These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data and assumptions for every question: μ₀ = 4π × 10⁻⁷ T·m/A; “the direction of B” means the direction the north end of a small compass needle points; all experimental readings are invented for practice. A scientific calculator is assumed. Round only at the end.
Question 1 (multiple choice · foundation)
A bar magnet lies along a table with its north pole at the east end. It is cut in half at its midpoint. Which statement describes the two pieces?
- (A) The east piece is a north pole only and the west piece is a south pole only.
- (B) Each piece is a dipole. The east piece has a south pole at the cut, and the west piece has a north pole at the cut.
- (C) Each piece is a dipole. The east piece has a north pole at the cut, and the west piece has a south pole at the cut.
- (D) Neither piece is magnetic, because cutting the magnet separates the poles and their fields cancel.
Answer and explanation
Answer: (B). There are no magnetic monopoles, so each piece is a complete dipole. Inside the original magnet the field ran from S (west) to N (east), and each piece keeps that alignment. The east piece still has N at its east end, so its other end (the cut) is S. The west piece still has S at its west end, so its cut end is N. The two cut faces are opposite poles and would attract if pushed back together.
- (A) assumes the poles can be separated. No isolated north or south pole has ever been found.
- (C) gets the dipoles right but reverses them: each piece would then have two north poles or two south poles at its ends, which is impossible for a dipole.
- (D) is wrong because the dipoles inside each piece are still aligned. Cutting does not scramble them.
Question 2 (multiple choice · core)
Which statement about magnetic permeability is correct?
- (A) Every material has a single fixed value of μ, which is listed in tables like its density.
- (B) μ₀ is a constant, but the permeability of a material depends on factors such as its temperature and the strength of the external field.
- (C) A diamagnetic material has a permeability much greater than μ₀.
- (D) Permeability measures how much electric charge a material can store in a magnetic field.
Answer and explanation
Answer: (B). The vacuum permeability μ₀ = 4π × 10⁻⁷ T·m/A is a constant of nature. The permeability of matter comes from its composition and structure, and it changes with temperature, orientation and the strength of the external field.
- (A) treats a material’s μ as a constant. That is true only of μ₀.
- (C) describes a ferromagnetic material. A diamagnetic material has μ slightly less than μ₀.
- (D) confuses permeability with capacitance. Permeability measures how strongly a material becomes magnetised in an external field.
Question 3 (multiple choice · core)
A bar magnet lies along a line running east–west, with its south pole at the west end. A small compass is placed on the same line, a short distance west of the magnet. The magnet’s field there is much stronger than Earth’s. Which way does the north end of the needle point?
- (A) East, towards the magnet
- (B) West, away from the magnet
- (C) North
- (D) It does not settle in any direction, because the field on the axis is zero
Answer and explanation
Answer: (A). Outside a magnet, field lines go into the south pole. On the axis just west of the S end, B points east, into the magnet. The needle lines up with B, so its north end points east. This also fits “unlike poles attract”: the needle’s N end is pulled towards the magnet’s S pole.
- (B) is the direction near a north pole, where the field points away from the magnet.
- (C) ignores the magnet. Earth’s field is much weaker here, so it has little effect.
- (D) is wrong: the field on the axis of a magnet is not zero. It points along the axis.
Question 4 (multiple choice · core)
A small sample hangs from a thread. When either the N pole or the S pole of a very strong magnet is brought close, the sample is pushed slightly away. When the magnet is removed, the sample has no magnetic effect on a compass. What is the sample?
- (A) A ferromagnetic material
- (B) A paramagnetic material
- (C) A diamagnetic material
- (D) A permanent magnet
Answer and explanation
Answer: (C). Weak repulsion from both poles means the sample’s response is an alignment opposite to the external field, whichever way that field points. That is diamagnetism. Nothing remains afterwards.
- (A) A ferromagnet would be strongly attracted to either pole, and might stay magnetised.
- (B) A paramagnet would be weakly attracted to either pole.
- (D) A permanent magnet would be attracted by one pole and repelled by the other. It would also deflect the compass after the magnet is removed.
Question 5 (calculation · core)
At a place where the horizontal part of Earth’s field is 1.8 × 10⁻⁵ T pointing north, a magnet is placed so that its own field at a compass points due east. The compass needle settles at 35° east of north.
(a) Calculate the magnet’s field at the compass. (b) Calculate the size of the total horizontal field. (c) The magnet is moved until its field at the compass is half the value in (a). A student says the needle will now point 17.5° east of north. Calculate the correct angle and explain the student’s error.
Worked solution
- The needle lines up with the total field, which is the vector sum of a northward part (Earth) and an eastward part (magnet). The angle east of north satisfies tan θ = B_magnet ÷ B_Earth.
- (a) B_magnet = B_Earth × tan 35° = (1.8 × 10⁻⁵ T)(0.700) = 1.26 × 10⁻⁵ T (≈ 1.3 × 10⁻⁵ T).
- (b) B_total = B_Earth ÷ cos 35° = 1.8 × 10⁻⁵ T ÷ 0.819 = 2.20 × 10⁻⁵ T. (Check: √[(1.8)² + (1.26)²] × 10⁻⁵ T = 2.20 × 10⁻⁵ T.)
- (c) New B_magnet = 0.630 × 10⁻⁵ T. θ = tan⁻¹(0.630 ÷ 1.8) = 19.3° east of north.
- The student halved the angle. The angle depends on the ratio of the two fields through tan θ, which is not proportional to θ. Halving one component halves tan θ, not θ.
Suggested mark points (4): 1 for setting up the right-angled vector sum with tan θ = B_magnet/B_Earth; 1 for 1.26 × 10⁻⁵ T; 1 for 2.20 × 10⁻⁵ T; 1 for 19.3° with the reason that tan θ, not θ, is halved.
Common errors: using sin 35° (gives 1.03 × 10⁻⁵ T) or cos 35° (gives 1.47 × 10⁻⁵ T) in (a), or dividing by tan 35° (gives 2.57 × 10⁻⁵ T), which puts the magnet’s field on the wrong side of the triangle.
Question 6 (constructed response · core)
A student models Earth’s magnetic field as the field of a dipole inside Earth.
(a) Sketch Earth with its rotation axis, and draw at least four field lines of the dipole model outside Earth, with arrows. Label which magnetic pole is in the northern hemisphere. (b) Explain, using the interaction between poles, why the north end of a compass points roughly towards geographic north. (c) In the dipole model, state the direction of the field at Earth’s centre, and justify your answer using a property of magnetic field lines. (d) A satellite moves from 500 km to 5000 km above Earth’s surface. State how the size of Earth’s field at the satellite changes, and justify.
Worked solution and suggested Marlbridge rubric
(a) Lines leave Earth from the southern hemisphere, curve round outside Earth and enter the ground in the northern hemisphere. The magnetic pole in the northern hemisphere is a magnetic south pole. (A sketch with the dipole slightly tilted from the rotation axis is fine but not required.)
(b) Unlike poles attract. The magnetic pole near geographic north is a south pole, so it attracts the north end of the needle. Equivalently, the needle lines up with Earth’s field, which points towards that pole.
(c) Field lines form closed loops. Outside Earth they run from the southern hemisphere to the northern hemisphere, so inside they must run back from north to south to close each loop. At the centre the field points roughly towards geographic south (from the magnetic S pole to the magnetic N pole, as inside any magnet).
(d) The field gets weaker. The field of a dipole decreases with distance from the dipole, and the field lines are further apart at greater distances.
| Point | What earns it |
|---|---|
| 1 | Sketch: closed field lines leaving the southern hemisphere and entering the northern hemisphere, with consistent arrows |
| 1 | Labels the northern-hemisphere pole as a magnetic south pole |
| 1 | (b) Links “unlike poles attract” (or alignment with B) to the north end of the needle pointing north |
| 1 | (c) Field at the centre points from north to south because field lines are closed loops |
| 1 | (d) Weaker, justified by the field of a dipole decreasing with distance (or line spacing) |
Do not award (c) for a direction with no reference to closed loops.
Question 7 (constructed response · stretch)
A student holds a small magnet on a line along its axis and measures the magnet’s field at several distances r from its centre with a probe. The probe was zeroed with the magnet absent, so the readings show the magnet’s field only. (Invented data.)
| r (cm) | 3.0 | 4.0 | 6.0 | 8.0 |
|---|---|---|---|---|
| B (μT) | 1190 | 500 | 148 | 62.5 |
(a) Use two pairs of readings to find the factor by which B changes when r doubles. (b) The student claims “B is inversely proportional to r”. Evaluate this claim using the data. (c) Assuming the same pattern continues, predict B at r = 12.0 cm. (d) The magnet lies flat on a table. At the student’s location the horizontal part of Earth’s field is about 35 μT. At r = 12.0 cm, would a flat compass on the axis line up mainly with the magnet’s field or with Earth’s? Justify.
Worked solution and suggested Marlbridge rubric
(a) From 4.0 cm to 8.0 cm: 500 ÷ 62.5 = 8. From 3.0 cm to 6.0 cm: 1190 ÷ 148 = 8.0. Doubling r divides B by 8 (= 2³).
(b) If B ∝ 1/r, doubling r would halve B, giving 250 μT at 8.0 cm, not 62.5 μT. The field falls much faster than 1/r (and faster than 1/r², which would give 125 μT). The claim is not supported. The data fit B ∝ 1/r³.
(c) From 4.0 cm to 12.0 cm, r triples, so B falls by 3³ = 27: 500 μT ÷ 27 = 18.5 μT.
(d) At 12.0 cm the magnet’s field (about 18.5 μT, horizontal) is about half the horizontal part of Earth’s field (about 35 μT). A flat compass responds only to the horizontal field. The compass lines up with the vector sum, which is dominated by Earth’s field. The needle would be pulled partly towards the magnet’s direction but would point mainly along Earth’s field.
| Point | What earns it |
|---|---|
| 1 | (a) Factor of 8 from at least one pair of readings with r doubled |
| 1 | (a) Confirms it with a second pair (consistency) |
| 1 | (b) Shows that 1/r predicts 250 μT (or a factor of 2) and compares with the data |
| 1 | (b) Clear conclusion: claim rejected; B falls faster (consistent with 1/r³) |
| 1 | (c) 18.5 μT (accept 18–19 μT) with a method based on the factor of change |
| 1 | (d) Compares 18.5 μT with the 35 μT horizontal part of Earth’s field and concludes Earth’s field dominates the vector sum |
You do not need to memorise the 1/r³ pattern; this question tests reading it from data. Carry forward an error in (c) into (d) once.
How did you do?
- Q1 or Q3 wrong: re-read “Dipoles, poles and field lines” and Figure 1 in the study guide.
- Q2 or Q4 wrong: revisit “How different materials respond” and “Magnetic permeability”, then Worked example 2.
- Q5 wrong: work through Worked example 1 again: the needle follows the vector sum.
- Q6 or Q7 incomplete: revisit “Earth as a magnet” and the closed-loop rule.
Then tick off the topic checklist.