Specification: AQA A-level Physics (7408) 3.7.5.1 | Paper: 2 | Data and formulae booklet: given in the exam
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Magnetic flux density, B: the force per unit length per unit current on a current-carrying conductor placed at right angles to the magnetic field, B = F ÷ Il. It is a vector, measured in teslas (T).
The tesla: a magnetic flux density of 1 T produces a force of 1 N on a 1 m length of wire carrying a current of 1 A at right angles to the field. So 1 T = 1 N A⁻¹ m⁻¹.
Fleming’s left-hand rule: hold the thumb, first finger and second finger of the left hand at right angles. The First finger points along the Field (north to south), the seCond finger along the Current (conventional, + to −) and the thuMb gives the direction of the force (Motion). A wire parallel to the field feels no force.
The equation
force = magnetic flux density × current × length F = BIl (wire at right angles to the field; at an angle θ to the field, F = BIl sin θ)
| Quantity | Symbol | Unit |
|---|---|---|
| force on the wire | F | newtons (N) |
| magnetic flux density | B | teslas (T), equivalent to N A⁻¹ m⁻¹ |
| current | I | amperes (A) |
| length of wire in the field | l | metres (m) |
| angle between the wire and the field | θ | degrees (°) |
Rearranged: B = F ÷ Il | I = F ÷ Bl | l = F ÷ BI | Only the length of wire inside the field counts. Convert mT to T (× 10⁻³) and cm to m before you substitute. The data booklet gives F = BIl for a wire perpendicular to the field; at an angle, use the component of the field perpendicular to the wire, B sin θ.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) A straight wire of length 0.15 m carries a current of 4.0 A at right angles to a uniform magnetic field of flux density 0.25 T. Calculate the force on the wire.
Show solution
F: F = BIl
I: F = 0.25 × 4.0 × 0.15
F: F is already the subject
A: 0.15 N
Q2 (Core) An 8.0 cm length of wire carrying a current of 6.0 A lies at right angles to a uniform magnetic field. The force on the wire is 0.024 N. Calculate the magnetic flux density in mT.
Show solution
Convert first: 8.0 cm = 0.080 m
F: F = BIl
I: 0.024 = B × 6.0 × 0.080
F: B = 0.024 ÷ (6.0 × 0.080) = 0.050 T
A: 50 mT
Q3 (Core) A 4.0 cm length of wire lies at right angles to a magnetic field of flux density 85 mT. Calculate the current needed to produce a force of 6.8 mN on the wire.
Show solution
Convert first: 4.0 cm = 0.040 m; 85 mT = 0.085 T; 6.8 mN = 6.8 × 10−3 N
F: F = BIl
I: 6.8 × 10−3 = 0.085 × I × 0.040
F: I = 6.8 × 10−3 ÷ (0.085 × 0.040)
A: 2.0 A
Q4 (Core) A 0.40 m length of wire carries a current of 6.0 A in a uniform magnetic field of flux density 0.25 T. The wire makes an angle of 30° with the field lines. Calculate the force on the wire.
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F: F = BIl sin θ (only the component of B perpendicular to the wire, B sin θ, produces a force)
I: F = 0.25 × 6.0 × 0.40 × sin 30°
F: F is already the subject
A: 0.60 × 0.5 = 0.30 N
Watch out: θ is the angle between the wire and the field. Using cos 30° gives 0.52 N. If the wire were parallel to the field (θ = 0°) the force would be zero.
Q5 (Exam) In a top-pan balance experiment, a U-shaped magnet sits on an electronic balance and a horizontal wire is clamped between its poles, at right angles to the field. The length of wire in the field is 5.0 cm. When a current of 2.0 A is switched on, the balance reading increases by 1.8 g. (a) Calculate the magnetic flux density between the poles, to 2 significant figures. (b) State the direction of the force on the wire. (g = 9.81 N kg⁻¹)
Show solution
Convert first: 1.8 g = 1.8 × 10−3 kg; 5.0 cm = 0.050 m
Step 1 – force from the change in reading
F = Δm × g = 1.8 × 10−3 × 9.81 = 0.017658 N
Step 2 – flux density
F: F = BIl
I: 0.017658 = B × 2.0 × 0.050
F: B = 0.017658 ÷ 0.10
A: (a) 0.17658 T = 0.18 T (2 s.f.)
(b) The reading increases, so the wire pushes the magnet down. By Newton’s third law the magnet pushes the wire upwards with an equal force.
Watch out: the balance reads in grams, not newtons. Convert the change in mass to kg and multiply by g to get the force.
Q6 (Exam) A horizontal overhead power cable runs from west to east and carries a current of 400 A towards the east. The horizontal component of the Earth’s magnetic field is 4.4 × 10−5 T, directed north. (a) Use Fleming’s left-hand rule to find the direction of the force on the cable. (b) Calculate the magnitude of this force on a 250 m span of the cable, to 2 significant figures.
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(a) Direction: first finger (field) points north, second finger (current) points east, so the thumb points vertically upwards
(b)
F: F = BIl
I: F = 4.4 × 10−5 × 400 × 250
F: F is already the subject
A: 4.4 N upwards (2 s.f.)
Watch out: Fleming’s left-hand rule uses conventional current (positive to negative), not the direction of electron flow.
Q7 (Exam) A horizontal metal rod of mass 12 g and length 0.20 m lies at right angles to a horizontal magnetic field of flux density 0.25 T. Calculate the current needed for the magnetic force on the rod to support its weight, to 2 significant figures. (g = 9.81 N kg⁻¹)
Show solution
Convert first: 12 g = 0.012 kg
Step 1 – weight
W = mg = 0.012 × 9.81 = 0.11772 N
Step 2 – current (magnetic force upwards = weight)
F: F = BIl
I: 0.11772 = 0.25 × I × 0.20
F: I = 0.11772 ÷ (0.25 × 0.20)
A: 2.3544 A = 2.4 A (2 s.f.)
Q8 (Exam) A student repeats the top-pan balance experiment for several currents and plots the change in balance reading (in g) against current (in A). The graph is a straight line through the origin with a gradient of 0.41 g A⁻¹. The length of wire in the field is 5.0 cm. Calculate the magnetic flux density, to 2 significant figures. (g = 9.81 N kg⁻¹)
Show solution
Convert first: 0.41 g A⁻¹ = 0.41 × 10−3 kg A⁻¹; 5.0 cm = 0.050 m
Step 1 – force per unit current
F ÷ I = 0.41 × 10−3 × 9.81 = 4.0221 × 10−3 N A⁻¹
Step 2 – flux density
F: F = BIl, so F ÷ I = Bl
I: 4.0221 × 10−3 = B × 0.050
F: B = 4.0221 × 10−3 ÷ 0.050
A: 0.080442 T = 0.080 T (2 s.f.)
Watch out: the gradient is mass per amp, not force per amp. Multiply by g before dividing by l.
Q9 (Stretch) A metal rod of mass 0.030 kg rests across two horizontal rails 0.12 m apart. A vertical magnetic field of flux density 0.30 T acts over the rails. A constant current of 5.0 A flows through the rod, which starts from rest. Ignoring friction, calculate the speed of the rod after it has moved 0.50 m along the rails, to 2 significant figures.
Show solution
Step 1 – force on the rod (the rod is at right angles to the field)
F: F = BIl
I: F = 0.30 × 5.0 × 0.12
A: F = 0.18 N, horizontal and along the rails
Step 2 – acceleration
a = F ÷ m = 0.18 ÷ 0.030 = 6.0 m s⁻²
Step 3 – speed (suvat, u = 0)
v² = u² + 2as = 0 + 2 × 6.0 × 0.50 = 6.0
A: v = √6.0 = 2.449… m s⁻¹ = 2.4 m s⁻¹ (2 s.f.)
Watch out: the rails are 0.12 m apart, so l = 0.12 m (the length of rod carrying current in the field), not the 0.50 m it travels.
Q10 (Stretch) A rectangular coil of 50 turns measures 5.0 cm by 3.0 cm. It carries a current of 0.40 A and is pivoted about an axis through its centre, parallel to the 5.0 cm sides. It sits in a uniform magnetic field of flux density 0.20 T, with the 5.0 cm sides at right angles to the field. (a) With the plane of the coil parallel to the field, calculate the force on each 5.0 cm side and the torque on the coil. (b) Calculate the torque when the plane of the coil has turned to 60° to the field.
Show solution
Convert first: 5.0 cm = 0.050 m; 3.0 cm = 0.030 m
Step 1 – (a) force on each long side (50 turns, each with a 0.050 m side in the field)
F = BIlN = 0.20 × 0.40 × 0.050 × 50 = 0.20 N, in opposite directions on the two sides
Step 2 – (a) torque
the two forces form a couple; perpendicular distance between them = 0.030 m
torque = F × d = 0.20 × 0.030 = 6.0 × 10−3 N m
Step 3 – (b) torque at 60°
the forces are still 0.20 N (the sides stay at right angles to the field), but the perpendicular distance between them is now 0.030 × cos 60° = 0.015 m
A: torque = 0.20 × 0.015 = 3.0 × 10−3 N m
Watch out: the forces on the 3.0 cm sides do not cause any turning about this axis, because they act along the axis. The torque falls to zero when the plane of the coil is at right angles to the field.
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