OxfordAQA International AS and A-level Physics: Force on conductors and moving charges in magnetic fields (9630) – 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 sections 3.10.1 (Magnetic flux density) and 3.10.2 (Moving charges in a magnetic field) of the OxfordAQA International AS and A-level Physics (9630) specification, version 4.4. Section 3.10 is International A-level only. Data: electron and proton charge magnitude 1.60 × 10⁻¹⁹ C; masses of electron 9.11 × 10⁻³¹ kg, proton 1.67 × 10⁻²⁷ kg and deuteron 3.34 × 10⁻²⁷ kg; 1 u = 1.66 × 10⁻²⁷ kg; g = 9.81 m s⁻². See the study guide, revision notes, the subject hub, the printable checklist and the free 10-minute diagnostics.

Questions

1. (a) State the definition of the tesla. [2] (b) Show that the base units of the tesla are kg s⁻² A⁻¹. [2]

2. A straight copper rod 0.14 m long lies horizontally along a west-east line inside a uniform field of 0.12 T that points vertically upward. It carries a current of 7.5 A towards the east. (a) Calculate the force on the rod. [2] (b) State the direction of the force, using compass directions or up/down. [2]

3. A student places a magnet assembly on a balance. A stiff wire, with 0.050 m of its length in the gap, runs at right angles to the field. The balance is zeroed, then the current is increased. The increase in reading is recorded.

I / A	1.0	2.0	3.0	4.0	5.0
increase in reading / g	0.54	1.07	1.60	2.15	2.68

(a) Explain why a graph of the increase in mass against current is a straight line through the origin. [2] (b) Find the gradient of the graph. [1] (c) Determine the flux density. [2]

4. A proton moves at 5.5 × 10⁵ m s⁻¹ perpendicular to a uniform field of 0.30 T. (a) Calculate the force on it. [2] (b) Calculate its acceleration. [2]

5. Electrons are accelerated from rest through 350 V and then enter a uniform field of 1.5 × 10⁻³ T at right angles to their velocity. (a) Calculate their speed on entering the field. [2] (b) Calculate the radius of their path. [2] (c) Explain why their kinetic energy does not change inside the field. [2]

6. (a) Show that the time for one orbit of a charged particle moving at right angles to a uniform field is independent of its speed. [3] (b) A proton in a field of 0.46 T moves perpendicular to it. Calculate the time for one orbit. [2] (c) The proton’s speed is 3.1 × 10⁵ m s⁻¹. Calculate the radius of its orbit, and state the radius and orbit time if the speed doubled. [3]

7. Singly charged chlorine ions of mass 35 u and 37 u are accelerated from rest through 1500 V. They enter a uniform field of 0.20 T at right angles and travel through semicircles to a detector plate. (a) Calculate the speed of the 35 u ions. [2] (b) Calculate the radius of each semicircle. [3] (c) Calculate the distance between the points where the two types of ion land. [2]

8. A cyclotron has a uniform field of 1.1 T and dees of radius 0.30 m. It accelerates deuterons (charge +e) and the potential difference across the gap has a peak value of 45 kV, so each gap crossing adds 45 keV. (a) Calculate the frequency of the alternating potential difference. [2] (b) Calculate the maximum speed of the deuterons. [2] (c) Calculate their maximum kinetic energy in joules and in MeV. [2] (d) Explain why the radius of the path increases while the time for each semicircle stays the same. [2]

9. A beam of electrons travelling to the right across the page enters a region where a uniform field points out of the page. Their speed is 1.8 × 10⁷ m s⁻¹ and the flux density is 4.0 × 10⁻³ T. (a) State the direction in which the beam is deflected. [2] (b) Calculate the radius of the circle. [2] (c) The flux density is doubled. State the new radius and the effect on the kinetic energy of the electrons. [2]

10. A rod of mass 18 g and length 0.22 m lies horizontally across the page, free to move, in a horizontal uniform field of 0.12 T directed into the page. (a) State the direction of the current that gives an upward force on the rod. [2] (b) Calculate the current for which the magnetic force exactly balances the weight. [3] (c) The current is raised to 9.0 A. Calculate the initial upward acceleration. [3]

Answers

1. (a) The flux density of a uniform field that exerts a force of 1 N [1] on 1 m of wire carrying 1 A and lying across the field [1]. (b) B = F / (IL), so 1 T = 1 N A⁻¹ m⁻¹ [1]. With 1 N = 1 kg m s⁻², 1 T = 1 kg m s⁻² A⁻¹ m⁻¹ = 1 kg s⁻² A⁻¹ [1]. [4] Examiner insight: A definition needs right angles, current, length and force. Leaving out the right angle is a common loss.

2. (a) F = BIL = 0.12 × 7.5 × 0.14 [1] = 0.126 N, so 0.13 N [1]. (b) Second finger east (current), first finger up (field) [1]. The thumb points south [1]. [4] Examiner insight: Name the digits you used for a direction.

3. (a) The wire feels F = BIL, and the magnets feel an equal and opposite force, which changes the balance reading by Δm g [1]. So Δm = (BL / g) I, which is proportional to I and is zero when I is zero [1]. (b) Gradient = (2.68 − 0.54) / 4.0 = 0.535 g A⁻¹ = 5.35 × 10⁻⁴ kg A⁻¹ [1]. (c) B = gradient × g / L = 5.35 × 10⁻⁴ × 9.81 / 0.050 [1] = 0.105 T, so 0.10 T to 2 significant figures [1]. [5] Examiner insight: Convert grams to kilograms before using g. Showing the conversion keeps the method visible.

4. (a) F = BQv = 0.30 × 1.60 × 10⁻¹⁹ × 5.5 × 10⁵ [1] = 2.6 × 10⁻¹⁴ N [1]. (b) a = F / m = 2.64 × 10⁻¹⁴ / 1.67 × 10⁻²⁷ [1] = 1.6 × 10¹³ m s⁻² [1]. [4] Examiner insight: Use the unrounded force in part (b) to avoid rounding drift.

5. (a) ½mv² = eV, so v = √(2eV / m) = √(2 × 1.60 × 10⁻¹⁹ × 350 / 9.11 × 10⁻³¹) [1] = 1.1 × 10⁷ m s⁻¹ [1]. (b) r = mv / (BQ) = 9.11 × 10⁻³¹ × 1.109 × 10⁷ / (1.5 × 10⁻³ × 1.60 × 10⁻¹⁹) [1] = 0.042 m [1]. (c) The magnetic force always acts at right angles to the velocity, so it does no work on the electrons [1]. No work means no change in kinetic energy, so the speed is constant and only the direction changes [1]. [6] Examiner insight: Explain “no work” from the force being at right angles to the motion.

6. (a) The magnetic force supplies the centripetal force: BQv = mv² / r [1]. So r = mv / (BQ), and T = 2πr / v [1]. Then T = 2πm / (BQ), and v has cancelled [1]. (b) T = 2πm / (BQ) = 2π × 1.67 × 10⁻²⁷ / (0.46 × 1.60 × 10⁻¹⁹) [1] = 1.4 × 10⁻⁷ s [1]. (c) r = mv / (BQ) = 1.67 × 10⁻²⁷ × 3.1 × 10⁵ / (0.46 × 1.60 × 10⁻¹⁹) [1] = 7.0 × 10⁻³ m [1]. At twice the speed the radius doubles to 1.4 × 10⁻² m and the time is unchanged at 1.4 × 10⁻⁷ s [1]. [8] Examiner insight: In a “show that” question, every line of the derivation should appear. Going straight to T = 2πm / (BQ) without the centripetal step leaves the argument incomplete.

7. (a) m = 35 × 1.66 × 10⁻²⁷ = 5.81 × 10⁻²⁶ kg [1]. v = √(2eV / m) = √(2 × 1.60 × 10⁻¹⁹ × 1500 / 5.81 × 10⁻²⁶) = 9.1 × 10⁴ m s⁻¹ [1]. (b) For 37 u: m = 6.142 × 10⁻²⁶ kg and v = 8.84 × 10⁴ m s⁻¹ [1]. r = mv / (eB) gives, for 35 u, 5.81 × 10⁻²⁶ × 9.089 × 10⁴ / (1.60 × 10⁻¹⁹ × 0.20) = 0.165 m [1], and for 37 u, 0.170 m [1]. (c) The landing points are one diameter from the entry point, so the separation is 2(r₃₇ − r₃₅) [1] = 2 × (0.16968 − 0.16503) = 9.3 × 10⁻³ m [1]. [7] Examiner insight: Keep extra digits until the final subtraction, and remember the separation is twice the radius difference.

8. (a) f = BQ / (2πm) = 1.1 × 1.60 × 10⁻¹⁹ / (2π × 3.34 × 10⁻²⁷) [1] = 8.4 × 10⁶ Hz [1]. (b) v = BQR / m = 1.1 × 1.60 × 10⁻¹⁹ × 0.30 / 3.34 × 10⁻²⁷ [1] = 1.6 × 10⁷ m s⁻¹ [1]. (c) Ek = ½mv² = ½ × 3.34 × 10⁻²⁷ × (1.581 × 10⁷)² [1] = 4.2 × 10⁻¹³ J, which is 2.6 MeV [1]. (d) After each crossing the deuteron is faster, and r = mv / (BQ) is proportional to v, so the radius grows [1]. The time for a semicircle is πm / (BQ), which does not depend on v, so it stays the same and the pd frequency can be fixed [1]. [8] Examiner insight: Say that the electric field in the gap speeds the particle up, not the magnetic field.

9. (a) Equivalent conventional current is to the left [1]. With the field out of the page, the thumb points up the page [1]. (b) r = mv / (BQ) = 9.11 × 10⁻³¹ × 1.8 × 10⁷ / (4.0 × 10⁻³ × 1.60 × 10⁻¹⁹) [1] = 0.026 m [1]. (c) r is inversely proportional to B, so the radius halves to 0.013 m [1]. The kinetic energy is unchanged because the magnetic force does no work [1]. [6] Examiner insight: Give the new numerical value and the reason, not just “the radius decreases”.

10. (a) Second finger must point along the current, first finger into the page, with the thumb up [1]. The current must flow to the right [1]. (b) BIL = mg [1]. I = mg / (BL) = 0.018 × 9.81 / (0.12 × 0.22) [1] = 6.7 A [1]. (c) F = BIL = 0.12 × 9.0 × 0.22 = 0.2376 N, and the weight is 0.1766 N [1]. Net force = 0.0610 N [1]. a = 0.0610 / 0.018 = 3.4 m s⁻² upward [1]. [8] Examiner insight: Convert grams to kilograms. Use the net force for the acceleration, not the magnetic force alone.

Where marks are usually lost

• Using conventional current directions for an electron without reversing them.
• Leaving out the right angle when defining the tesla or flux density.
• Forgetting that grams, centimetres and u must be converted to kilograms and metres.
• Dropping units or leaving the answer at the wrong number of significant figures.

Next steps

• Revise with the revision notes
• Read the study guide
• See the subject hub and printable checklist
• Find weak areas: all free 10-minute diagnostics
• Ready for help? Book a free trial class

Official syllabus

OxfordAQA International AS and A-level Physics (9630), version 4.4, for International AS and A-level exams from May/June 2020 onwards, published by Oxford International AQA Examinations. Sections 3.10.1 and 3.10.2.