AQA A-Level Physics: Fields and their consequences (7408) – 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 section 3.7 Fields and their consequences of the AQA AS and A-level Physics specification (7407/7408, version 1.4, AS and A-level exams June 2016 onwards): 3.7.1 Fields to 3.7.5 Magnetic fields, including Required practicals 9, 10 and 11. All of 3.7 is A-level only, assessed in Paper 2, and is core content: the options (3.9 to 3.13, one studied for Paper 3 Section B) are separate.

Take G = 6.67 × 10⁻¹¹ N m² kg⁻², ε₀ = 8.85 × 10⁻¹² F m⁻¹, e = 1.60 × 10⁻¹⁹ C, proton mass 1.67 × 10⁻²⁷ kg and g = 9.81 N kg⁻¹ at Earth’s surface.

Questions

1. Define a force field. State one similarity and one difference between gravitational and electrostatic forces. [3]

2. A planet has mass 4.2 × 10²⁴ kg and radius 5.8 × 10⁶ m.

(a) Calculate the gravitational field strength at its surface. [2] (b) Calculate the escape velocity from its surface. [2] (c) Explain why gravitational potential near the planet is negative. [2]

3. For the planet in Question 2, V = −2.80 × 10⁷ J kg⁻¹ at r = 1.0 × 10⁷ m and V = −2.33 × 10⁷ J kg⁻¹ at r = 1.2 × 10⁷ m. Use these values to estimate g at r = 1.1 × 10⁷ m. [2]

4. An 800 kg satellite orbits the planet in Question 2 at radius 2.0 × 10⁷ m.

(a) Show that T² ∝ r³ for a circular orbit. [3] (b) Calculate the orbital period in hours. [2] (c) Calculate the total energy of the satellite. [2]

5. Small spheres X (+3.0 nC) and Y (−5.0 nC) are 4.0 cm apart in air.

(a) Calculate the force between them and state whether it is attractive or repulsive. [2] (b) Calculate the electric field strength at the midpoint of XY and state its direction. [3] (c) Calculate the electric potential at the midpoint. [2]

6. Parallel plates 4.0 cm apart and 6.0 cm long have a p.d. of 600 V. A proton enters midway between them at 1.5 × 10⁶ m s⁻¹, moving parallel to the plates.

(a) Show, using the work done moving a charge between the plates, that E = V/d. [2] (b) Calculate the proton’s sideways displacement as it leaves the plates. [4]

7. A parallel-plate capacitor has square plates of side 0.15 m, 2.0 mm apart. With a dielectric filling the gap, C = 450 pF.

(a) Calculate the relative permittivity of the dielectric. [2] (b) Explain, in terms of polar molecules, why the dielectric increases the capacitance. [3]

8. In Required practical 9, a capacitor discharges through a 47 kΩ resistor. A graph of ln V against t is a straight line of gradient −0.125 s⁻¹.

(a) Calculate the capacitance. [3] (b) Calculate the time for V to halve. [1] (c) Calculate the energy released as V falls from 9.0 V to 3.0 V. [2]

9. A 100 μF capacitor, initially uncharged, is charged from a 6.0 V supply through 22 kΩ. Calculate the charge on it and the current 3.0 s after the switch is closed. [4]

10. In Required practical 10, a horizontal wire passes through the field of a magnet sitting on a top pan balance. 5.0 cm of wire is in the field. A current of 2.5 A increases the reading by 1.53 g.

(a) Explain why the balance reading changes. [2] (b) Calculate the magnetic flux density. [3]

11. Alpha particles (mass 6.64 × 10⁻²⁷ kg, charge +3.2 × 10⁻¹⁹ C) move at 2.0 × 10⁶ m s⁻¹ perpendicular to a uniform 0.50 T magnetic field.

(a) Explain why their path is a circle. [2] (b) Calculate the radius of the path. [2] (c) Show that the time for one orbit does not depend on speed, and explain why this matters in a cyclotron. [2] (d) Calculate the frequency of the alternating p.d. needed for a cyclotron using this field. [1]

12. A generator coil of 150 turns and area 0.012 m² rotates at 25 revolutions per second in a uniform 0.20 T field.

(a) Calculate the maximum flux linkage, and the flux linkage when the normal to the coil is at 40° to the field. [2] (b) Calculate the peak emf and the rms emf. [3] (c) Use Lenz’s law to explain why more work is needed to turn the coil when a current is drawn. [2] (d) 50 kW is sent through cables of total resistance 0.80 Ω. Calculate the power lost in the cables at 400 V and at 10 kV. [3]

Answers

1. A region in which a body experiences a non-contact force [1]. Similarity: both obey an inverse-square law (or both use field lines, potential, equipotentials) [1]. Difference: masses always attract, but charges may attract or repel [1]. [3] Examiner insight: “Both are forces” or “both have fields” is too vague; name a shared property such as the inverse-square law.

2. (a) g = GM/R² = 6.67 × 10⁻¹¹ × 4.2 × 10²⁴ / (5.8 × 10⁶)² [1] = 8.3 N kg⁻¹ [1] (b) v = √(2GM/R) [1] = 9.8 × 10³ m s⁻¹ [1] (c) V is zero at infinity [1]; the force is attractive, so work must be done on a mass to move it to infinity, so V is less than zero everywhere else [1]. Examiner insight: In (c), give both ideas: zero at infinity and an attractive force.

3. g = −ΔV/Δr = −(−2.33 × 10⁷ − (−2.80 × 10⁷)) / (2.0 × 10⁶) [1] = 2.35 ≈ 2.4 N kg⁻¹ (magnitude) [1] [2] Examiner insight: Keep the signs consistent: ΔV is positive moving outwards, so the minus sign gives g pointing inwards.

4. (a) GMm/r² = mv²/r [1]; v = 2πr/T [1]; so T² = (4π²/GM) r³, and 4π²/GM is constant [1]. (b) T = √(4π² × (2.0 × 10⁷)³ / (6.67 × 10⁻¹¹ × 4.2 × 10²⁴)) = 3.36 × 10⁴ s [1] = 9.3 h [1] (c) E = −GMm/2r = −6.67 × 10⁻¹¹ × 4.2 × 10²⁴ × 800 / (2 × 2.0 × 10⁷) [1] = −5.6 × 10⁹ J [1] Examiner insight: On a “show that” in (a), write every step, from the force equation to the constant 4π²/GM; jumping straight to the result shows nothing.

5. (a) F = 8.99 × 10⁹ × 3.0 × 10⁻⁹ × 5.0 × 10⁻⁹ / 0.040² = 8.4 × 10⁻⁵ N [1]; attractive (unlike charges) [1] (b) E from X = 6.74 × 10⁴ V m⁻¹ and from Y = 1.12 × 10⁵ V m⁻¹, both pointing towards Y [1]; add: E = 1.8 × 10⁵ V m⁻¹ [1] towards Y [1] (c) V = 8.99 × 10⁹ × (3.0 × 10⁻⁹ − 5.0 × 10⁻⁹) / 0.020 [1] = −9.0 × 10² V [1] Examiner insight: Field strengths add as vectors, potentials as signed scalars; subtracting the two field strengths here gives a wrong answer.

6. (a) Work done moving Q between the plates: W = Fd = QV [1]; so E = F/Q = V/d [1]. (b) E = 600/0.040 = 1.5 × 10⁴ V m⁻¹ [1]; a = eE/m = 1.44 × 10¹² m s⁻² [1]; t = 0.060 / 1.5 × 10⁶ = 4.0 × 10⁻⁸ s [1]; y = ½at² = 1.1 × 10⁻³ m [1] Examiner insight: In (b), find the time from the horizontal motion only; the field does not change the speed along the plates.

7. (a) εᵣ = Cd / (Aε₀) = 450 × 10⁻¹² × 2.0 × 10⁻³ / (0.0225 × 8.85 × 10⁻¹²) [1] = 4.5 [1] (b) The polar molecules rotate so their positive ends face the negative plate [1]. This produces a field opposing the applied field [1]. For the same charge the p.d. is lower, so C = Q/V is larger [1]. Examiner insight: εᵣ has no unit; writing F m⁻¹ after it is a common unit error.

8. (a) Gradient = −1/RC, so RC = 1/0.125 = 8.0 s [1]; C = 8.0 / 47 000 [1] = 1.7 × 10⁻⁴ F [1] (b) T½ = 0.69 × 8.0 = 5.5 s [1] (c) ΔE = ½C(9.0² − 3.0²) = ½ × 1.7 × 10⁻⁴ × 72 [1] = 6.1 × 10⁻³ J [1] Examiner insight: In (c), ½C × 6.0² is a common slip: energy is not ½C(ΔV)², so find the stored energy at each voltage and subtract.

9. RC = 22 000 × 100 × 10⁻⁶ = 2.2 s; Q₀ = CV = 6.0 × 10⁻⁴ C [1]; Q = Q₀(1 − e^(−3.0/2.2)) [1] = 4.5 × 10⁻⁴ C [1]; I = (6.0/22 000) e^(−3.0/2.2) = 7.0 × 10⁻⁵ A [1] [4] Examiner insight: Current decays during charging too; do not use 1 − e^(−t/RC) for current.

10. (a) The magnet exerts a force on the current-carrying wire [1]; by Newton’s third law the wire exerts an equal and opposite force on the magnet, changing the reading [1]. (b) F = Δm × g = 1.53 × 10⁻³ × 9.81 = 1.50 × 10⁻² N [1]; B = F/Il = 1.50 × 10⁻² / (2.5 × 0.050) [1] = 0.12 T [1] Examiner insight: Convert grams to kilograms before multiplying by g.

11. (a) The force BQv is always perpendicular to the velocity [1], so the speed is constant and the force acts as a centripetal force [1]. (b) r = mv/BQ = 6.64 × 10⁻²⁷ × 2.0 × 10⁶ / (0.50 × 3.2 × 10⁻¹⁹) [1] = 8.3 × 10⁻² m [1] (c) T = 2πr/v = 2πm/BQ, which has no v in it [1]; so a fixed-frequency alternating p.d. stays in step with the particles as they speed up [1]. (d) f = BQ/2πm = 3.8 × 10⁶ Hz [1] Examiner insight: In (c), substitute r = mv/BQ to show v cancels; a bare statement of the result is not a “show”.

12. (a) NΦ max = BAN = 0.20 × 0.012 × 150 = 0.36 Wb-turns [1]; at 40°: 0.36 cos 40° = 0.28 Wb-turns [1] (b) ω = 2π × 25 = 157 rad s⁻¹ [1]; ε₀ = BANω = 56.5 ≈ 57 V [1]; ε_rms = 56.5/√2 = 40 V [1] (c) The induced current produces a force opposing the coil’s rotation [1], so work must be done against it, supplying the electrical energy [1]. (d) At 400 V: I = 125 A, loss = I²R = 12.5 kW [1]; at 10 kV: I = 5.0 A [1], loss = 20 W [1] Examiner insight: In (d), use I²R with the line current; V²/R with 400 V or 10 kV uses the wrong p.d.

Where marks are usually lost

• Using the height above a surface instead of the distance from the centre.
• Explaining negative potential without mentioning the zero at infinity.
• Adding field strengths as scalars when they point in different directions.
• Confusing the time constant with the half-time on a decay graph.
• Using a charging equation for current, which always decays.
• Forgetting to convert grams to newtons with g in the top pan balance practical.

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Official syllabus

AQA AS and A-level Physics specification (7407/7408), version 1.4, July 2026, AS and A-level exams June 2016 onwards, published by AQA. Section 3.7 Fields and their consequences.