QCAA Physics Alternative Sequence Electromagnetism

15 sample questions with marking guides and sample answers

Q19
2024
QCAA
Paper 1
1 mark
Q19
1 mark

A 7 μC7 \text{ }\mu\text{C} charge requires 1.5×10−8 J1.5 \times 10^{-8} \text{ J} of energy to be moved between two points in an electric field.

What is the order of magnitude of the potential difference between the two points?

A

10−2 V10^{-2} \text{ V}

B

10−3 V10^{-3} \text{ V}

C

10−9 V10^{-9} \text{ V}

D

10−13 V10^{-13} \text{ V}

Reveal Answer
A

10−2 V10^{-2} \text{ V}

This order of magnitude is too large. It likely results from a math error with exponents or ignoring the coefficient of the charge during calculation.

B

10−3 V10^{-3} \text{ V}

Correct Answer

Using the formula V=WqV = \frac{W}{q}, the potential difference is 1.5×10−8 J7×10−6 C≈2.1×10−3 V\frac{1.5 \times 10^{-8} \text{ J}}{7 \times 10^{-6} \text{ C}} \approx 2.1 \times 10^{-3} \text{ V}, which has an order of magnitude of 10−3 V10^{-3} \text{ V}.

C

10−9 V10^{-9} \text{ V}

This order of magnitude is too small and does not match the correct calculation of V=WqV = \frac{W}{q}.

D

10−13 V10^{-13} \text{ V}

This is the result of incorrectly multiplying the energy and charge (W×q≈10−13W \times q \approx 10^{-13}) instead of dividing them.

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 the thickness of the wire does not directly increase the magnetic field strength and could actually decrease the current due to higher electrical resistance.

B

increasing the length of the solenoid

Increasing the length of the solenoid while keeping the number of turns constant decreases the turn density (n=N/Ln = N/L), which would decrease the magnetic field strength.

C

adding more turns of wire to the solenoid

Correct Answer

The magnetic field inside a solenoid is directly proportional to the number of turns per unit length (B=μ0nIB = \mu_0 n I). Adding more turns increases this density, resulting in a stronger magnetic field.

D

using an alternating current instead of a direct current

Using alternating current causes the magnetic field to continuously change direction and magnitude, but it does not inherently increase the maximum strength of the field compared to a direct current.

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 7.5×10−4 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=7.5×10−4 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=20×7.5×10−40.0050\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=20×7.5×10−40.0050\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.

Q5
2024
QCAA
Paper 1
1 mark
Q5
1 mark

The magnitude of the electrostatic force between two positively charged particles

A

is inversely proportional to the square of the distance between the particles.

B

increases as the square of the distance between the particles increases.

C

is proportional to the square of the distance between the particles.

D

is unrelated to the square of the distance between the particles.

Reveal Answer
A

is inversely proportional to the square of the distance between the particles.

Correct Answer

According to Coulomb's Law, the electrostatic force is given by F=k∣q1q2∣r2F = k \frac{|q_1 q_2|}{r^2}, meaning the force is inversely proportional to the square of the distance (r2r^2) between the particles.

B

increases as the square of the distance between the particles increases.

Because the force is inversely proportional to the square of the distance, the force actually decreases as the distance between the particles increases.

C

is proportional to the square of the distance between the particles.

The force is inversely proportional, not directly proportional, to the square of the distance. Direct proportionality would mean the force grows as distance grows, which violates Coulomb's Law.

D

is unrelated to the square of the distance between the particles.

The force is fundamentally related to the square of the distance, as defined by the r2r^2 term in the denominator of Coulomb's Law.

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.

Incorrect. Magnetic flux is a measure of the total magnetic field passing through a specific area, but it is not the definition of the magnetic field itself.

B

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

Incorrect. This definition better describes a gravitational field, which only exerts attractive forces. Magnetic fields can exert both attractive and repulsive forces.

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.

Incorrect. This is the definition of an electric field. A magnetic field specifically requires moving charges, electric currents, or magnetic materials to exert a force.

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

Correct. A magnetic field is defined by the region where magnetic forces are exerted, which are created by and act upon magnets, electric currents, and moving charges.

Q13
2022
QCAA
Paper 1
1 mark
Q13
1 mark

A rectangular coil of 3000 turns and dimensions 0.1 m × 0.2 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

Incorrect. A frequency of 1 rev/s would only produce an average EMF of 0.480.48 V, which is insufficient.

B

3

Incorrect. A frequency of 3 rev/s would produce an average EMF of 1.441.44 V, falling short of the required 66 V.

C

13

Correct Answer

Correct. Using the formula for average EMF Eavg=4NABfE_{avg} = 4NABf, we find f=12.5f = 12.5 rev/s. The minimum whole number of revolutions per second to achieve at least 66 V is 13.

D

50

Incorrect. A frequency of 50 rev/s would produce an average EMF of 2424 V, which is much higher than the minimum required to reach 66 V.

Q21
2020
QCAA
Paper 1
3 marks
Q21
3 marks

An electron is 5.29×10−115.29 \times 10^{-11} m away from a proton. Calculate the net electrostatic force (to 3 significant figures) and direction of force that the electron will experience.

Reveal Answer

F=kQqr2F = k \frac{Qq}{r^2}

F=9.0×109 N m2C−2+1.6×10−19C×−1.6×10−19C(5.29×10−11 m)2F = 9.0 \times 10^9 \text{ N m}^2\text{C}^{-2} \frac{+1.6 \times 10^{-19}\text{C} \times -1.6 \times 10^{-19}\text{C}}{(5.29 \times 10^{-11} \text{ m})^2}

F=8.23×10−8 N towards the protonF = 8.23 \times 10^{-8} \text{ N towards the proton}

Marking Criteria
DescriptorMarks

indicates an understanding of the physical scenario in relation to Coulomb's Law by substituting in relevant values

1

provides pertinent mathematical operation/s correctly performed and arriving at a consequentially correct value for force

1

includes the correct direction

1
Q10
2024
QCAA
Paper 1
1 mark
Q10
1 mark

An experiment was conducted to determine the force experienced by an 85 cm85 \text{ cm} wire with a 2.4 A2.4 \text{ 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=0.0306sin⁡θF = 0.0306 \sin\theta.

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

A

10−4 T10^{-4} \text{ T}

B

10−2 T10^{-2} \text{ T}

C

102 T10^2 \text{ T}

D

104 T10^4 \text{ T}

Reveal Answer
A

10−4 T10^{-4} \text{ T}

This order of magnitude is too small. It may result from failing to convert the wire's length from centimeters to meters and making a calculation error.

B

10−2 T10^{-2} \text{ T}

Correct Answer

Using the magnetic force formula F=BILsin⁡θF = BIL \sin\theta, we can set BIL=0.0306BIL = 0.0306. Solving for BB gives B=0.03062.4×0.85=0.015 TB = \frac{0.0306}{2.4 \times 0.85} = 0.015 \text{ T}, which has an order of magnitude of 10−2 T10^{-2} \text{ T}.

C

102 T10^2 \text{ T}

This order of magnitude is too large. It likely results from incorrectly dividing the ILIL product by the force coefficient (2.04/0.0306≈66.7 T2.04 / 0.0306 \approx 66.7 \text{ T}).

D

104 T10^4 \text{ T}

This value is extremely large and incorrect, possibly resulting from a combination of inverted division and unit conversion errors.

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.

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.

Reveal Answer

F=14πε0Qqr2F = \frac{1}{4\pi\varepsilon_0}\frac{Qq}{r^2}

0.025=9×109×Q20.3020.025 = \frac{9\times 10^{9}\times Q^2}{0.30^2}

Q=0.025×0.3029×109=5.0×10−7Q = \sqrt{\frac{0.025\times 0.30^2}{9\times 10^{9}}} = 5.0\times 10^{-7}

Charge = 5.0×10−7 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 5.0×10−7 C5.0\times 10^{-7}\ \text{C}

1
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 (qv⃗×B⃗q\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.

Q1
2024
QCAA
Paper 2
4 marks
Q1
4 marks

A coil of wire with 100 turns and a radius of 1.4 cm is placed perpendicular to a magnetic field of strength 0.510 T. The magnetic field strength is then changed to 0.030 T in 0.020 s.

Calculate the magnitude of electromotive force (emf) induced in the coil. Show your working.

Reveal Answer

r=1.4 cm=0.014 mr = 1.4 \text{ cm} = 0.014 \text{ m}

A=πr2=π×0.0142≈6.16×10−4 m2A = \pi r^2 = \pi \times 0.014^2 \approx 6.16 \times 10^{-4} \text{ m}^2

emf=−nΔ(BA⊥)Δt=−100×(0.030−0.510)×(6.16×10−4)0.020=1.48\begin{align*} \text{emf} &= -\frac{n\Delta(BA_{\perp})}{\Delta t}\\ &= -\frac{100 \times (0.030 - 0.510) \times (6.16 \times 10^{-4})}{0.020}\\ &= 1.48 \end{align*}

Magnitude of emf = 1.5 V

Marking Criteria
DescriptorMarks

converts the radius to SI units (from cm to m)

1

determines the area of the coil

1

recognises the scenario relates to induction of an electromotive force by using the equation

1

calculates emf

1
Q6
2020
SCSA
4 marks
Q6
4 marks

Calculate the electric field strength 2.25×10−32.25 \times 10^{-3} m from a point charge of 4.00×10−184.00 \times 10^{-18} C.

Reveal Answer

F=q24πεr2F = \frac{q^2}{4\pi \varepsilon r^2}

E=Fq=q4πεr2E = \frac{F}{q} = \frac{q}{4\pi \varepsilon r^2}

E=4.00×10−18(4π×8.85×10−12)(2.25×10−3)2E = \frac{4.00 \times 10^{-18}}{(4\pi \times 8.85 \times 10^{-12})(2.25 \times 10^{-3})^2}

E=7.11×10−3 N C−1E = 7.11 \times 10^{-3}\ \mathrm{N\,C^{-1}}

Marking Criteria
DescriptorMarks

Uses Coulomb's Law

1

Divides both sides by qq to get EE on LHS and eliminate qq on RHS

1

Uses correct constant and squares rr

1

Correct/consistent numerical answer

1
Q1
2022
QCAA
Paper 1
1 mark
Q1
1 mark

Electromotive force is

A

the production of voltage across an electrical conductor due to its dynamic interaction with a magnetic field.

B

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

C

the repulsion experienced by two negatively charged particles.

D

one of the four fundamental forces.

Reveal Answer
A

the production of voltage across an electrical conductor due to its dynamic interaction with a magnetic field.

Incorrect. This describes electromagnetic induction, which is just one specific method of generating an electromotive force, rather than the general definition of the term itself.

B

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

Correct Answer

Correct. Electromotive force (EMF) is defined as the electrical potential difference (voltage) provided by a source, such as a battery, that drives an electric current through a circuit.

C

the repulsion experienced by two negatively charged particles.

Incorrect. This describes the electrostatic force (Coulomb force) between like charges, not the potential difference that drives a current.

D

one of the four fundamental forces.

Incorrect. Despite the word "force" in its name, electromotive force is actually a potential difference measured in volts (energy per unit charge), not a physical force measured in Newtons.

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