 marks)

(e) Apply Kirchhoff’s voltage law to the closed loop containing the \(2.0\,\Omega\) and \(6.0\,\Omega\) resistors. Show that the potential changes around the loop are consistent with energy conservation. (2 marks)

Question 3 (8 marks)

(a) A resistor has a potential difference of \(4.0\,\mathrm V\) across it and a current of \(0.80\,\mathrm A\) through it.

Calculate its resistance and state the condition under which a conductor obeys Ohm’s law. (2 marks)

(b) A filament lamp produces the following measurements.

Calculate the lamp’s resistance at \(0.50\,\mathrm V\) and at \(2.50\,\mathrm V\). (2 marks)

(c) Plot current against potential difference using the data above. Describe the shape of the graph and determine whether the lamp obeys Ohm’s law over the measured range. (2 marks)

(d) A wire has resistance \(R_0\), length \(L\), and cross-sectional area \(A\). A second wire is made from the same material at the same temperature. Its length is \(3L\), and its cross-sectional area is \(1.5A\).

Determine the resistance of the second wire in terms of \(R_0\). (2 marks)

Question 4 (12 marks)

Use Coulomb’s law and the supplied value of \(k\). Assume air is a vacuum unless a different medium is specified.

(a) Two point charges, \(+3.0\,\mathrm{\mu C}\) and \(-2.0\,\mathrm{\mu C}\), are separated by \(0.20\,\mathrm m\).

Calculate the magnitude of the electrostatic force between them and state whether the force is attractive or repulsive. (3 marks)

(b) Consider each change separately. State how the force magnitude in part (a) changes if:

• The separation doubles.
• The magnitude of one charge doubles.
• The charges are placed in a medium with greater permittivity than air.

(3 marks)

(c) Calculate the electric field strength at a point \(0.30\,\mathrm m\) from a stationary \(+4.0\,\mathrm{\mu C}\) point charge. State the field direction at that point. (2 marks)

(d) A charge of \(-2.0\,\mathrm{nC}\) is placed in a uniform electric field of \(1.5\times10^4\,\mathrm{N\,C^{-1}}\), directed to the right.

Calculate the magnitude and direction of the force on the charge. (2 marks)

(e) A positive charge of \(5.0\,\mathrm{\mu C}\) moves from a point at electric potential \(180\,\mathrm V\) to a point at \(60\,\mathrm V\).

Calculate:

• The work done by the electric field on the charge.
• The change in the charge’s electric potential energy.

(2 marks)

Question 5 (7 marks)

Two large parallel plates have a potential difference of \(600\,\mathrm V\) and a separation of \(0.040\,\mathrm m\). The left plate is positive, and the right plate is negative. Ignore edge effects when considering the central region between the plates.

(a) Calculate the electric field strength between the plates and state its direction. (2 marks)

(b) A positively charged particle has charge \(+4.0\,\mathrm{nC}\) and mass \(2.0\times10^{-6}\,\mathrm{kg}\). It enters the central region moving vertically upwards at \(2.0\times10^3\,\mathrm{m\,s^{-1}}\), perpendicular to the horizontal electric field.

Ignoring gravity, calculate the electric force and acceleration, then describe the particle’s trajectory. (3 marks)

(c) Explain why the electric field is approximately uniform in the central region but not near the plate edges. (2 marks)

Question 6 (8 marks)

(a) Explain how charging by friction differs from charging by contact. Include the movement of electrons and the typical final charge signs. (2 marks)

(b) A negatively charged rod is brought close to a neutral conducting sphere without touching it.

Explain why the sphere is attracted to the rod, even though the sphere’s net charge remains zero. (2 marks)

(c) Describe how a neutral conducting sphere can be charged positively by induction using a negatively charged rod and an earth connection. State the correct order for removing the earth connection and the rod. (2 marks)

(d) Explain how a lightning discharge can form between a charged cloud and the ground. Then explain why a fully enclosed, metal-bodied vehicle can reduce the risk to its occupants during a lightning strike. (2 marks)

Question 7 (10 marks)

Use the supplied value of \(\mu_0\). Assume air-core magnetic fields can be modelled using the ideal equations provided by the syllabus scope.

(a) A long, straight wire carries a current of \(8.0\,\mathrm A\). Calculate the magnetic field strength at a point \(0.040\,\mathrm m\) from the wire using

\[ B=\frac{\mu_0 I}{2\pi r} \]

(3 marks)

(b) An ideal air-core solenoid has \(800\) turns, a length of \(0.40\,\mathrm m\), and a current of \(1.5\,\mathrm A\).

Calculate the magnetic field strength inside the solenoid using

\[ B=\frac{\mu_0 NI}{L} \]

where \(N\) is the number of turns and \(L\) is the solenoid’s length. (3 marks)

(c) Describe how the right-hand grip rule determines:

• The magnetic field direction around a straight current-carrying wire.
• The north end of a current-carrying solenoid.

(2 marks)

(d) Compare the magnetic field patterns around a long, straight current-carrying wire and inside an ideal solenoid. Include one relevant feature of field strength for each. (2 marks)

Question 8 (7 marks)

(a) An idealised electric field map around a single point charge consists of radial lines pointing away from the charge. The lines become more widely spaced as the distance from the charge increases.

Identify the sign of the charge and explain what the spacing indicates about the electric field strength. (2 marks)

(b) An electric dipole consists of two point charges of equal magnitude: \(+2.0\,\mathrm{nC}\) on the left and \(-2.0\,\mathrm{nC}\) on the right.

Describe three features a correct electric field-line sketch of this dipole should show, including the direction of the resultant electric field at the midpoint. (3 marks)

(c) Calculate the resultant electric field strength at the midpoint between the charges in part (b). State its direction. (2 marks)

Question 9 (8 marks)

An electric kettle operates at its rated voltage of \(230\,\mathrm V\) and rated power of \(2.0\,\mathrm{kW}\). It is used for \(4.0\,\mathrm{min}\). Assume its resistance remains constant during this period.

(a) Calculate the current drawn by the kettle. (2 marks)

(b) Calculate the kettle’s resistance at its operating temperature. (2 marks)

(c) Calculate the electrical energy transferred during the \(4.0\,\mathrm{min}\) operating period. Give your answer in kilojoules. (2 marks)

(d) Explain how resistive heating occurs in the kettle’s element. Distinguish electrical power from electrical energy. (2 marks)

Question 10 (8 marks)

A student investigates an unknown resistor by measuring the potential difference across it and the current through it. The stated measurement uncertainties are \(\pm0.02\,\mathrm V\) for voltage and \(\pm0.01\,\mathrm A\) for current.

The results are shown below.

(a) Calculate the resistance for each trial. (3 marks)

(b) Calculate the mean resistance and comment on the consistency of the results. (2 marks)

(c) Using trial 2, calculate the approximate percentage uncertainty in the resistance. When dividing measured quantities, add their percentage uncertainties. (2 marks)

(d) Suggest one practical improvement that would reduce random uncertainty in this investigation. (1 mark)

Question 11 (12 marks)

(a) A field-line map of a bar magnet shows lines emerging from pole A, curving around outside the magnet, and entering pole B. The arrows outside the magnet point from A to B.

Identify poles A and B, describe the field lines inside the magnet, and explain how line density represents field strength. (3 marks)

(b) Two identical bar magnets face one another across a small gap, with their north poles facing each other.

Describe the expected field pattern in the gap and the force between the magnets. State how the pattern and force would differ if a north pole faced a south pole. (3 marks)

(c) Distinguish a permanent magnet from an unmagnetised ferromagnetic object, such as an iron nail. In your answer, distinguish producing a persistent magnetic field from being attracted by a magnet. (2 marks)

(d) An electromagnet must switch on and off repeatedly. Explain why a soft iron core is generally more suitable than a steel core for this purpose. (2 marks)

(e) State two changes that would increase the magnetic field strength inside an ideal solenoid. (2 marks)

03Worked solutions and marking guide

Solution 1

(a) Calculate the current.

Current is the rate of charge flow:

\[ I=\frac{q}{t} \]

Convert the time to seconds:

\[ t=2.0\times60=120\,\mathrm s \]

Substitute:

\[ I=\frac{360}{120} =3.0\,\mathrm A \]

Solution 2

(a) Equivalent resistance.

For the parallel section:

\[ \frac{1}{R_p} =\frac{1}{6.0}+\frac{1}{3.0} \]

\[ \frac{1}{R_p}=\frac{3}{6} =0.50 \]

Therefore,

\[ R_p=2.0\,\Omega \]

The parallel section is in series with the \(2.0\,\Omega\) resistor:

\[ R_{\mathrm{total}}=2.0+2.0 =4.0\,\Omega \]

Solution 3

(a) Resistance and Ohm’s law.

Using Ohm’s law:

\[ R=\frac{V}{I} \]

\[ R=\frac{4.0}{0.80} =5.0\,\Omega \]

A conductor obeys Ohm’s law when its current is directly proportional to its potential difference, provided its temperature and other relevant physical conditions remain constant.

Answer: \(2.5\,\Omega\) at \(0.50\,\mathrm V\), and approximately \(4.6\,\Omega\) at \(2.50\,\mathrm V\).

Marking (2):

• 1 mark for each correct resistance.

(c) Current-voltage graph and Ohm’s law.

Plot potential difference on the horizontal axis and current on the vertical axis. The graph rises, but its gradient decreases as voltage increases.

• Filament lamp

The lamp does not obey Ohm’s law over the measured range. Its resistance increases as the filament heats up, so current is no longer directly proportional to voltage.

Marking (2):

• 1 mark for correctly plotting the data with labelled axes and units.
• 1 mark for identifying the nonlinear relationship and explaining the changing resistance.

(d) Resistance and wire geometry.

For a uniform wire:

\[ R=\frac{\rho L}{A} \]

where \(\rho\) is the material’s resistivity, \(L\) is length, and \(A\) is cross-sectional area.

For the second wire, the material and temperature are unchanged, so \(\rho\) remains constant.

\[ \frac{R_{\mathrm{new}}}{R_0} =\frac{3L}{1.5A}\times\frac{A}{L} \]

\[ \frac{R_{\mathrm{new}}}{R_0}=2 \]

Therefore,

\[ R_{\mathrm{new}}=2R_0 \]

Solution 4

(a) Coulomb’s law.

Coulomb’s law gives the magnitude of the force between two point charges:

\[ F=k\frac{|q_1q_2|}{r^2} \]

Substitute the charge magnitudes and separation:

\[ F=8.99\times10^9 \frac{(3.0\times10^{-6})(2.0\times10^{-6})} {(0.20)^2} \]

\[ F=\frac{8.99\times10^9 \times6.0\times10^{-12}}{0.040} \]

\[ F\approx1.35\,\mathrm N \]

The charges have opposite signs, so the force is attractive.

Solution 5

(a) Electric field between parallel plates.

For an approximately uniform field between parallel plates:

\[ E=\frac{V}{d} \]

Substitute:

\[ E=\frac{600}{0.040} =1.5\times10^4\,\mathrm{V\,m^{-1}} \]

The field points from the positive plate to the negative plate, so it is directed from left to right.

Solution 6

(a) Charging by friction and contact.

During charging by friction, rubbing two different materials can transfer electrons from one material to the other. The objects typically acquire opposite net charges.

During charging by contact, a charged object touches another object and electrons move between them. A previously neutral conductor will typically acquire the same sign as the original charged object.

Marking (2):

• 1 mark for electron transfer and opposite charge signs in friction charging.
• 1 mark for electron transfer through contact and the typical final sign.

(b) Attraction between a charged rod and a neutral sphere.

The negative rod repels electrons within the conducting sphere. Electrons shift towards the far side, leaving the near side relatively positive.

This separation of charge is called polarisation. The sphere remains neutral overall, but the closer positive charge experiences a stronger attraction to the rod than the repulsion from the more distant negative charge. The net force is attractive.

Marking (2):

• 1 mark for explaining induced charge separation while the sphere remains neutral overall.
• 1 mark for explaining why the net force is attractive.

(c) Charging by induction.

Bring the negative rod near the neutral conducting sphere without touching it. Connect the sphere to Earth while the rod remains nearby, allowing electrons to flow from the sphere into Earth.

Disconnect the earth connection first, while the rod is still nearby. Then remove the rod. The sphere is left with a net positive charge.

Marking (2):

• 1 mark for explaining that electrons leave the sphere through the earth connection.
• 1 mark for disconnecting Earth before removing the rod and identifying the final positive charge.

(d) Lightning and vehicle protection.

Charge separation in a cloud and between the cloud and ground can create a very large potential difference. If the electric field becomes strong enough to ionise the air, a conductive path forms and a rapid electrical discharge can occur.

A fully enclosed metal-bodied vehicle can act approximately as a Faraday cage. Charge travels around the conducting exterior rather than through the enclosed space, reducing the electric field inside and helping protect occupants, provided they remain inside and avoid contact with exposed exterior conductors.

Marking (2):

• 1 mark for linking charge separation, a strong electric field, air ionisation, and discharge.
• 1 mark for explaining the protective role of the conducting vehicle body.

Solution 7

(a) Magnetic field around a straight wire.

The magnetic field strength at distance \(r\) from a long, straight wire is:

\[ B=\frac{\mu_0 I}{2\pi r} \]

Substitute:

\[ B=\frac{(4\pi\times10^{-7})(8.0)} {2\pi(0.040)} \]

Simplifying gives:

\[ B=4.0\times10^{-5}\,\mathrm T \]

Solution 8

(a) Point-charge field lines.

Field lines point away from a positive point charge and towards a negative point charge. Since the arrows point away from the source, the charge is positive.

Field-line density represents relative field strength. Lines that are closer together indicate a stronger field, while wider spacing indicates a weaker field.

Marking (2):

• 1 mark for identifying the positive charge.
• 1 mark for relating line spacing to field strength.

(b) Electric dipole field pattern.

A correct field-line sketch for a dipole should show:

1. Lines beginning on the positive charge and ending on the negative charge.
2. Arrowheads pointing from positive to negative, with a pattern that reflects the symmetry of the equal and opposite charges.
3. At the midpoint, the electric field contributions reinforcing rather than cancelling.

At the midpoint, the field due to the positive charge points away from that charge, towards the right. The field due to the negative charge points towards that charge, also towards the right. The resultant field is therefore directed from left to right.

Marking (3):

• 1 mark for the correct starting and ending points of field lines.
• 1 mark for the correct arrow direction and dipole symmetry.
• 1 mark for explaining that the midpoint field contributions reinforce.

(c) Resultant field at the midpoint.

The charges are separated by \(0.12\,\mathrm m\), so the midpoint is \(0.060\,\mathrm m\) from each charge.

The field due to either charge has magnitude:

\[ E_1=E_2=\frac{k|q|}{r^2} \]

\[ E_1= \frac{(8.99\times10^9)(2.0\times10^{-9})} {(0.060)^2} \]

\[ E_1\approx5.0\times10^3\,\mathrm{N\,C^{-1}} \]

Both contributions point to the right, so they add:

\[ E_{\mathrm{resultant}}=E_1+E_2 \]

\[ E_{\mathrm{resultant}} \approx1.0\times10^4\,\mathrm{N\,C^{-1}} \]

Solution 9

(a) Current drawn by the kettle.

Electrical power is related to voltage and current by:

\[ P=VI \]

Rearrange:

\[ I=\frac{P}{V} \]

Convert power to watts:

\[ P=2.0\,\mathrm{kW}=2000\,\mathrm W \]

Then:

\[ I=\frac{2000}{230} \approx8.7\,\mathrm A \]

Solution 10

(a) Resistance for each trial.

Use:

\[ R=\frac{V}{I} \]

Trial 1:

\[ R_1=\frac{2.00}{0.40} =5.0\,\Omega \]

Trial 2:

\[ R_2=\frac{3.00}{0.60} =5.0\,\Omega \]

Trial 3:

\[ R_3=\frac{4.00}{0.80} =5.0\,\Omega \]

Solution 11

(a) Bar-magnet field lines.

Magnetic field lines emerge from the north pole and enter the south pole outside a magnet. Therefore, pole A is north and pole B is south.

The field lines form closed loops: outside the magnet they run from north to south, while inside the magnet they return from south to north.

Closer line spacing indicates a stronger magnetic field. The field is typically strongest near the poles, where the lines are most concentrated.

Marking (3):

• 1 mark for identifying A as north and B as south.
• 1 mark for describing the closed-loop pattern and the field direction inside the magnet.
• 1 mark for explaining the relationship between line density and field strength.

(b) Like and unlike magnetic poles.

When two north poles face each other, the magnets repel. The field lines bend away from the gap instead of joining directly across it.

If a north pole faces a south pole, the magnets attract. The field lines connect across the gap from the north pole to the south pole.

Marking (3):

• 1 mark for identifying repulsion between like poles.
• 1 mark for describing the field pattern when north poles face each other.
• 1 mark for identifying attraction and connecting field lines when unlike poles face each other.

(c) Permanent magnets and ferromagnetic materials.

A permanent magnet produces a persistent magnetic field because its magnetic domains have a substantial overall alignment.

An unmagnetised ferromagnetic object, such as an iron nail, can have little or no net magnetic field because its domains are not aligned overall. However, the domains can respond strongly to an external field, making the object strongly attracted to a magnet and capable of becoming magnetised.

Marking (2):

• 1 mark for distinguishing a permanent magnet’s persistent field from the negligible net field of an unmagnetised object.
• 1 mark for explaining the strong response and ability of ferromagnetic material to become magnetised.

(d) Choosing a core material.

Soft iron is generally more suitable for a switchable electromagnet because it becomes strongly magnetised when current flows and loses much of its magnetism when the current is switched off.

Steel tends to retain more of its magnetisation. That makes it useful when persistent magnetism is wanted, but less suitable when the electromagnet must switch off cleanly.

Marking (2):

• 1 mark for identifying the ease of magnetisation of soft iron.
• 1 mark for explaining its easier demagnetisation compared with steel.

(e) Increasing solenoid field strength.

Two valid changes are:

• Increase the current through the solenoid.
• Increase the number of turns per unit length.
• Insert a suitable ferromagnetic core, such as soft iron, to strengthen the field.

Any two of these earn full marks.

Marking (2):

• 1 mark for each valid factor that increases field strength.

End of paper.