Senior 5 practice book · Unit 9 of 11

Magnetic Field and Magnetic Forces

50 questions that complete the Senior 5 quiz for this unit: 26 core and 24 challenge. Easy and hard questions are mixed. Try each question, then tap “Show answer”.

Common misconceptions

Magnetic field lines are real lines that start at the north pole and stop at the south pole.Field lines are a drawing tool; they form closed loops, continuing inside the magnet from S back to N.
A magnetic field pushes a current-carrying wire along the field lines.The force F = BIL sin θ is at right angles to both the field and the current; a wire along the field feels no force at all.
A charge sitting still in a magnetic field feels a magnetic force.Only a MOVING charge (with a velocity component across the field) feels a magnetic force; at rest F = qvB sin θ = 0.
If one wire carries more current, it pushes the other wire harder than it is pushed back.By Newton's third law the forces between two parallel wires are equal and opposite, μ₀I₁I₂L/(2πd) on each.
The solenoid formula B = μ₀nI uses the total number of turns.n is the number of turns PER METRE (N/L); a long solenoid and a short one with the same N give different fields.

What this unit covers

Topics marked new are not tested much in the quiz, so this book gives them extra questions.

Magnetic fields and field lines: magnets, neutral points, the Earth's field
Field of a long straight wire B = μ₀I/(2πr) and the right-hand grip rule
Field inside a solenoid B = μ₀nI
Field at the centre of a flat circular coil B = μ₀NI/(2r)
Magnetic flux Φ = BA cos θ and its unit, the weber
Force on a current-carrying conductor F = BIL sin θ and Fleming's left-hand rule
Force between parallel currents and the definition of the ampere
Force on a moving charge F = qvB sin θ and its direction
Circular and helical motion of charges: radius, period, momentum
Applications of forces on moving charges: velocity selector, mass spectrometer, cyclotron, cathode-ray tube, aurora
Torque on a coil and the DC motor
Moving-coil meters (galvanometer, ammeter) and radial fields
Electromagnets and their uses: cranes, relays, electric bells

Go to the questions

Questions (1–50)

Easier and harder questions are mixed, just like in a real exam. The 24 harder ones are marked ★ Challenge.

1True or false
An electric bell keeps ringing because its electromagnet switches itself off each time it pulls the hammer across.
Show answer
Answer: True
Pulling the armature breaks the contact, the magnet switches off, the spring returns the armature, the contact closes again, and the cycle repeats.
Common mistakeSome think the current reverses to push the hammer back; a spring returns it, and the circuit is simply broken and re-made.
2True or false · ★ Challenge
When a large current flows through a loosely wound coil (like a spring), the turns are pulled closer together.
Show answer
Answer: True
Neighbouring turns carry current in the same direction, and like currents attract.
Common mistakeSome expect the turns to push apart because "like repels"; for currents, like directions attract.
3Fill in the blank
Placing a soft-iron core inside a current-carrying solenoid ______ the magnetic field.
Show answer
Answer: greatly increases (strengthens)
The iron becomes magnetised and adds its own field to that of the coil.
Common mistakeSome think iron "blocks" the field; a magnetic material concentrates and strengthens it.
4True or false · ★ Challenge
Doubling the radius of a flat circular coil while keeping the number of turns and the current the same doubles the field at its centre.
Show answer
Answer: False
B = μ₀NI/(2r): doubling r HALVES the field.
Common mistakeLearners sometimes think a bigger coil has "more wire" so a stronger field; the wire is further from the centre, so the field there is weaker.
5True or false
The field inside a long solenoid depends on the number of turns per metre and the current, but not on the diameter of the solenoid.
Show answer
Answer: True
B = μ₀nI contains no radius, so a wide and a narrow long solenoid with the same n and I give the same field.
Common mistakeSome expect a wider coil to give a weaker field as for a single loop; for a long solenoid the diameter does not appear.
6True or false · ★ Challenge
Doubling the number of turns on the coil of a galvanometer (with the same springs and magnet) makes it more sensitive, giving a larger deflection for each ampere.
Show answer
Answer: True
Torque = NBIA, so twice the turns gives twice the torque per ampere and so twice the deflection.
Common mistakeSome think extra turns only add resistance; they also double the torque, which is what sensitivity measures.
7Fill in the blank
At the centre of a flat circular coil carrying a current, the magnetic field is ______ to the plane of the coil.
Show answer
Answer: perpendicular (at right angles)
By the right-hand grip rule every part of the loop gives a field through the centre along the axis, at right angles to the plane of the coil.
Common mistakeSome think the field lies in the plane of the coil; that is true near a straight wire, but at the centre of a loop the field is along the axis.
8Fill in the blank · ★ Challenge
In a cyclotron, the time for each half-circle does not depend on the particle's ______, so an alternating voltage of fixed frequency can keep accelerating it.
Show answer
Answer: speed (velocity)
T = 2πm/(qB) contains no v: faster particles move on bigger circles but take the same time.
Common mistakeMany expect faster particles to come round sooner; the bigger radius exactly cancels the higher speed.
9Multiple choice
In a moving-coil galvanometer, the coil turns until
1. Athe current in the coil falls to zero
2. Bthe commutator reverses the current
3. Cthe coil plane is perpendicular to the field
4. Dthe magnetic torque is balanced by the springs
Show answer
Answer: D. the magnetic torque is balanced by the springs
The hair springs give a restoring torque proportional to the angle turned; the pointer stops where this equals NBIA.
Common mistakeChoosing the commutator confuses a meter with a motor; a meter has no commutator and turns only part of a revolution.
10Multiple choice · ★ Challenge
A vertical wire carries current upwards in a horizontal magnetic field pointing north. In which direction is the force on the wire?
1. AEast
2. BNorth
3. CDownward
4. DWest
Show answer
Answer: D. West
Fleming's left-hand rule: first finger (field) north, second finger (current) up, thumb (force) points west.
Common mistakeChoosing east comes from using the right hand; the motor (force) rule uses the LEFT hand.
11Multiple choice
Two long parallel wires carry currents in opposite directions. The wires
1. Aattract each other
2. Bfeel no force at all
3. Crepel each other
4. Dturn to be at right angles
Show answer
Answer: C. repel each other
Unlike (opposite) currents repel; like currents attract.
Common mistakeChoosing 'attract' carries over the rule for magnetic poles or charges; for currents, LIKE currents attract.
12Multiple choice · ★ Challenge
Why does a moving-coil meter have curved pole pieces and a soft-iron cylinder that make a radial field?
1. ASo that the coil can spin round continuously like a motor
2. BSo that the deflection is proportional to current, giving an even scale
3. CSo that the meter can measure alternating current directly
4. DSo that no springs are needed to stop the moving coil
Show answer
Answer: B. So that the deflection is proportional to current, giving an even scale
In a radial field the coil plane is always along the field, so the torque NBIA does not change with angle and the deflection θ ∝ I.
Common mistakeChoosing 'continuous spinning' describes a motor; the meter is designed to turn through a set angle against springs.
13True or false
A current-carrying wire laid along the direction of the magnetic field lines feels no magnetic force.
Show answer
Answer: True
F = BIL sin θ and θ = 0°, so F = 0.
Common mistakeSome think any wire in a field is pushed; the force needs a component of current across the field.
14Fill in the blank · ★ Challenge
The magnetic field 4.0 cm from a long straight wire is 3.0 × 10⁻⁵ T. The current in the wire is ______ A. (μ₀ = 4π × 10⁻⁷ T m/A)
Show answer
Answer: 6.0
I = 2πrB/μ₀ = 2π × 0.040 × 3.0 × 10⁻⁵ ÷ 4π × 10⁻⁷ = 6.0 A.
Common mistakeLeaving r as 4.0 instead of 0.040 m gives 600 A; distances must be in metres.
15True or false
The radius of the circular path of a charged particle in a uniform magnetic field is proportional to its momentum.
Show answer
Answer: True
qvB = mv²/r gives r = mv/(qB) = p/(qB).
Common mistakeSome think r ∝ kinetic energy; it is proportional to momentum mv.
16Multiple choice
A straight wire carries current into the page. Looking at the page, the magnetic field lines around it are
1. Aanticlockwise circles centred on the wire
2. Bclockwise circles centred on the wire
3. Cstraight lines pointing into the page
4. Dstraight lines spreading out from the wire
Show answer
Answer: B. clockwise circles centred on the wire
Right-hand grip rule: point the thumb into the page; the fingers curl clockwise.
Common mistakeChoosing anticlockwise comes from pointing the thumb out of the page; the thumb must follow the current.
17Fill in the blank · ★ Challenge
An alpha particle (charge 3.2 × 10⁻¹⁹ C) moves at 1.5 × 10⁶ m/s at 30° to a 0.40 T field. The magnetic force on it is ______ × 10⁻¹⁴ N.
Show answer
Answer: 9.6
F = qvB sin θ = 3.2 × 10⁻¹⁹ × 1.5 × 10⁶ × 0.40 × sin 30° = 9.6 × 10⁻¹⁴ N.
Common mistakeForgetting sin 30° = 0.5 gives 19.2 × 10⁻¹⁴ N; the force uses only the part of v across the field.
18Multiple choice
In a simple DC motor, at which position is the torque on the coil zero, and what carries the coil past it?
1. APlane parallel to the field; the commutator pushes it past
2. BPlane perpendicular to the field; the brushes push it past
3. CPlane at 45° to the field; the magnets pull it past
4. DPlane perpendicular to the field; momentum carries it past
Show answer
Answer: D. Plane perpendicular to the field; momentum carries it past
When the plane is perpendicular to B the forces on the sides act along the same line through the axle and give no turning effect; the rotating coil coasts past by its momentum, and the commutator then reverses the current.
Common mistakeChoosing "plane parallel to the field" is the position of MAXIMUM torque, where the forces have the largest lever arm.
19Short answer · ★ Challenge
Explain how a relay allows a small current from a light-sensor circuit to switch on a large water-pump motor.
Show answer
Model answer: The small sensor current flows through the coil of the relay, making it an electromagnet. This attracts a soft-iron armature, which closes a separate pair of heavy-duty contacts in the motor circuit. The two circuits are separate, so the sensor circuit is safe while the large motor current is switched.
Common mistakeSome think the sensor current flows through the motor; the relay links the circuits only magnetically.
20Multiple choice · ★ Challenge
A velocity selector in a research laboratory uses E = 3.0 × 10⁴ V/m and B = 0.15 T. At what speed do ions pass straight through?
1. A4.5 × 10³ m/s
2. B5.0 × 10⁻⁶ m/s
3. C2.0 × 10⁵ m/s
4. D2.0 × 10³ m/s
Show answer
Answer: C. 2.0 × 10⁵ m/s
v = E/B = 3.0 × 10⁴ ÷ 0.15 = 2.0 × 10⁵ m/s.
Common mistakeChoosing 4.5 × 10³ m/s multiplies E by B; the balance qE = qvB gives v = E/B.
21True or false
The Earth's geographic North Pole is near a magnetic south pole, which is why the north end of a compass needle points north.
Show answer
Answer: True
The N end of a needle is attracted by a S pole, so the magnetic pole in the Arctic must be a south pole.
Common mistakeMany learners think the Arctic magnetic pole is a north pole because of its name; unlike poles attract, so it is magnetically south.
22Fill in the blank
A charged particle entering a uniform magnetic field at an angle that is neither 0° nor 90° follows a ______ path.
Show answer
Answer: helical (spiral)
The part of v across the field gives circular motion, the part along the field is unchanged, so the particle moves on a helix.
Common mistakeAnswering "circular" only applies when v is exactly perpendicular to B.
23Short answer · ★ Challenge
Explain why practical motors use several coils set at different angles, and curved pole pieces that make a radial field.
Show answer
Model answer: With one coil the torque rises and falls to zero twice per turn, giving jerky motion. Several coils at different angles, each switched on by the commutator when it is near its best position, give a nearly steady, large torque. Curved poles with a radial field keep the field parallel to the plane of each active coil, so it always gets the maximum torque NBIA.
Common mistakeSome think more coils just make the motor "stronger"; the main gain is a smoother, nearly constant torque.
24Multiple choice
In a laboratory, positive ions travel horizontally towards the north through a field pointing straight up. Give the direction of the sideways push on the ions.
1. AWest
2. BNorth
3. CDownward
4. DEast
Show answer
Answer: D. East
Left-hand rule with the current in the direction the positive ions move: field up, current north → thumb points east.
Common mistakeChoosing west comes from treating the ions like electrons; for a positive charge the current is in the direction it moves.
25Short answer · ★ Challenge
Using the idea of combining fields ("catapult field"), explain why a current-carrying wire between the poles of a magnet is pushed sideways.
Show answer
Model answer: The circular field of the wire adds to the magnet's field on one side (lines crowded, strong field) and opposes it on the other (lines spread, weak field). The field lines behave like stretched elastic and push the wire from the strong side towards the weak side, at right angles to both field and current.
Common mistakeSaying the wire is attracted to a pole is wrong: the force is sideways, not towards N or S.
26Multiple choice · ★ Challenge
A solenoid has 2000 turns per metre. What current is needed to make a field of 0.010 T inside it? (μ₀ = 4π × 10⁻⁷ T m/A)
1. A50 A
2. B0.40 A
3. C4.0 A
4. D40 A
Show answer
Answer: C. 4.0 A
I = B/(μ₀n) = 0.010 ÷ (4π × 10⁻⁷ × 2000) = 0.010 ÷ 2.51 × 10⁻³ ≈ 4.0 A.
Common mistakeChoosing 50 A forgets the 4π in μ₀ (using 10⁻⁷ alone); 0.40 A and 40 A come from slips in powers of ten.
27Fill in the blank
One ampere is the current which, in two very long parallel wires 1 m apart in a vacuum, gives a force of ______ N on each metre of wire.
Show answer
Answer: 2 × 10⁻⁷
F/L = μ₀ × 1 × 1 ÷ (2π × 1) = 2 × 10⁻⁷ N/m.
Common mistakeAnswering 4π × 10⁻⁷ gives μ₀ itself; the force per metre is μ₀/2π for 1 A and 1 m.
28Multiple choice
Which particle feels NO magnetic force when it moves at right angles through a magnetic field?
1. AA neutron
2. BAn alpha particle
3. CA proton
4. DAn electron
Show answer
Answer: A. A neutron
F = qvB sin θ; the neutron has q = 0, so F = 0.
Common mistakeChoosing the electron because it is light confuses mass with charge; the magnetic force depends on charge, not mass.
29Multiple choice · ★ Challenge
Two long parallel wires 2.0 cm apart carry 10 A and 15 A. What is the force on each metre of either wire? (μ₀ = 4π × 10⁻⁷ T m/A)
1. A9.4 × 10⁻³ N
2. B1.5 × 10⁻⁵ N
3. C1.5 × 10⁻³ N
4. D3.0 × 10⁻³ N
Show answer
Answer: C. 1.5 × 10⁻³ N
F/L = μ₀I₁I₂/(2πd) = 2 × 10⁻⁷ × 10 × 15 ÷ 0.020 = 1.5 × 10⁻³ N per metre.
Common mistakeChoosing 9.4 × 10⁻³ N forgets the 2π; 1.5 × 10⁻⁵ N leaves d in centimetres.
30Multiple choice
Why are the aurora (polar lights) seen mainly near the North and South Poles?
1. AThe Earth's field is weakest there, so particles slow down
2. BCharged solar particles spiral down field lines there
3. CSunlight is reflected strongly from ice near the poles
4. DMagnetic forces speed the particles up near the equator
Show answer
Answer: B. Charged solar particles spiral down field lines there
Charged particles spiral along the Earth's field lines, which come down into the atmosphere near the magnetic poles; there they hit air atoms and make them glow.
Common mistakeChoosing 'the field is weakest' is wrong: the field is strongest near the poles, and it is the shape of the field lines that guides the particles down there.
31Short answer · ★ Challenge
In a mass spectrometer, ions of carbon-12 and carbon-14 with the same charge enter the same magnetic field at the same speed. Which has the larger path radius, by what factor, and how does this separate the isotopes?
Show answer
Model answer: r = mv/(qB) with v, q and B the same, so r ∝ m. Carbon-14 has the larger radius, 14/12 ≈ 1.17 times that of carbon-12. The two kinds of ion land at different places on the detector, so they are separated and their amounts can be measured.
Common mistakeSome think the heavier ion bends more; a larger mass means more momentum, so it bends LESS (larger radius).
32Multiple choice
In a velocity selector the electric field E and magnetic field B are at right angles. Which charged particles pass straight through without being deflected?
1. AThose with speed v = E/B
2. BThose with speed v = B/E
3. CThose with the largest charge
4. DThose with the smallest mass
Show answer
Answer: A. Those with speed v = E/B
They go straight when qE = qvB, so v = E/B, whatever their charge or mass.
Common mistakeChoosing the largest charge forgets that q cancels: both forces are proportional to q.
33Multiple choice · ★ Challenge
A cable under a road carries a steady 50 A. A compass is held 2.0 cm directly above it. What magnetic field does the cable produce at the compass? (μ₀ = 4π × 10⁻⁷ T m/A)
1. A1.0 × 10⁻³ T
2. B1.6 × 10⁻³ T
3. C5.0 × 10⁻⁶ T
4. D5.0 × 10⁻⁴ T
Show answer
Answer: D. 5.0 × 10⁻⁴ T
B = μ₀I/(2πr) = 4π × 10⁻⁷ × 50 ÷ (2π × 0.020) = 5.0 × 10⁻⁴ T.
Common mistakeChoosing 1.6 × 10⁻³ T uses μ₀I/(2r), the formula for the centre of a loop; 5.0 × 10⁻⁶ T leaves r in centimetres.
34Fill in the blank
An electric motor converts electrical energy mainly into ______ energy.
Show answer
Answer: kinetic (mechanical)
The magnetic forces turn the coil, doing work to make it and its load rotate.
Common mistakeAnswering "magnetic energy" names the means, not the useful output.
35Multiple choice · ★ Challenge
At a school, the vertical part of the Earth's magnetic field is 3.0 × 10⁻⁵ T. What is the magnetic flux through the flat horizontal roof of a hall 20 m long and 10 m wide?
1. A0 Wb
2. B6.0 × 10⁻³ Wb
3. C9.0 × 10⁻⁴ Wb
4. D1.5 × 10⁻⁷ Wb
Show answer
Answer: B. 6.0 × 10⁻³ Wb
Φ = BA = 3.0 × 10⁻⁵ × (20 × 10) = 6.0 × 10⁻³ Wb (the vertical field is perpendicular to a horizontal roof).
Common mistakeChoosing 0 Wb assumes the field is parallel to the roof; it is the VERTICAL component that passes through a horizontal surface. 9.0 × 10⁻⁴ Wb adds 20 + 10 instead of multiplying.
36Fill in the blank
A flux of 2.4 × 10⁻⁴ Wb passes at right angles through a ring of area 8.0 cm². The magnetic flux density is ______ T.
Show answer
Answer: 0.30
B = Φ/A = 2.4 × 10⁻⁴ ÷ 8.0 × 10⁻⁴ = 0.30 T (8.0 cm² = 8.0 × 10⁻⁴ m²).
Common mistakeUsing 8.0 × 10⁻² m² (treating cm² like cm) gives 3.0 × 10⁻³ T; 1 cm² = 10⁻⁴ m².
37Multiple choice · ★ Challenge
In a current-balance experiment, 4.0 cm of wire lies at right angles to a uniform field. The measured force is 1.0 mN at 0.50 A, 2.0 mN at 1.0 A and 3.0 mN at 1.5 A. What is the flux density?
1. A0.050 T
2. B2.0 × 10⁻³ T
3. C5.0 × 10⁻⁴ T
4. D8.0 × 10⁻⁵ T
Show answer
Answer: A. 0.050 T
Gradient F/I = 2.0 × 10⁻³ N/A = BL, so B = 2.0 × 10⁻³ ÷ 0.040 = 0.050 T.
Common mistakeChoosing 2.0 × 10⁻³ T takes the gradient as B; the gradient is BL, so divide by the length in metres.
38Multiple choice
A flat circular coil of 50 turns and radius 5.0 cm carries 2.0 A. Using B = μ₀NI/(2r), what is the field at its centre? (μ₀ = 4π × 10⁻⁷ T m/A)
1. A1.3 × 10⁻³ T
2. B6.3 × 10⁻⁴ T
3. C2.5 × 10⁻⁵ T
4. D4.0 × 10⁻⁴ T
Show answer
Answer: A. 1.3 × 10⁻³ T
B = 4π × 10⁻⁷ × 50 × 2.0 ÷ (2 × 0.050) = 1.26 × 10⁻³ T.
Common mistakeChoosing 6.3 × 10⁻⁴ T uses the diameter instead of the radius; 2.5 × 10⁻⁵ T forgets the 50 turns.
39Multiple choice
A plotting compass is placed just beyond the north pole of a bar magnet, on its axis. Which way does the north end of the compass needle point?
1. ATowards the magnet's north pole
2. BAt right angles to the magnet axis
3. CTowards geographic north only
4. DAway from the magnet's north pole
Show answer
Answer: D. Away from the magnet's north pole
The needle lines up with the field, and outside a magnet the field points away from the N pole.
Common mistakeChoosing 'towards the north pole' forgets that like poles repel: the compass's N end is pushed away from the magnet's N pole.
40Multiple choice · ★ Challenge
A proton (m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C) moves in a circle in a 0.50 T field. How long does one revolution take? (T = 2πm/(qB))
1. A2.1 × 10⁻⁸ s
2. B1.3 × 10⁻⁷ s
3. C7.2 × 10⁻¹¹ s
4. D7.6 × 10⁶ s
Show answer
Answer: B. 1.3 × 10⁻⁷ s
T = 2π × 1.67 × 10⁻²⁷ ÷ (1.6 × 10⁻¹⁹ × 0.50) = 1.3 × 10⁻⁷ s.
Common mistakeChoosing 7.6 × 10⁶ s gives the frequency (1/T) with the wrong unit; 2.1 × 10⁻⁸ s forgets the 2π.
41Short answer · ★ Challenge
Two bar magnets lie on a table in a line with their north poles facing each other 10 cm apart. Describe the field pattern between them and explain what a compass would do at the neutral point.
Show answer
Model answer: The field lines from each N pole curve away from each other and turn sideways, so no lines cross the middle. Midway, the two equal fields are equal and opposite, so the resultant field is zero (neutral point). A compass placed there feels only the weak Earth field, so it points to magnetic north (or may turn freely if the Earth field is also cancelled).
Common mistakeDrawing lines crossing at the neutral point is wrong: field lines never cross, and at the neutral point there is no field to draw.
42Short answer
Explain how the electron beam in an old cathode-ray TV tube is moved across the screen using magnetic fields.
Show answer
Model answer: Coils around the neck of the tube make magnetic fields at right angles to the beam. The force F = qvB on the moving electrons bends the beam; changing the coil currents changes the field and so moves the spot left–right and up–down to trace the picture.
Common mistakeSaying the field attracts the electrons towards a pole is wrong; the force is sideways, perpendicular to both the beam and the field.
43Short answer · ★ Challenge
In a cloud-chamber photograph, two tracks made by the same kind of particle in the same magnetic field have radii of 2.0 cm and 6.0 cm. Compare the momenta of the particles, and explain why a track that slowly curls into a spiral shows the particle losing energy.
Show answer
Model answer: r = p/(qB), so p ∝ r: the 6.0 cm track has 3 times the momentum of the 2.0 cm one. As a particle loses energy by ionising the gas, its momentum falls, so its radius gets smaller and smaller and the path curls into a spiral.
Common mistakeThinking the particle on the tighter circle is faster reverses the relation; a faster (higher-momentum) particle bends LESS.
44Short answer
A galvanometer gives full-scale deflection with only a few milliamperes. Explain how it can be made into an ammeter reading up to 5 A, and why the added resistor must have a very small resistance.
Show answer
Model answer: A shunt (a resistor of very small resistance) is connected in PARALLEL with the galvanometer. Most of the current passes through the shunt and only a small, fixed fraction through the coil. A small resistance is needed so that nearly all the current goes through it and the ammeter itself has a very low resistance and does not change the current being measured.
Common mistakePutting the resistor in series makes a voltmeter, not an ammeter.
45Short answer · ★ Challenge
A learner winds 200 turns evenly on a tube 10 cm long, and another 200 turns on a tube 20 cm long. Both carry 1.5 A. Calculate the field inside each and compare them. (μ₀ = 4π × 10⁻⁷ T m/A)
Show answer
Model answer: Short tube: n = 200 ÷ 0.10 = 2000 m⁻¹, B = 4π × 10⁻⁷ × 2000 × 1.5 ≈ 3.8 × 10⁻³ T. Long tube: n = 1000 m⁻¹, B ≈ 1.9 × 10⁻³ T. Same number of turns, but the shorter solenoid has twice the turns per metre and twice the field.
Common mistakeUsing N = 200 in place of n gives the same answer for both tubes; the formula needs turns per metre.
46Short answer
Explain why the magnetic flux through a loop is zero when the plane of the loop is parallel to the field, even if the field is very strong.
Show answer
Model answer: Flux measures how many field lines pass THROUGH the area: Φ = BA cos θ, where θ is between B and the normal. When the plane is parallel to B, θ = 90° and cos θ = 0; the lines skim along the loop without passing through it.
Common mistakeThinking "strong field means large flux" ignores the angle; flux depends on the part of B perpendicular to the area.
47Multiple choice
Which is the best core material for the electromagnet of a scrap-metal crane?
1. ASteel, as it keeps its magnetism after switching off
2. BCopper, as it is the very best conductor of electricity
3. CSoft iron, as it is easily magnetised and demagnetised
4. DAluminium, as it is light and it does not rust
Show answer
Answer: C. Soft iron, as it is easily magnetised and demagnetised
The crane must drop the scrap when the current is switched off, so the core must lose its magnetism at once: soft iron.
Common mistakeChoosing steel is wrong: steel stays magnetised, so the crane could not release the load.
48Short answer · ★ Challenge
Wire A carries 2.0 A and wire B carries 6.0 A, parallel and 3.0 cm apart. A learner says: 'B pushes A three times harder than A pushes B.' Evaluate this, and find the force per metre on each wire. (μ₀ = 4π × 10⁻⁷ T m/A)
Show answer
Model answer: The claim is wrong. Each wire sits in the other's field: A feels B_B × I_A, B feels B_A × I_B, and both equal μ₀I_AI_B/(2πd) = 2 × 10⁻⁷ × 2.0 × 6.0 ÷ 0.030 = 8.0 × 10⁻⁵ N per metre. The forces are equal and opposite (Newton's third law).
Common mistakeThe bigger current makes the bigger field, but the force on the other wire also depends on that wire's smaller current, so the products are equal.
49Short answer
Describe how to show the shape and direction of the magnetic field around a straight vertical wire carrying a current, using a card, iron filings and plotting compasses.
Show answer
Model answer: Pass the wire vertically through a horizontal card. Pass a large current, sprinkle iron filings and tap the card: they form concentric circles round the wire. Place plotting compasses on a circle; their needles show the direction of the field. Reverse the current: the compasses reverse, showing the direction depends on the current (right-hand grip rule).
Common mistakeLearners often expect the filings to point away from the wire; the field lines are circles round it, not radial lines.
50Short answer · ★ Challenge
Compare the field 10 cm from a long straight wire carrying 3.5 A with the field at the centre of a single circular loop of radius 10 cm carrying 3.5 A. (μ₀ = 4π × 10⁻⁷ T m/A)
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Model answer: Wire: B = μ₀I/(2πr) = 4π × 10⁻⁷ × 3.5 ÷ (2π × 0.10) = 7.0 × 10⁻⁶ T. Loop: B = μ₀I/(2r) = 4π × 10⁻⁷ × 3.5 ÷ 0.20 = 2.2 × 10⁻⁵ T. The loop gives π times the field, because all of its wire is at the same distance from the centre.
Common mistakeMixing the two formulas (using 2πr for the loop) gives equal fields; the loop formula has 2r, not 2πr.