The shape of the lesson
By the end of the lesson, learners can
- Change one thing at a time: Vary the flux density B, the current I, the length L in the field and the angle θ.
- Defining the tesla: The force law is the route to a definition of field strength.
- On a conductor: F = BIL sinθ; the tesla is defined from B = F/(IL) (N A⁻¹ m⁻¹).
- On a charge: F = BQv sinθ, always perpendicular to v, so the path is a circle with r = mv/BQ.
- Induction: Φ = BA, flux linkage = NΦ, induced e.m.f. = rate of change of flux linkage (Faraday); direction opposes the change (Lenz).
One picture every idea returns to
A current-carrying conductor in a magnetic field feels a force F = BIL sinθ, where θ is the angle between the current and the field. The force is perpendicular to both, given by Fleming's left-hand rule. Rearranging the perpendicular case defines the magnetic flux density: B = F / (IL), measured in tesla (T).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, The push on a wire.. Put a current in a magnetic field and the wire is pushed sideways. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Change one thing at a time | Vary the flux density B, the current I, the length L in the field and the angle θ. Project tab 3 of the lab (Force on a conductor) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Defining the tesla | The force law is the route to a definition of field strength. | Connect-Extend-Challenge |
| 30 to 43 min | Check and plenary | The quick-check questions from the student lesson, answered on mini whiteboards one at a time; re-teach the one most get wrong, then a random call on the last. | Individual, then whole class |
| 43 to 45 min | Exit and homework | Set the worksheet below (3 questions, 36 marks) from the topic practice set. Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Defining the tesla, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Connect-Extend-Challenge
After the new idea, each student writes three lines: how it connects to what they knew, how it extends their thinking, and what still challenges them. Pairs compare, then the challenges become the plenary.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| The angle factor is sinθ, not cosθ, and θ is between the current and the field: the force is a maximum when they are perpendicular and zero when parallel. | Use the left hand for the motor effect (force on a current); the right hand is for induction. The flux density B is defined from F = BIL, so its unit, the tesla, is N A⁻¹ m⁻¹. |
Force on a current-carrying conductor: topic worksheet
3 questions, 36 marks. Written in the style of the papers; no past paper question is reproduced. These questions cover the whole topic, so sibling lessons share them.
A straight wire passes at right angles through the uniform field between the poles of a magnet, as shown in Fig. 1.1. The flux density is 0.25 T and the length of wire in the field is 4.0 cm.
(a) State two rules that magnetic field lines always obey. [2]
(b) Define magnetic flux density. [2]
(c) The wire carries a current of 3.0 A. Calculate the force on the wire. [2]
(d) State the direction of this force, and the rule you used. [2]
(e) The wire is now turned so that the current makes an angle of 30° with the field. Calculate the new force. [2]
(f) State the angle between the current and the field for which the force is a maximum. [1]
(g) State the value of the force when the current is parallel to the field. [1]
A horizontal wire of length 5.0 cm rests on a top-pan balance and lies at right angles to a horizontal magnetic field of flux density 0.18 T. When a current of 4.5 A passes through the wire, the magnetic force on it acts vertically.
(a) Calculate the magnetic force on the wire. [2]
(b) Calculate the change in the balance reading, in grams, that this force produces. [2]
(c) Define the tesla. [2]
(d) Express the tesla in SI base units. [2]
(e) State and explain the effect on the force of doubling the current. [2]
(f) State what reading change would be seen if the wire were rotated to lie along the field. [1]
A metal rod of length 0.25 m rests on two horizontal rails and is pulled along them at a steady 4.0 m s⁻¹. A uniform magnetic field of flux density 0.30 T is perpendicular to the plane of the rails, as in Fig. 9.1. The rails are joined by a resistor of resistance 0.50 Ω.
(a) Show that the e.m.f. induced across the rod is given by ε = BLv, and calculate its value. [3]
(b) Calculate the current driven through the resistor. [2]
(c) The current in the rod lies in the magnetic field. Calculate the force on the rod due to this current. [2]
(d) State the direction of this force relative to the motion, and name the law that gives it. [2]
(e) Explain, using energy, why a force must be applied to keep the rod moving at constant speed. [2]
(f) State and explain the effect on this opposing force of doubling the speed. [2]
Answer key and marking notes
- (a)Any two: they run from north to south outside a magnet; they never cross; they are closer where the field is stronger ✓✓
- (b)The force per unit current per unit length on a conductor at right angles to the field ✓✓
- (c)F = BIL = 0.25 × 3.0 × 0.040 ✓ = 0.030 N ✓
- (d)The force is vertical (up or down depending on the current sense) ✓; from Fleming's left-hand rule ✓
- (e)F = BIL sinθ = 0.25 × 3.0 × 0.040 × sin30° ✓ = 0.015 N ✓
- (f)90° (current perpendicular to the field) ✓
- (g)Zero ✓
- (a)F = BIL = 0.18 × 4.5 × 0.050 ✓ = 0.041 N ✓
- (b)Δm = F/g = 0.0405 / 9.81 ✓ = 4.1 × 10⁻³ kg = 4.1 g ✓
- (c)One tesla is the flux density giving a force of 1 N on each metre of a wire carrying 1 A at right angles to the field ✓✓
- (d)From B = F/(IL): T = N A⁻¹ m⁻¹ = kg s⁻² A⁻¹ ✓✓
- (e)The force doubles ✓, because F is proportional to I ✓
- (f)No change: the force becomes zero (current parallel to the field) ✓
- (a)In a time Δt the rod sweeps area ΔA = L v Δt, so ΔΦ = B L v Δt ✓; ε = ΔΦ/Δt = BLv ✓. ε = 0.30 × 0.25 × 4.0 = 0.30 V ✓
- (b)I = ε/R = 0.30 / 0.50 ✓ = 0.60 A ✓
- (c)F = BIL = 0.30 × 0.60 × 0.25 ✓ = 0.045 N ✓
- (d)It opposes the motion (acts backwards) ✓; this follows from Lenz's law ✓
- (e)Work done against this opposing force ✓ is transferred to electrical energy and then heat in the resistor; with no applied force the rod would slow down ✓
- (f)It doubles ✓; the e.m.f. and current both double, and F = BIL is proportional to the current, hence to the speed ✓
The lesson's own checks as a quiz
The quick-check and exam-style questions from the student lesson, printable with the key.
The force F = BIL sinθ on a conductor is greatest when the angle θ between the current and the field is:
- A90°
- B0°
- C45°
- D180°
The tesla, the unit of magnetic flux density, is equivalent to:
- AN A⁻¹ m⁻¹
- BN A m
- CJ A⁻¹
- DN C⁻¹
In Fleming's left-hand rule the first finger, second finger and thumb represent, in that order:
- Afield, current, force
- Bcurrent, field, force
- Cforce, field, current
- Dfield, force, current
A straight wire of length 0.20 m carries a current of 5.0 A at right angles to a uniform field of flux density 0.40 T. The force on the wire is:
- A0.40 N
- B2.0 N
- C0.040 N
- D10 N
Quiz key
sinθ is largest at 90°, so the force is maximum when the current is perpendicular to the field, and zero when parallel.
From B = F / (IL), the tesla is one newton per ampere per metre, N A⁻¹ m⁻¹.
First finger = Field, seCond finger = Current, thuMb = Force (Motion), held mutually at right angles.
F = BIL = 0.40 × 5.0 × 0.20 = 0.40 N, since the wire is perpendicular to the field (sin90° = 1).
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.