Magnetism · 18.1

Magnets & Magnetic Fields

Every magnet has two inseparable poles. Field lines map the force it would exert — emerging from north, entering south — and the Earth itself carries a field that compasses follow.

Open the lab ↓Download lesson PDF

Build the model

Connect the measurement to the mechanism.

Like poles repel, unlike attract, and cutting a magnet always yields two new complete magnets. The field concept replaces action-at-a-distance: a magnet fills space with a field, and other magnets or moving charges respond to the field where they are. Field strength B is measured in tesla.

Simple definition: A magnetic field is the region around a magnet or current where magnetic materials and moving charges experience a force.
Example: Iron filings sprinkled around a bar magnet line up along the field, drawing its shape from pole to pole.

Field directionN → S outside the magnet

Field lines leave the north pole, curve round to the south, and continue inside the magnet — they never start, stop, or cross.

Closed loops overall

Field strengthB in tesla (T)

Closer field lines mean a stronger field — the tesla is the unit that will connect to force in the next lessons.

Earth ≈ 5 × 10⁻⁵ T; fridge magnet ≈ 10⁻² T

No monopolespoles always come in pairs

There is no isolated north or south — every cut produces a fresh pair of poles.

Halving a magnet makes two magnets

Magnetic materials

Iron, nickel, cobalt and their alloys are ferromagnetic: their atomic magnets can align into domains. Induced magnetism explains why an unmagnetised nail sticks to a magnet.

Hard and soft

Steel (hard) keeps its magnetism — good for permanent magnets. Soft iron magnetises and demagnetises easily — good for electromagnet cores.

Earth's field

A compass needle is a small magnet aligning with Earth's field. The geographic north pole is magnetically a south pole — that is why the needle's north end points there.

Change one variable at a time

Make the relationship visible.

Flip the magnet

Lines always leave the north pole and enter the south — flipping the magnet reverses every arrow but keeps the shape.

Strongest fieldat the poles — lines closest

Compass beside N polepoints away from the magnet

Catch the common trap

Explain before calculating.

A bar magnet is cut in half between its poles. What results?

Choose an answer to test the model.

Worked examples

State the rule, substitute, then check units.

EasyTwo north poles are brought together. What happens, and what rule does it illustrate?

They push apart.
Like poles repel; unlike poles attract.

AnswerRepulsion — like poles repel

MediumA compass placed east of a bar magnet's north pole points away from the magnet. Sketch why, and state where the field is strongest.

Field lines emerge from the north pole, radiating outward; the compass aligns along the local line.
Pointing away from N confirms lines leave north poles.
Lines crowd closest at the poles — the field is strongest there.

AnswerThe needle follows the outgoing field line; strongest field at the poles

HardA compass needle points north. A bar magnet is placed east of it with its north pole facing the compass. Sketch-reason the needle's final direction for a magnet field at the compass equal in strength to Earth's horizontal field.

The magnet's field at the compass points west (away from its N pole, toward the compass).
Earth pulls the needle north with equal strength.
The vector sum points northwest — the needle settles at 45° west of north.

AnswerIt swings to point northwest (45°)

ChallengingA steel ship's hull becomes magnetised during construction. Explain how, and why warships 'deperm' by wrapping current-carrying cables around the hull.

Hammering and vibration in Earth's field lets iron domains settle aligned with it — the hull acquires a permanent magnetic signature.
That signature can trigger magnetic mines and betray the ship's heading history.
Deperming drives decreasing alternating fields through the hull, randomising the domains layer by layer — erasing the signature like degaussing a tape.

AnswerEarth-field-aligned domains from construction; alternating decaying fields scramble them

Exam diagrams for this topic2 figures to inspect and practiseQuestions, hints and marking points in one compact subsection.

See it. Read it. Work it.

These figures come from GioPhysics practice papers. Open one, decode the drawing, work the guided questions, then follow its link to the full paper question.

01
InspectRead the figure comment.
02
TraceFollow labels, arrows and axes.
03
AnswerWork one part at a time.
04
CheckReveal hints and marking points.

A Level

01Fig. 5.1Magnetic fieldsA Level

Figure comment

Fig. 5.1An electron, drawn as a small circle marked with a negative sign, travels horizontally to the right, and an arrow labelled v gives its velocity. Ahead of it lies a large rectangular region with a dashed boundary, filled with a regular array of crosses and labelled as a uniform magnetic field directed into the page. The electron is shown outside that region, at the instant before it crosses the boundary, with its velocity lying in the plane of the page.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

20 Magnetic fields Open A Level question · A Level extension Q5

Guided questions 5 parts

Reading cue. Crosses mean B points into the page, and the particle is negative, so the conventional current is opposite to the arrow v — apply the left-hand rule to that reversed direction.

aState State the direction of the force on the electron at the instant it crosses the dashed boundary in Fig. 5.1.
recall2 marks
Check answer 2 marks
1. vertically downwards in the plane of the page
2. perpendicular to both v and B, from the left-hand rule applied with the conventional current opposite to v
bExplain Explain why the electron travels at constant speed inside the field region even though its velocity is changing.
routine3 marks
Check answer 3 marks
1. the magnetic force is always perpendicular to the velocity
2. so no work is done on the electron and its kinetic energy is unchanged
3. a constant-magnitude force perpendicular to the motion changes only the direction of the velocity, giving a circular path
cDetermine The electron crosses the boundary at 2.4 × 10⁷ m s⁻¹ into a field of flux density 8.5 mT. Determine the radius of its path, and hence the least distance the field region must extend beyond the boundary if the electron is to leave the region travelling at right angles to its original direction.
demanding3 marks
Check answer 3 marks
1. r = mv/(eB) = (9.11 × 10⁻³¹ × 2.4 × 10⁷) / (1.60 × 10⁻¹⁹ × 8.5 × 10⁻³)
2. r = 1.6 × 10⁻² m
3. after turning through 90° the electron has advanced one radius along its original direction, so the region must extend at least 1.6 cm
dDeduce A proton crosses the boundary at the same point and with the same velocity as the electron. Deduce how its path differs from that of the electron.
top of the paper3 marks
Check answer 3 marks
1. the proton is positive, so the magnetic force acts in the opposite direction and the path curves upwards instead of downwards
2. r = mv/(qB) and the charge magnitudes are equal, so the radius is greater in the ratio of the masses, about 1.8 × 10³
3. r ≈ 29 m, so within the region drawn the proton's path is almost straight

Transfer challenge

A beam of protons travelling at 3.0 × 10⁵ m s⁻¹ passes undeflected through a region containing a uniform magnetic field of flux density 0.12 T perpendicular to the beam together with a uniform electric field. Determine the electric field strength, and state its direction relative to the magnetic force on the protons.

Check answer 3 marks

for no deflection the electric force balances the magnetic force: qE = qvB
E = 3.0 × 10⁵ × 0.12 = 3.6 × 10⁴ V m⁻¹
the electric force must oppose the magnetic force, so E acts in the direction opposite to that magnetic force

02Fig. 10.1Electric fields · CapacitanceA Level

Figure comment

Fig. 10.1Two long horizontal metal plates are drawn one directly above the other and connected by wires at their left-hand ends to a battery labelled 2.0 kV; the upper plate carries a plus sign and the lower plate a minus sign. The space between the plates is empty and labelled vacuum, and a dimension at the right-hand end gives the separation of the plates as 5.0 mm. No field lines are drawn between the plates.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

18 Electric fields 19 Capacitance 20 Magnetic fields Open A Level question · A Level extension Q10

Guided questions 5 parts

Reading cue. 5.0 mm is the gap, not a plate length, and no field lines are drawn — the plus and minus signs on the plates are the only thing in the figure that fixes the field direction.

aState State the direction of the electric field in the gap in Fig. 10.1, and the direction of the force it exerts on an electron placed there.
recall2 marks
Check answer 2 marks
1. the field points vertically downwards, from the positive upper plate to the negative lower plate
2. the force on an electron is vertically upwards, towards the positive plate
bCalculate Calculate the work done on an electron that moves from the lower plate to the upper plate, and the speed with which it arrives.
routine3 marks
Check answer 3 marks
1. W = eV = 1.60 × 10⁻¹⁹ × 2.0 × 10³ = 3.2 × 10⁻¹⁶ J
2. ½mv² = 3.2 × 10⁻¹⁶ J
3. v = 2.7 × 10⁷ m s⁻¹
cDetermine A charged dust particle of weight 1.28 × 10⁻¹³ N is held at rest midway between the plates. Determine the magnitude and sign of its charge, and the number of excess electrons it carries.
demanding4 marks
Check answer 4 marks
1. E = V/d = 2.0 × 10³ / 5.0 × 10⁻³ = 4.0 × 10⁵ V m⁻¹
2. for equilibrium qE = weight, so q = 1.28 × 10⁻¹³ / 4.0 × 10⁵ = 3.2 × 10⁻¹⁹ C
3. the electric force must act upwards while the field points downwards, so the charge is negative
4. 3.2 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2 excess electrons
dDeduce The battery p.d. is suddenly reduced to 1.0 kV while the particle is still midway between the plates. Deduce the acceleration of the particle, and calculate the time it takes to reach a plate.
top of the paper4 marks
Check answer 4 marks
1. new field = 2.0 × 10⁵ V m⁻¹, so the electric force = 3.2 × 10⁻¹⁹ × 2.0 × 10⁵ = 6.4 × 10⁻¹⁴ N upwards
2. resultant = 1.28 × 10⁻¹³ − 6.4 × 10⁻¹⁴ = 6.4 × 10⁻¹⁴ N downwards
3. mass = 1.28 × 10⁻¹³ / 9.81 = 1.30 × 10⁻¹⁴ kg, so a = 4.9 m s⁻² downwards, that is g/2
4. falling the 2.5 mm to the lower plate: t = √(2 × 2.5 × 10⁻³ / 4.9) = 3.2 × 10⁻² s

Transfer challenge

An electron travelling at 2.0 × 10⁷ m s⁻¹ enters midway between two parallel plates 5.0 cm long, moving parallel to them, in a uniform field of strength 1.2 × 10⁴ V m⁻¹. Calculate the sideways deflection of the electron as it leaves the plates.

Check answer 3 marks

a = eE/m = (1.60 × 10⁻¹⁹ × 1.2 × 10⁴) / 9.11 × 10⁻³¹ = 2.1 × 10¹⁵ m s⁻²
time between the plates t = 0.050 / 2.0 × 10⁷ = 2.5 × 10⁻⁹ s
deflection = ½at² = 6.6 × 10⁻³ m

Extra challenging questions

Take the ideas further with course-specific extension questions, supplied-data graphs and worked solutions.