QCAA Physics Electromagnetism

15 sample questions with marking guides and sample answers

Q5
2021
QCAA
Paper 2
4 marks
Q5
4 marks

An alpha particle with a charge of +3.2×1019+3.2 \times 10^{-19} C moves through an electric field, accelerating from rest through a potential difference of 240 V.

Determine the velocity of the particle at the end of its acceleration, expressing your answer in scientific notation. (m/s to 2 significant figures)

Reveal Answer

The change in potential energy of an electric charge moving through an electric field is equivalent to the work done on the charge.
V=ΔUqV = \frac{\Delta U}{q}
ΔU=Vq\Delta U = Vq
=240×3.2×1019= 240 \times 3.2 \times 10^{-19}
=7.68×1017J=W= 7.68 \times 10^{-17} J = W

The work done on an object is equal to the change in kinetic energy.
Ek=12mv2E_k = \frac{1}{2}mv^2
7.68×1017=12×6.64×1027×v27.68 \times 10^{-17} = \frac{1}{2} \times 6.64 \times 10^{-27} \times v^2
v2=7.68×101712×6.64×1027v^2 = \frac{7.68 \times 10^{-17}}{\frac{1}{2} \times 6.64 \times 10^{-27}}
v=7.68×101712×6.64×1027v = \sqrt{\frac{7.68 \times 10^{-17}}{\frac{1}{2} \times 6.64 \times 10^{-27}}}
Velocity = 1.5×105 m s11.5 \times 10^5 \text{ m s}^{-1} (to 2 significant figures)

Marking Criteria
DescriptorMarks

Recognises the scenario relates to work done on a moving charge in an electric field

1

Identifies that work done on the charge equates to its kinetic energy

1

Provides appropriate mathematical reasoning

1

Determines the velocity

1
Q17
2020
QCAA
Paper 1
1 mark
Q17
1 mark

The definition of magnetic field is

A

a region of space through which the total magnetic flux is measured.

B

a region of space surrounding a body in which another body experiences a force of attraction.

C

a region of space around an electrically charged particle or object within which a force would be exerted on other electrically charged particles or objects.

D

a region of space near a magnet, electric current or moving electrically charged particle in which a magnetic force acts on any other magnet, electric current or moving electrically charged particle.

Reveal Answer
A

a region of space through which the total magnetic flux is measured.

This describes the domain for calculating magnetic flux, which is the integral of the magnetic field over an area, rather than defining the field itself.

B

a region of space surrounding a body in which another body experiences a force of attraction.

This is a generic definition that could apply to gravitational or electric fields; it fails to specify that the force is magnetic or that it acts on magnets and moving charges.

C

a region of space around an electrically charged particle or object within which a force would be exerted on other electrically charged particles or objects.

This is the definition of an electric field, which exerts forces on stationary or moving electric charges, whereas magnetic fields specifically affect moving charges or magnetic materials.

D

a region of space near a magnet, electric current or moving electrically charged particle in which a magnetic force acts on any other magnet, electric current or moving electrically charged particle.

Correct Answer

This accurately defines a magnetic field by identifying its sources (magnets, currents, moving charges) and its effect (exerting magnetic force on similar entities).

Q12
2025
VCAA
7 marks
Q12

Denzil is using a demonstration hand-cranked generator.

A schematic diagram of the generator is shown in Figure 17. The generator contains a rectangular coil with side lengths of 5.0 cm and 2.5 cm, consisting of 20 turns of insulated copper wire. The coil is rotated between two bar magnets that provide a field strength of 0.60 T between the magnets.

Denzil rotates the coil at a frequency of 50 Hz.

Q12a
1 mark

State why the flux through the coil changes as the coil rotates.

Reveal Answer

The flux through the coil is determined by the angle between the plane of the coil and the magnetic field.

Marking Criteria
DescriptorMarks

States that the angle or orientation between the plane of the coil and the magnetic field changes

1
Q12b
1 mark

Show that the change in flux as the coil rotates from a horizontal to a vertical position is 7.5×104 Wb7.5 \times 10^{-4} \text{ Wb}.

Reveal Answer

Δϕ=BA\Delta\phi = BA

Δϕ=0.60×(0.05×0.025)\Delta\phi = 0.60 \times (0.05 \times 0.025)

Δϕ=7.5×104 Wb\Delta\phi = 7.5 \times 10^{-4} \text{ Wb}

Marking Criteria
DescriptorMarks

Demonstrates correct substitution into the magnetic flux formula (e.g., Δϕ=0.60×(0.05×0.025)\Delta\phi = 0.60 \times (0.05 \times 0.025))

1
Q12c
3 marks

Calculate the average EMF induced as the coil is rotated through a quarter turn from a horizontal to a vertical position.

Reveal Answer

ε=nΔϕΔt\varepsilon = n\frac{\Delta\phi}{\Delta t}

ε=20×7.5×1040.0050\varepsilon = 20 \times \frac{7.5 \times 10^{-4}}{0.0050}

ε=3.0 V\varepsilon = 3.0 \text{ V}

Marking Criteria
DescriptorMarks

Calculates the correct time for a quarter turn (Δt=0.0050 s\Delta t = 0.0050 \text{ s})

1

Substitutes values correctly into Faraday's law formula (ε=20×7.5×1040.0050\varepsilon = 20 \times \frac{7.5 \times 10^{-4}}{0.0050})

1

Calculates the correct average EMF of 3.0 V3.0 \text{ V}

1
Q12d
2 marks

State a change to the set-up in Figure 17 that could produce a DC output from the generator. Give a reason for your choice.

Reveal Answer

Replace the slip rings with a split-ring commutator. The split-ring commutator will reverse the connections to the loop every half turn to ensure a DC output.

Marking Criteria
DescriptorMarks

Identifies replacing the slip rings with a split-ring commutator

1

Explains that the split-ring commutator reverses the connections to the loop every half turn to ensure a DC output

1
Q1
2025
NESA
1 mark
Q1
1 mark

Which of the following did Maxwell contribute to the understanding of the nature of light?

A

Explanation of atomic emission spectra

B

Prediction of the speed of electromagnetic waves

C

Experimental support for the particle model of light

D

Experimental confirmation of light beyond the visible spectrum

Reveal Answer
A

Explanation of atomic emission spectra

Incorrect. The explanation of atomic emission spectra was developed later by Niels Bohr and other quantum physicists using quantum mechanics, not by Maxwell.

B

Prediction of the speed of electromagnetic waves

Correct Answer

Correct. Maxwell formulated a set of equations that predicted the existence of electromagnetic waves traveling at the speed of light (cc), establishing that light itself is an electromagnetic wave.

C

Experimental support for the particle model of light

Incorrect. Maxwell's work firmly established the wave model of light; experimental support for the particle model came later from Einstein's explanation of the photoelectric effect.

D

Experimental confirmation of light beyond the visible spectrum

Incorrect. Maxwell provided the theoretical framework for electromagnetic waves, but experimental confirmation of invisible light like infrared and radio waves was achieved by scientists like Herschel and Hertz.

Q26
2024
QCAA
Paper 1
3 marks
Q26
3 marks

The centres of two small equally positively charged metallic spheres are separated by a distance of 0.30 m and experience a force of 0.025 N between them.

Calculate the charge on each of the metallic spheres. Show your working.
Charge = _______ C

Reveal Answer

F=14πϵ0Qqr2F = \frac{1}{4\pi\epsilon_0} \frac{Qq}{r^2}
0.025=9×109×Q20.3020.025 = \frac{9 \times 10^9 \times Q^2}{0.30^2}
Q=0.025×0.3029×109Q = \sqrt{\frac{0.025 \times 0.30^2}{9 \times 10^9}}
=5.0×107= 5.0 \times 10^{-7}
Charge = 5.0×107 C5.0 \times 10^{-7} \text{ C}

Marking Criteria
DescriptorMarks

recognises the scenario relates to Coulomb's Law

1

recognises the charges have the same value in the equation

1

calculates the charge of the metallic spheres to be 5.0×107 C5.0 \times 10^{-7} \text{ C}

1
Q10
2024
QCAA
Paper 1
1 mark
Q10
1 mark

An experiment was conducted to determine the force experienced by an 85 cm wire with a 2.4 A current flowing through it in an external magnetic field. It was rotated through varying angles within the magnetic field such that data analysis identified the relationship F=0.0306sinθF = 0.0306 \sin \theta.

What is the order of magnitude of the strength of the external magnetic field?

A

10410^{-4} T

B

10210^{-2} T

C

10210^{2} T

D

10410^{4} T

Reveal Answer
A

10410^{-4} T

This value is too small. The calculated magnetic field strength is approximately 0.015 T0.015 \text{ T}, which is two orders of magnitude larger than 104 T10^{-4} \text{ T}.

B

10210^{-2} T

Correct Answer

Comparing the experimental relationship F=0.0306sinθF = 0.0306 \sin \theta to the theoretical formula F=ILBsinθF = ILB \sin \theta, we see that ILB=0.0306ILB = 0.0306. Solving for BB gives B=0.03062.4×0.850.015 TB = \frac{0.0306}{2.4 \times 0.85} \approx 0.015 \text{ T}, which is on the order of 102 T10^{-2} \text{ T}.

C

10210^{2} T

This value is significantly larger than the actual magnetic field. The calculation yields B0.015 TB \approx 0.015 \text{ T}, whereas 102 T10^2 \text{ T} represents an extremely strong magnetic field not supported by the data.

D

10410^{4} T

This value is far too large. The calculated field strength is approximately 1.5×102 T1.5 \times 10^{-2} \text{ T}, which is six orders of magnitude smaller than 104 T10^4 \text{ T}.

Q13
2022
QCAA
Paper 1
1 mark
Q13
1 mark

A rectangular coil of 3000 turns and dimensions 0.1 m×0.2 m0.1 \text{ m} \times 0.2 \text{ m} is rotated in a uniform magnetic field of 2 mT.
Calculate the minimum number of revolutions per second required to produce an average EMF of 6 V.

A

1

B

3

C

13

D

50

Reveal Answer
A

1

This rotation speed is too slow. Using the formula εavg=4NBAf\varepsilon_{avg} = 4NBAf, a frequency of 1 rev/s1 \text{ rev/s} would only produce an average EMF of 0.48 V0.48 \text{ V}.

B

3

This value is insufficient. Substituting f=3f=3 into the average EMF equation yields approximately 1.44 V1.44 \text{ V}, which is less than the required 6 V6 \text{ V}.

C

13

Correct Answer

The average EMF for a rotating coil is εavg=4NBAf\varepsilon_{avg} = 4NBAf. Solving for frequency: f=64(3000)(2×103)(0.02)=12.5 rev/sf = \frac{6}{4(3000)(2 \times 10^{-3})(0.02)} = 12.5 \text{ rev/s}. The closest integer option is 13.

D

50

This frequency is too high. At 50 rev/s50 \text{ rev/s}, the generated average EMF would be 24 V24 \text{ V}, far exceeding the required 6 V6 \text{ V}.

Q10
2022
QCAA
Paper 1
1 mark
Q10
1 mark

Electric field strength refers to the

A

intensity of an electric field at a particular location.

B

change in electrical potential energy between two defined points.

C

sum of electrically charged particles passing a point in a given time.

D

physical property of an object experiencing a force in an electromagnetic field.

Reveal Answer
A

intensity of an electric field at a particular location.

Correct Answer

Electric field strength is a measure of the intensity or magnitude of the electric field at a specific point, defined as the force exerted per unit positive charge (E=F/qE = F/q).

B

change in electrical potential energy between two defined points.

The change in electrical potential energy per unit charge between two points defines electric potential difference (voltage), not electric field strength.

C

sum of electrically charged particles passing a point in a given time.

The rate at which charged particles pass a specific point defines electric current (I=Q/tI = Q/t), not electric field strength.

D

physical property of an object experiencing a force in an electromagnetic field.

This describes electric charge, which is the property of matter that causes it to experience a force, whereas electric field strength describes the field itself.

Q7
2022
QCAA
Paper 1
1 mark
Q7
1 mark

Which change would produce the greatest increase in magnetic field strength inside a current-carrying solenoid?

A

decreasing the thickness of the wire

B

increasing the length of the solenoid

C

adding more turns of wire to the solenoid

D

using an alternating current instead of a direct current

Reveal Answer
A

decreasing the thickness of the wire

Decreasing wire thickness increases electrical resistance, which would reduce the current II (for a fixed voltage) and consequently decrease the magnetic field strength.

B

increasing the length of the solenoid

The magnetic field strength B=μ0NLIB = \mu_0 \frac{N}{L} I is inversely proportional to the length LL; increasing the length while keeping the number of turns constant decreases the turn density, weakening the field.

C

adding more turns of wire to the solenoid

Correct Answer

The magnetic field strength inside a solenoid is directly proportional to the number of turns (NN); increasing the number of turns increases the turn density nn, which increases the field strength according to B=μ0nIB = \mu_0 n I.

D

using an alternating current instead of a direct current

Alternating current produces a magnetic field that fluctuates in magnitude and direction, and inductive reactance often reduces the current compared to DC, which would not increase the field strength.

Q7
2021
VCAA
7 marks
Q7

The generator of an electrical power plant delivers 500 MW500 \text{ MW} to external transmission lines when operating at 25 kV25 \text{ kV}. The generator's voltage is stepped up to 500 kV500 \text{ kV} for transmission and stepped down to 240 V240 \text{ V} 100 km100 \text{ km} away (for domestic use). The overhead transmission lines have a total resistance of 30.0 Ω30.0 \text{ } \Omega. Assume that all transformers are ideal.

Q7a
2 marks

Explain why the voltage is stepped up for transmission along the overhead transmission lines.

Reveal Answer

Students were required to identify that stepping up the voltage allowed the current to be reduced while maintaining constant power. The reason for reducing the current is that the power lost is related to the transmission current by: P=I2RP = I^2R.

Marking Criteria
DescriptorMarks

Identifies that stepping up the voltage allows the current to be reduced while maintaining constant power

1

Relates the reduced current to a reduction in power lost during transmission (P=I2RP = I^2R)

1
Q7b
2 marks

Calculate the current in the overhead transmission lines. Show your working.

Reveal Answer

P=VIP = VI
I=500×106500×103I = \frac{500 \times 10^6}{500 \times 10^3}
I=1000 A or 1.0 kAI = 1000 \text{ A or } 1.0 \text{ kA}

Marking Criteria
DescriptorMarks

Correct substitution into P=VIP = VI

1

Correct final answer of 1000 A1000 \text{ A} or 1.0 kA1.0 \text{ kA}

1
Q7c
3 marks

Determine the maximum power available for domestic use at 240 V240 \text{ V}. Show all your working.

Reveal Answer

This solution has two steps. The first is to calculate the power lost:
P=I2RP = I^2R
P=10002×30P = 1000^2 \times 30
P=30×106 W (30 MW)P = 30 \times 10^6 \text{ W (30 MW)}
This was then subtracted from the power delivered by the generator:
Pavail=500×10630×106P_{avail} = 500 \times 10^6 - 30 \times 10^6
Pavail=470 MWP_{avail} = 470 \text{ MW}

Marking Criteria
DescriptorMarks

Calculates the power lost in the lines (30 MW30 \text{ MW})

1

Subtracts the power lost from the total power delivered by the generator

1

Calculates the correct available power (470 MW470 \text{ MW})

1
Q19
2023
QCAA
Paper 1
1 mark
Q19
1 mark

Calculate the electric field strength experienced at a distance of 2.8×10112.8\times10^{-11} m from the centre of a helium nucleus.

A

1.0×102 N C11.0\times10^2\text{ N C}^{-1}

B

2.0×102 N C12.0\times10^2\text{ N C}^{-1}

C

3.7×1012 N C13.7\times10^{12}\text{ N C}^{-1}

D

7.3×1012 N C17.3\times10^{12}\text{ N C}^{-1}

Reveal Answer
A

1.0×102 N C11.0\times10^2\text{ N C}^{-1}

This is the value for the electric potential (V=kQrV = \frac{kQ}{r}), not the electric field strength (E=kQr2E = \frac{kQ}{r^2}).

B

2.0×102 N C12.0\times10^2\text{ N C}^{-1}

This value represents the electric potential calculated incorrectly by using the mass number (4) instead of the atomic number (2) for the charge.

C

3.7×1012 N C13.7\times10^{12}\text{ N C}^{-1}

Correct Answer

Using the formula E=kQr2E = \frac{kQ}{r^2} with the charge of a helium nucleus (Q=2e=3.2×1019Q = 2e = 3.2 \times 10^{-19} C), the result is 3.7×1012 N C13.7 \times 10^{12} \text{ N C}^{-1}.

D

7.3×1012 N C17.3\times10^{12}\text{ N C}^{-1}

This result comes from incorrectly using the mass number (4) instead of the atomic number (2) to calculate the total charge (Q=4eQ = 4e).

Q14
2021
QCAA
Paper 1
1 mark
Q14
1 mark

Moving electric charges in a magnetic field experience

A

a decrease in charge.

B

an increase in charge.

C

a force parallel to the direction of the magnetic field.

D

a force perpendicular to the direction of the magnetic field.

Reveal Answer
A

a decrease in charge.

Electric charge is a fundamental conserved property and does not decrease due to motion or the presence of a magnetic field.

B

an increase in charge.

The magnitude of an electric charge is invariant; it does not increase regardless of its speed or the external fields applied.

C

a force parallel to the direction of the magnetic field.

The magnetic force is the result of a cross product (vecF=qvecvtimesvecB\\vec{F} = q\\vec{v} \\times \\vec{B}), which produces a vector perpendicular to the magnetic field, not parallel to it.

D

a force perpendicular to the direction of the magnetic field.

Correct Answer

According to the Lorentz force law, the magnetic force exerted on a moving charge is always perpendicular to both the velocity of the charge and the direction of the magnetic field.

Q11
2024
VCAA
1 mark
Q11
1 mark

In Victoria, the electrical energy generated at the Loy Yang A power station is transmitted to Melbourne, approximately 170 km170 \text{ km} away, using 500 kV500 \text{ kV} transmission lines.

Which one of the following best describes the reason for the use of high-voltage transmission of electrical energy over long distances?

A

Transformers can be used to increase the voltage.

B

High voltages reduce energy losses in the transmission lines.

C

High voltages can easily carry the large power required by cities.

D

High voltages reduce the overall total resistance in the transmission lines.

Reveal Answer
A

Transformers can be used to increase the voltage.

While transformers are indeed used to step up the voltage, this explains how high voltages are achieved, not why they are beneficial for long-distance transmission.

B

High voltages reduce energy losses in the transmission lines.

Correct Answer

For a given amount of power, transmitting at a higher voltage reduces the current (P=VIP=VI). A lower current significantly reduces the power lost as heat in the transmission lines (Ploss=I2RP_{\text{loss}} = I^2R).

C

High voltages can easily carry the large power required by cities.

While high voltages are used to transmit large amounts of power, the fundamental reason for stepping up the voltage is to minimize power loss during transmission, not just to increase capacity.

D

High voltages reduce the overall total resistance in the transmission lines.

The resistance of a transmission line is determined by its physical properties (material, length, and cross-sectional area), not by the voltage applied to it.

Q17
2021
QCAA
Paper 1
1 mark
Q17
1 mark

Electrical potential energy is the

A

intensity of an electric field at a particular location.

B

difference in potential that tends to give rise to an electric current.

C

capacity of electric charge carriers to do work due to their position in an electric circuit.

D

work done on an electron in accelerating it through an electrical potential difference of one volt.

Reveal Answer
A

intensity of an electric field at a particular location.

This describes electric field strength (EE), which is the force exerted per unit charge at a specific location, not the energy.

B

difference in potential that tends to give rise to an electric current.

This describes electric potential difference (voltage), which is the difference in electric potential energy per unit charge between two points.

C

capacity of electric charge carriers to do work due to their position in an electric circuit.

Correct Answer

Electrical potential energy is the stored energy a charge possesses due to its position in an electric field, representing the capacity to do work.

D

work done on an electron in accelerating it through an electrical potential difference of one volt.

This is the specific definition of an electron-volt (eV), a unit of energy, rather than the general definition of electrical potential energy.

Q19
2024
NESA
1 mark
Q19
1 mark

In a vacuum chamber there is a uniform electric field and a uniform magnetic field.

A proton having a velocity, vv, enters the chamber. Its velocity remains unchanged as it travels through the chamber.

A second proton having a velocity, 2v2v, in the same direction as the first proton, then enters the chamber at the same point as the first proton.

In the chamber, the acceleration of the second proton

A

is zero.

B

is constant in magnitude and direction.

C

changes in both magnitude and direction.

D

is constant in magnitude, but not direction.

Reveal Answer
A

is zero.

The magnetic force depends on velocity (FB=qvBF_B = qvB), so doubling the velocity doubles the magnetic force. This unbalances it with the constant electric force, resulting in a non-zero net force and acceleration.

B

is constant in magnitude and direction.

The initial net force changes the proton's velocity vector. Since the magnetic force depends on this changing velocity, the net force and acceleration cannot remain constant.

C

changes in both magnitude and direction.

Correct Answer

The unbalanced forces cause the proton's velocity to change. Because the magnetic force depends on the instantaneous velocity vector (qv×Bq\vec{v} \times \vec{B}), the net force and resulting acceleration will continuously change in both magnitude and direction.

D

is constant in magnitude, but not direction.

As the proton's velocity changes in both magnitude and direction relative to the magnetic field, the magnitude of the magnetic force (and thus the net acceleration) will also change, not just its direction.

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