AQA A-Level Physics Paper 2, June 2025: Question 6

📐 Step-by-Step Calculation

1. Calculate required centripetal acceleration (a):
   v = 74.2 km s⁻¹ = 74.2 × 10³ m s⁻¹
   a = v² / r = (74.2 × 10³)² / (4.22 × 10⁸) = 13.0 m s⁻²
2. Calculate Jupiter's gravitational field strength (g):
   g = GM / r² = (6.67 × 10⁻¹¹ × 1.90 × 10²⁷) / (4.22 × 10⁸)²
   g = 0.712 m s⁻²
3. Compare and conclude:
   Since g = 0.71 m s⁻² ≠ 13 m s⁻² (gravity is far too weak), the suggestion is not correct.

🧠 Alternative Approaches

You can also solve this by:

• Calculating the expected orbital speed if gravity alone provided the centripetal force: v = √(GM / r) = 17.3 km s⁻¹ , which is much lower than the actual speed of 74.2 km s⁻¹ .
• Calculating required centripetal force ( mv²/r ) and gravitational force ( GMm/r² ) assuming an arbitrary mass m .

✅ Mark Scheme Award

• Mark 1: Evaluation of centripetal acceleration 13.0 m s⁻² (or equivalent method).
• Mark 2: Evaluation of gravitational acceleration 0.712 m s⁻² (or equivalent method).
• Mark 3: Clear deduction showing 0.71 ≠ 13 and concluding that the suggestion is incorrect.

❌ Common Traps

• Prefix error: Forgetting to convert 74.2 km s⁻¹ to 74.2 × 10³ m s⁻¹ leads to a power-of-ten (POT) penalty.
• Missing conclusion: Simply calculating the numbers without explicitly stating whether the student's suggestion is correct or incorrect.

💡 Physical Reasoning

Use Fleming's Left-Hand Rule (FLHR) for a positive charge:

• Thumb (Force, F): Must point towards the centre of the circular path (towards Jupiter, horizontally to the left).
• Second Finger (Current, I / velocity of positive charge): Points along velocity vector v (towards the viewer / tangentially forward-right).
• First Finger (Field, B): Must point vertically down the page.

✅ Required Diagram Response

Description: An arrow starting at the position of the ion (the dot on the right side of the torus), pointing vertically downwards towards the bottom of the page.

📐 Step-by-Step Calculation

1. Equate magnetic force to centripetal force:
   BQv = mv² / r ⟹ B = mv / (Qr)
2. Substitute given values:
   m = 5.31 × 10⁻²⁶ kg
   v = 74.2 × 10³ m s⁻¹
   Q = 3.20 × 10⁻¹⁹ C
   r = 4.22 × 10⁸ m
   B = (5.31 × 10⁻²⁶ × 74.2 × 10³) / (3.20 × 10⁻¹⁹ × 4.22 × 10⁸)
   B = 2.92 × 10⁻¹¹ T ≈ 2.9 × 10⁻¹¹ T
3. Derive base units for B:
   From F = B I L ⟹ B = F / (I L)
   Units of force F = kg m s⁻²
   Units of I = A , units of L = m
   Unit of B = (kg m s⁻²) / (A × m) = kg A⁻¹ s⁻²

✅ Mark Scheme Award

• Mark 1: Correct formula setup: BQv = mv²/r (or BQv = ma using a = 13.0 from 06.1).
• Mark 2: Value of 2.9 × 10⁻¹¹ (condoning 1 or 2 sig figs: 2.9 × 10⁻¹¹ or 2.92 × 10⁻¹¹).
• Mark 3: Correct fundamental base unit: kg A⁻¹ s⁻² .

❌ Common Errors

• Writing Tesla (T): The question specifically asked for fundamental (base) units. Writing "T" gets 0 for the unit mark!
• Confusion with elementary charge e: Using e = 1.60 × 10⁻¹⁹ C instead of the explicitly stated charge Q = 3.20 × 10⁻¹⁹ C .

📐 Step-by-Step Calculation

1. Relate electric force to Newton's Second Law:
   F = QE and F = ma
   ma = QE ⟹ a = QE / m
2. Convert prefixes and substitute values:
   E = 371 μV m⁻¹ = 371 × 10⁻⁶ V m⁻¹
   Q = 3.20 × 10⁻¹⁹ C
   m = 5.31 × 10⁻²⁶ kg
   a = (3.20 × 10⁻¹⁹ × 371 × 10⁻⁶) / (5.31 × 10⁻²⁶)
3. Calculate final acceleration:
   a = 2.236 × 10³ m s⁻² ≈ 2.24 × 10³ m s⁻²

🧠 Examiner Insights & Guidance

• Mark 1: Use of F = ma and F = QE .
• Mark 2: Correct substitution with proper values.
• Mark 3: Final answer 2.24 × 10³ m s⁻² (or 2200 m s⁻² ).
• Condoned errors: Examiners allowed partial credit if you accidentally used 1.60 × 10⁻¹⁹ C instead of 3.20 × 10⁻¹⁹ C , but top answers preserved the stated ion properties.

📐 Step-by-Step Calculation

1. Determine charge values:
   Q_X = +2e = 2 × (1.60 × 10⁻¹⁹) = 3.20 × 10⁻¹⁹ C
   Q_Y = +1e = 1.60 × 10⁻¹⁹ C
   r = 2.0 cm = 0.020 m
2. Calculate component field strengths:
   Field from point charge: E = Q / (4πε₀r²)
   1 / (4πε₀) ≈ 8.99 × 10⁹ N m² C⁻²
   
   Due to X (directed rightwards):
   E_X = (8.99 × 10⁹ × 3.20 × 10⁻¹⁹) / (0.020)² = 7.19 × 10⁻⁶ V m⁻¹
   
   Due to Y (directed upwards):
   E_Y = (8.99 × 10⁹ × 1.60 × 10⁻¹⁹) / (0.020)² = 3.60 × 10⁻⁶ V m⁻¹
3. Combine perpendicular vectors using Pythagoras:
   E_total = √(E_X² + E_Y²)
   E_total = √[(7.19 × 10⁻⁶)² + (3.60 × 10⁻⁶)²]
   E_total = 8.04 × 10⁻⁶ V m⁻¹ ≈ 8.0 × 10⁻⁶ V m⁻¹ (or 8.1 × 10⁻⁶ V m⁻¹ )

❌ Severe Pitfall: Distance vs Field Vectors

Never combine distances with Pythagoras to find total field!

Electric field strength is a vector. You cannot find the distance between X and Y and plug that into Coulomb's law. You must:

• Find the individual vector field E_X at P.
• Find the individual vector field E_Y at P.
• Add the two perpendicular fields vectorially using Pythagoras: E_resultant = √(E_X² + E_Y²) .

✅ Mark Scheme Breakdown (Any 2 of the first 3 + Final Answer)

• Correct charges identified: 3.2 × 10⁻¹⁹ C and 1.6 × 10⁻¹⁹ C .
• Use of E = Q / (4πε₀r²) for either charge.
• Pythagoras used on the two field strengths: E_total = √(E_X² + E_Y²) .
• Final answer: 8.0 × 10⁻⁶ V m⁻¹ or 8.1 × 10⁻⁶ V m⁻¹.