OCR A Level Physics A: Charge, current, drift velocity, p.d. and e.m.f. (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 sections 4.1.1, 4.1.2, 4.2.1 and 4.2.2 of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017. H556 is the full A level. Use e = 1.6 × 10⁻¹⁹ C, electron mass 9.11 × 10⁻³¹ kg and proton mass 1.67 × 10⁻²⁷ kg.

Related: study guide, revision notes, OCR A Level Physics hub and checklist.

Questions

1. A heating element carries a steady current of 2.8 A for 25 minutes. (a) Calculate the charge that passes through the element. [2] (b) Calculate the number of electrons that pass through it in this time. [2]

2. Four oil droplets are found to carry charges of 3.2, 4.8, 5.6 and 8.0 (all × 10⁻¹⁹ C). Identify the charge that cannot be correct and explain your choice. [3]

3. State Kirchhoff’s first law and explain why it is a consequence of conservation of charge. Currents of 3.5 A and 1.2 A enter a junction, and 2.9 A leaves in one wire. Calculate the current in the only other wire leaving the junction. [4]

4. (a) A metal wire is connected across a cell. State the direction of electron flow relative to the direction of conventional current. [2] (b) Describe how charge is carried through an electrolyte. [2] (c) In an electrolyte, 3.0 × 10¹⁸ singly charged positive ions cross a plane each second in one direction and 2.0 × 10¹⁸ singly charged negative ions cross it each second in the other direction. Calculate the current. [2]

5. Show that the current I in a conductor is given by I = Anev, defining each symbol. [4]

6. A copper wire has a diameter of 1.2 mm and carries a current of 4.5 A. The number density of conduction electrons in copper is 8.5 × 10²⁸ m⁻³. (a) Calculate the mean drift velocity of the electrons. [3] (b) Calculate the time taken for an electron to drift 1.0 m along the wire. [2]

7. (a) Two wires X and Y of the same metal are joined in series. The diameter of Y is one third of the diameter of X. Compare the mean drift velocities in the two wires. [3] (b) A metal has n = 8.5 × 10²⁸ m⁻³ and a semiconductor sample of the same cross-sectional area has n = 4.0 × 10²² m⁻³ (both given). The two carry the same current. Calculate the ratio of the drift velocity in the semiconductor to that in the metal. [2] (c) Explain, in terms of n, why an insulator carries almost no current. [1]

8. Draw a circuit diagram for an experiment to find how the current in a filament lamp varies with the p.d. across it. The circuit must allow the current to be changed. [4]

9. A motor has a p.d. of 9.0 V across it and a current of 0.80 A for 3.0 minutes. Calculate the energy transferred to the motor. [3]

10. A battery of e.m.f. 4.5 V supplies a steady current of 0.60 A to a lamp for 5.0 minutes. The p.d. across the lamp is 4.1 V. (a) State the difference between e.m.f. and p.d. in terms of energy transfer. [2] (b) Calculate the charge that passes and the energy the battery supplies to this charge. [2] (c) Calculate the energy transferred in the lamp and state where the remaining energy goes. [2]

11. Electrons and protons are each released from rest and accelerated through a p.d. of 1.2 kV in a vacuum. (a) Calculate the final speed of an electron. [3] (b) Calculate the final speed of a proton and compare it with that of the electron. [2]

12. A beam of alpha particles (charge +2e, mass 6.64 × 10⁻²⁷ kg) is accelerated from rest through 1.5 kV. The beam current is 2.0 μA. (a) Calculate the kinetic energy gained by one alpha particle. [1] (b) Calculate its speed. [2] (c) Calculate the number of alpha particles arriving each second. [2] (d) Calculate the power carried by the beam, and check it using the beam current. [2]

Answers

1. (a) t = 25 × 60 = 1500 s [1]. Q = It = 2.8 × 1500 = 4.2 × 10³ C (4200 C) [1]. (b) N = Q/e [1] = 4200 / 1.6 × 10⁻¹⁹ = 2.6 × 10²² electrons [1]. [4] Examiner insight: Convert the time to seconds before substituting, and write that conversion as a separate line so it is easy to check.

2. Divide each charge by e: 3.2/1.6 = 2.0, 4.8/1.6 = 3.0, 5.6/1.6 = 3.5, 8.0/1.6 = 5.0 [1]. Charge is quantised, so every value must be a whole-number multiple of e [1]. 5.6 × 10⁻¹⁹ C cannot be correct because it is 3.5e [1]. [3] Examiner insight: Name the principle (whole-number multiples of e) and show the division, rather than quoting the value alone.

3. Total current into a junction equals total current out of it [1]. Charge cannot be created, destroyed or stored at the junction, so the charge entering per second equals the charge leaving per second [1]. 3.5 + 1.2 = 2.9 + I [1]. I = 1.8 A [1]. [4] Examiner insight: Mention charge not building up at the junction; restating the law does not explain it.

4. (a) Electrons are repelled from the negative terminal and travel through the wire towards the positive terminal [1]. Conventional current is drawn the opposite way, out of the positive terminal [1]. (b) Free ions carry the charge: positive ions move towards the negative electrode and negative ions move towards the positive electrode [1]. Both contribute to the current, so the current is the sum of the two [1]. (c) Charge crossing per second = (3.0 × 10¹⁸ + 2.0 × 10¹⁸) × 1.6 × 10⁻¹⁹ [1] = 0.80 A [1]. [6] Examiner insight: In (c), check that every carrier crossing the plane is counted, including both signs of ion.

5. In time t, carriers drift a distance vt, so the volume of conductor passed is Avt, where A is the cross-sectional area and v the mean drift velocity [1]. The number of carriers in that volume is nAvt, with n the number density [1]. The charge passing is nAvte, with e the charge on each carrier [1]. I = Q/t = nAvte / t = Anev [1]. [4] Examiner insight: In a “show that” question, give every step in order, ending on the stated result.

6. (a) A = π(0.60 × 10⁻³)² = 1.13 × 10⁻⁶ m² [1]. v = I / (Ane) [1] = 4.5 / (1.13 × 10⁻⁶ × 8.5 × 10²⁸ × 1.6 × 10⁻¹⁹) = 2.9 × 10⁻⁴ m s⁻¹ [1]. (b) t = distance / speed = 1.0 / 2.9 × 10⁻⁴ [1] = 3.4 × 10³ s (about 57 minutes) [1]. [5] Examiner insight: Substituting 1.2 mm where the radius belongs gives the wrong area and a drift velocity out by a factor of four, so write the radius on its own line first.

7. (a) The current is the same and n is the same, so v ∝ 1/A [1]. A ∝ d², so Y has one ninth of the area of X [1]. The drift velocity in Y is 9 times that in X [1]. (b) For the same I and A, v ∝ 1/n [1]. v(semiconductor) / v(metal) = 8.5 × 10²⁸ / 4.0 × 10²² = 2.1 × 10⁶ [1]. (c) An insulator has almost no free charge carriers, so n is extremely small and no useful current flows even with a large p.d. [1] [6] Examiner insight: Solve comparisons by proportion, stating which quantities stay constant.

8. Cell (or power supply) in series with a switch, a variable resistor and the filament lamp, forming a closed loop [1]. An ammeter in series with the lamp [1]. A voltmeter connected in parallel across the lamp only [1]. Standard symbols and ruled lines [1]. [4]

   +--[cell]--[switch]--[variable resistor]--+
   |                                          |
   |       +-----(V)-----+                    |
   |       |             |                    |
   +--(A)--+----(lamp)---+--------------------+

Examiner insight: Check each meter’s position against its job: ammeter in series, voltmeter across the lamp.

9. Q = It = 0.80 × 180 [1] = 144 C. W = VQ [1] = 9.0 × 144 = 1.3 × 10³ J (1296 J) [1]. [3] Examiner insight: Show the charge as an intermediate result and quote the unit.

10. (a) E.m.f. is the energy transferred to each coulomb of charge by the source, from chemical (or other) energy to electrical [1]. P.d. is the energy transferred from each coulomb of charge, from electrical to other forms, in a component [1]. (b) Q = It = 0.60 × 300 = 180 C [1]. W = EQ = 4.5 × 180 = 810 J [1]. (c) W = VQ = 4.1 × 180 = 738 J [1]. The remaining 72 J is transferred as thermal energy inside the battery itself [1]. [6] Examiner insight: Define both terms using the direction of the energy change. Saying “e.m.f. is the voltage of the battery” does not make the distinction.

11. (a) Electrical energy gained = kinetic energy: eV = ½mv² [1]. v = √(2eV/m) = √(2 × 1.6 × 10⁻¹⁹ × 1200 / 9.11 × 10⁻³¹) [1] = 2.05 × 10⁷ m s⁻¹ [1]. (b) v = √(2 × 1.6 × 10⁻¹⁹ × 1200 / 1.67 × 10⁻²⁷) = 4.8 × 10⁵ m s⁻¹ [1]. The electron is about 43 times faster, the square root of the mass ratio [1]. [5] Examiner insight: A comparison needs both speeds and the ratio. Check the mass used for each particle.

12. (a) KE = qV = 2 × 1.6 × 10⁻¹⁹ × 1500 = 4.8 × 10⁻¹⁶ J [1]. (b) ½mv² = 4.8 × 10⁻¹⁶, so v = √(2 × 4.8 × 10⁻¹⁶ / 6.64 × 10⁻²⁷) [1] = 3.8 × 10⁵ m s⁻¹ [1]. (c) Rate = I / (2e) = 2.0 × 10⁻⁶ / (3.2 × 10⁻¹⁹) [1] = 6.3 × 10¹² s⁻¹ [1]. (d) Power = rate × KE = 6.25 × 10¹² × 4.8 × 10⁻¹⁶ = 3.0 × 10⁻³ W [1]. Check: IV = 2.0 × 10⁻⁶ × 1500 = 3.0 × 10⁻³ W [1]. [7] Examiner insight: Each particle carries 2e, so the number arriving per second is half what it would be for singly charged particles at the same current.

Where marks are usually lost

• Not converting minutes into seconds.
• Using the diameter or radius in place of the area, or forgetting to square the radius.
• Leaving n in cm⁻³ or the area in mm² when other quantities are in SI units.
• Missing one of the two ion types when finding current in an electrolyte.
• Describing drift velocity as the speed of electrons.
• Defining p.d. and e.m.f. without energy per unit charge.
• Using the proton mass for an electron, or e instead of 2e for an alpha particle.

Next steps

• Re-read the revision notes and the study guide.
• Go to the OCR A Level Physics hub and tick off the checklist.
• Take the free 10-minute diagnostics to find gaps.
• Book a free trial class.

Official syllabus

OCR Advanced GCE in Physics A (H556), Version 3.0 (March 2026), for first assessment in 2017, published by OCR. Sections 4.1.1, 4.1.2, 4.2.1 and 4.2.2.