Selecting electrons with crossed fields

9 marks · A-level · original practice case

Write your working and explanations below or on paper. Your responses stay in this page; PhysicsHelp does not submit or save them. Copy anything you want to keep before closing it. Check calculations separately from method marks. Use unrounded working between parts.

The situation

Electrons accelerated from rest through 2.00 kV travel horizontally to the right into a velocity selector. The magnetic field is 1.50 mT into the page. The uniform electric field is adjusted so these electrons pass straight through. The electric plates are 8.00 mm apart. Electron mass is 9.11 × 10^-31 kg and charge magnitude is 1.60 × 10^-19 C. Use non-relativistic mechanics and ignore edge fields and collisions.

Selector data
Quantity	Value
Accelerating p.d.	2.00 kV
Magnetic field	1.50 mT into the page
Plate separation	8.00 mm

Part (a)

Calculate the speed of the electrons entering the selector. [2 marks]

Give the numerical answer to 3 significant figures.

Final value / m s⁻¹

Working

Part (b)

Determine the potential difference between the selector plates. [2 marks]

Give the numerical answer to 3 significant figures.

Final value / V

Working

Part (c)

Explain which selector plate must be at the higher potential. [3 marks]

Your explanation

Part (d)

Predict what happens to a slower electron entering in the same direction while both selector fields remain unchanged. [2 marks]

Your explanation

Marking rubric and worked answers

Indicative practice marks. Method marks assess valid working; accuracy marks require the associated valid method. Independent points stand on their own. Tick only points supported by your response. Do not award the same explanation twice. These ticks record your judgement, not an examiner’s result.

Part (a) marking points

Final answer: 2.65 × 10⁷ m s⁻¹. Numerical checks assess the final value only; they do not check working or the number of digits typed.

Use eV = ½mv². (1 method mark)
Convert 2.00 kV to 2000 V and obtain 2.65 × 10⁷ m s⁻¹. (1 accuracy mark)

No earlier numerical result is required. Credit a valid alternative method. Accuracy credit requires a physically valid method.

Part (b) marking points

Final answer: 318 V. Numerical checks assess the final value only; they do not check working or the number of digits typed.

For undeflected motion use eE = evB, so E = vB. (1 method mark)
Use plate p.d. = Ed and obtain 318 V. (1 accuracy mark)

Carry forward a numerical error in the speed from part (a) through Vplates = vBd, with correct SI conversions. Do not carry forward a model equating voltage directly to speed.

Part (c) marking points

For electrons moving right in B into the page, magnetic force is downward. (1 independent mark)
Electric force must be upward, so electric field must point downward. (1 independent mark)
The upper plate is at the higher potential because field points from higher to lower potential. (1 independent mark)

Credit equivalent correct explanations. Award each distinct point once; a contradiction invalidates the affected point.

Part (d) marking points

The electric force magnitude is unchanged, whereas magnetic force magnitude evB is smaller. (1 independent mark)
The resultant force is upward, so the electron is deflected towards the upper plate. (1 independent mark)

Credit equivalent correct explanations. Award each distinct point once; a contradiction invalidates the affected point.

Common errors: The two forces remain equal for all speeds: only one speed passes undeflected.

Report a problem with this case

Related skills

Scope and reference sources

AQA 3.7.3.2–3.7.3.3 and 3.7.5.2 (application combining electric and magnetic forces); OCR A 6.2.3 and 6.3.2. These are shared original exercises; wording, indicative marks and difficulty have not been endorsed by either board.

AQA specification · OCR A specification

Case reference: electron-velocity-selector · version 1.1.0 · seed 0. Choose another case