OCR A Level Physics A: Magnetic fields, force on conductors and moving charges (H556) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover 6.3.1 (Magnetic fields) and 6.3.2 (Motion of charged particles) of the H556 Physics A specification from OCR (Version 3.0, March 2026). Take g = 9.81 m s⁻², e = 1.60 × 10⁻¹⁹ C, electron mass 9.11 × 10⁻³¹ kg, proton mass 1.673 × 10⁻²⁷ kg and 1 u = 1.66 × 10⁻²⁷ kg. Worked explanations are in the study guide and the revision notes. Also see the hub and checklist. If several answers go wrong, try a free 10-minute diagnostic.

Questions

1. (a) Describe the field inside and outside a long solenoid carrying a current, and how to find its north end. [3] (b) Describe the field lines around a long straight wire carrying a current. [2]

2. A horizontal wire carries a conventional current towards the east. A uniform magnetic field points vertically downwards. (a) State the direction of the force on the wire. [2] (b) An electron travels east through the same field. State the direction of the force on it. [2]

3. A straight wire has 0.25 m of its length in a uniform field of 0.18 T. The force on this length is 0.027 N when the wire is at right angles to the field. (a) Calculate the current. [2] (b) The wire is rotated so it makes 40° with the field lines and carries 0.60 A. Calculate the force. [2]

4. Define the tesla and show that its base units are kg s⁻² A⁻¹. [3]

5. Magnets on a yoke sit on a digital balance, and a horizontal wire passes through the gap at right angles to the field. The length of wire in the field is 0.045 m. With the balance zeroed, a student records the increase in reading for several currents.

I / A	0.50	1.00	1.50	2.00	2.50
increase in reading / g	0.31	0.62	0.95	1.24	1.55

(a) Explain why the increase in reading is proportional to the current. [2] (b) Find the gradient of a graph of mass (in kg) against current. [2] (c) Determine the flux density. [2] (d) Suggest one source of uncertainty and one improvement. [1]

6. A horizontal rod of mass 12 g and length 0.15 m lies at right angles to a horizontal uniform field of 0.40 T. (a) Calculate the current that makes the magnetic force equal to the weight. [2] (b) The current is then doubled. Calculate the initial acceleration of the rod. [3]

7. An electron moves at 3.2 × 10⁶ m s⁻¹ at right angles to a uniform field of 0.080 T. (a) Calculate the force on the electron. [2] (b) Calculate its acceleration. [2] (c) Describe its path and say what happens to its speed. [2]

8. An alpha particle (mass 6.64 × 10⁻²⁷ kg, charge +2e) and a proton each enter a uniform field of 0.40 T at right angles at 1.5 × 10⁶ m s⁻¹. (a) Show that the radius of an orbit is r = mv / (BQ) and calculate it for the alpha particle. [3] (b) Compare it with the radius for the proton. [2] (c) Calculate the period of each orbit. [2]

9. (a) Show that the period of a charged particle in a uniform field at right angles to its velocity is independent of its speed. [3] (b) Two protons enter a field of 0.30 T at right angles at 2.0 × 10⁵ m s⁻¹ and 8.0 × 10⁵ m s⁻¹. Calculate the radius of each orbit. [2]

10. A velocity selector has parallel plates 4.0 mm apart with a p.d. of 240 V, and a uniform magnetic field of 0.015 T at right angles to the electric field. (a) Calculate the speed of particles that pass undeflected. [3] (b) A proton enters at 2.0 × 10⁶ m s⁻¹. Explain which force is larger and how the proton is deflected. [2] (c) Explain whether an alpha particle at 4.0 × 10⁶ m s⁻¹ would also pass through. [2]

11. Singly charged ions of neon-20 (20 u) and neon-22 (22 u) pass through a velocity selector with E = 3.0 × 10⁴ V m⁻¹ and B = 0.10 T. They then enter a region of uniform field 0.50 T only and move through semicircles. (a) Calculate the speed of the ions that leave the selector. [2] (b) Explain why both isotopes pass. [1] (c) Calculate the radius of each semicircle. [3] (d) Calculate the distance between the two landing points. [2] (e) Explain why the heavier ion lands further out. [2]

Answers

1. (a) Inside, away from the ends, the lines are straight, parallel and equally spaced, so the field is uniform [1]. Outside, the pattern matches a bar magnet, with lines leaving one end and returning to the other [1]. Curl the fingers of the right hand along the current: the thumb points to the north end [1]. (b) The lines are rings centred on the wire, each in a plane perpendicular to it [1], getting further apart with distance, so the field weakens [1]. Examiner insight: State inside and outside separately, and mention line spacing, not just “circles”.

2. (a) Thumb is force, first finger is field (downwards), second finger is conventional current (east) [1]. The force is towards the north [1]. (b) The conventional current for an electron moving east points west [1], so the force is reversed, towards the south [1]. Examiner insight: Say explicitly that you reversed the current direction for the electron.

3. (a) F = BIL, so I = F / (BL) [1] = 0.027 / (0.18 × 0.25) = 0.60 A [1] (b) F = BIL sinθ = 0.18 × 0.60 × 0.25 × sin 40° [1] = 0.017 N [1] Examiner insight: Write the equation before substituting, and show sin 40° so the angle used is visible.

4. Flux density is force per unit current per unit length, for a conductor at right angles to the field [1]. One tesla is the flux density for which 1 m of conductor carrying 1 A perpendicular to the field feels 1 N [1]. B = F/(IL), so 1 T = 1 N A⁻¹ m⁻¹ and with 1 N = 1 kg m s⁻², T = kg s⁻² A⁻¹ [1] Examiner insight: Write each step of a units derivation, including the base units of the newton.

5. (a) The force on the wire is F = BIL, and L and B are constant [1]. The magnets feel an equal and opposite force, so Δm g = F and Δm is proportional to I [1]. (b) Using the end points, gradient = (1.55 − 0.31) × 10⁻³ / (2.50 − 0.50) [1] = 6.2 × 10⁻⁴ kg A⁻¹ [1] (c) B = gradient × g / L [1] = 6.2 × 10⁻⁴ × 9.81 / 0.045 = 0.135 T [1] (d) Any one: the field is not uniform near the edges, so the effective length is uncertain; use larger pole faces. [1] Examiner insight: Convert grams to kilograms first, and make the improvement fit the error you named.

6. (a) W = 0.012 × 9.81 = 0.118 N [1]. I = W / (BL) = 0.118 / (0.40 × 0.15) = 2.0 A [1] (b) The new current is 3.92 A, so the magnetic force is BIL = 0.40 × 3.92 × 0.15 = 0.235 N [1]. Net force = 0.235 − 0.118 = 0.118 N upwards [1]. a = 0.118 / 0.012 = 9.8 m s⁻² upwards [1] Examiner insight: Subtract the weight before using F = ma, and state the direction.

7. (a) F = BQv = 0.080 × 1.60 × 10⁻¹⁹ × 3.2 × 10⁶ [1] = 4.1 × 10⁻¹⁴ N [1] (b) a = F/m = 4.1 × 10⁻¹⁴ / 9.11 × 10⁻³¹ [1] = 4.5 × 10¹⁶ m s⁻² [1] (c) The path is a circle in the plane at right angles to the field [1]. The speed is constant, because the force is perpendicular to the velocity and does no work [1] Examiner insight: Give the reason the speed is constant, not just the statement.

8. (a) The magnetic force is the centripetal force: BQv = mv²/r [1], so r = mv / (BQ) [1]. For the alpha: r = 6.64 × 10⁻²⁷ × 1.5 × 10⁶ / (2 × 1.60 × 10⁻¹⁹ × 0.40) = 0.078 m [1] (b) For the proton, r = 1.673 × 10⁻²⁷ × 1.5 × 10⁶ / (1.60 × 10⁻¹⁹ × 0.40) = 0.039 m [1]. The alpha’s radius is twice as large (four times the mass, twice the charge) [1] (c) T = 2πm / (BQ): alpha 3.3 × 10⁻⁷ s [1]; proton 1.6 × 10⁻⁷ s [1] Examiner insight: Use Q = 2e for the alpha; omitting the 2 changes every later answer.

9. (a) BQv = mv²/r gives r = mv / (BQ) [1]. T = 2πr / v [1], so T = 2πm / (BQ), in which v has cancelled [1] (b) r = mv/(eB) = 1.673 × 10⁻²⁷ × 2.0 × 10⁵ / (1.60 × 10⁻¹⁹ × 0.30) = 7.0 × 10⁻³ m [1], and for 8.0 × 10⁵ m s⁻¹, 2.8 × 10⁻² m [1] Examiner insight: Start the derivation from the force equation and show the cancellation.

10. (a) E = V/d = 240 / 4.0 × 10⁻³ = 6.0 × 10⁴ V m⁻¹ [1]. For no deflection QE = BQv [1], so v = E/B = 4.0 × 10⁶ m s⁻¹ [1] (b) The magnetic force per unit charge is Bv = 0.015 × 2.0 × 10⁶ = 3.0 × 10⁴ V m⁻¹, which is less than E [1]. So the electric force is larger and the proton is deflected in the direction of the electric force, towards the negative plate [1] (c) Yes. The condition v = E/B does not contain charge or mass [1], and 4.0 × 10⁶ m s⁻¹ equals E/B, so the alpha is undeflected [1] Examiner insight: Compare the two forces with numbers, then state the direction.

11. (a) v = E/B [1] = 3.0 × 10⁴ / 0.10 = 3.0 × 10⁵ m s⁻¹ [1] (b) Speed v = E/B does not depend on mass or charge [1] (c) r = mv/(eB). Neon-20: 20 × 1.66 × 10⁻²⁷ × 3.0 × 10⁵ / (1.60 × 10⁻¹⁹ × 0.50) [1] = 0.1245 m [1]. Neon-22: 0.1370 m [1] (d) A semicircle’s landing point is 2r from the entry point [1], so the separation is 2 × (0.1370 − 0.1245) = 0.025 m [1] (e) For equal speed, charge and field, r is proportional to m [1], so the heavier ion has the larger radius and lands further from the entry point [1] Examiner insight: Keep four significant figures in the radii and round only the separation.

Where marks are usually lost

• Using grams in the balance calculation, or forgetting g.
• Taking θ from the wrong line.
• Not reversing the current direction for an electron.
• Stating v = E/B without saying the two forces are equal and opposite.

Next steps

• Revisit the revision notes and the study guide.
• See the checklist to print and the subject hub.
• Find gaps with the free 10-minute diagnostics.
• Want a tutor? Book a free trial class.

Official syllabus

Source: the OCR Physics A (H556) specification, Version 3.0 (March 2026), first assessment 2017. Section 6.3 Electromagnetism; outcomes 6.3.1(a)-(f) and 6.3.2(a)-(c).