AQA GCSE Physics Physics Paper 2 (Higher), November 2021: Question 9

13 marks · High Demand difficulty · Extended Answer

Explain Fleming's left-hand rule for a current-carrying copper rod in a magnetic field, suggest changes to increase the force, and calculate the maximum velocity of the accelerated rod.

Practise this question

Question

A three-part physics question about a copper rod accelerating on brass rails in a magnetic field. Figure 16 shows a horseshoe magnet with a copper rod resting on two parallel brass rails connected in a circuit with a DC power supply, a switch, a variable resistor, and an ammeter. Part 09.1 asks to explain how Fleming's left-hand rule predicts the motion (5 marks). Part 09.2 asks for two changes to increase the force on the rod (2 marks). Part 09.3 gives numerical values for time, current, mass, length, and magnetic flux density, and asks to calculate the maximum velocity (6 marks).
Question text

09 A teacher demonstrated how a magnetic field can cause a copper rod to accelerate.

The teacher placed the copper rod on two brass rails in a magnetic field.

The copper rod was able to move.

Figure 16 shows the equipment used.

Figure 16

09.1 The teacher closes the switch and the copper rod accelerates.

Explain how Fleming’s left hand rule can be used to predict the direction in which the

copper rod will move.

[5 marks]

*0349*2 Suggest two changes to the equipment that would increase the force on the

.

copper rod.

[2 marks]

09.3 The teacher closed the switch and the copper rod accelerated uniformly from rest

for 0.15 s.

The current in the copper rod was 1.7 A.

mass of copper rod = 4.0 g

length of copper rod in the magnetic field = 0.050 m

magnetic flux density = 0.30 T

Calculate the maximum possible velocity of the copper rod when it left the

magnetic field.

[6 marks]

Maximum velocity = m/s

Mark scheme

Show the mark scheme The mark scheme for Question 9 provides detailed point-by-point allocation for all three sub-questions, totaling 13 marks. For 09.1, it lists applying the left-hand rule with specific finger orientations and directions. For 09.2, it accepts decreasing resistance or using a stronger magnet. For 09.3, it outlines step-by-step calculations involving magnetic force, Newton's second law, and kinematics.

Question 9

AO /

Question Answers Extra information Mark

Spec. Ref.

09.1 hold thumb first finger and allow first two fingers/index and 1 AO1

second finger (of left hand) at middle for first and second

right angles to each other finger throughout

second finger represents the 1 AO1

current pointing out of the paper

first finger represents the field 1 AO3

pointing downwards

thumb points in the direction of 1 AO3

the force / thrust / acceleration

(therefore) the rod moves left to allow correct description (eg 1 AO3

right away from the magnet)

dependent on scoring marking

point 3 or 4 4.7.2.2

09.2 decrease the resistance of the allow increase the current/pd 1 AO3

variable resistor 4.7.2.2

use a stronger magnet allow use a magnet with a 1

greater flux density

09.3 F = 0.30 × 1.7 × 0.050 1 AO2

4.5.6.2.2

F = 0.0255 (N) 1 4.5.6.1.5

4.7.2.2

m = 0.004(0 kg) 1

0.0255 = 0.0040 × a this mark may be awarded if m 1

is incorrectly / not converted

and / or F is incorrectly

calculated

a = 0.0255 / 0.0040 this mark may be awarded if m 1

or is incorrectly / not converted

a = 6.375 and / or F is incorrectly 21

calculated

Δv = 6.375 × 0.15 = 0.95625 allow a correct calculation using 1

(m/s) an incorrectly / not converted m

and / or an incorrectly

calculated F

allow 0.96 or 0.956 (m/s)

alternative method

F = 0.30 × 1.7 × 0.050 (1)

F = 0.0255 (N) (1)

m = 0.004(0 kg) (1)

this mark may be awarded if m

0.0255 = 0.0040 × Δv (1) is incorrectly / not converted

0.15

and / or F is incorrectly

calculated

Δv = 0.0255 × 0.15 (1) this mark may be awarded if m

0.0040 is incorrectly / not converted

and / or F is incorrectly

calculated

Δv = 0.95625 (m/s) (1) allow a correct calculation using

an incorrectly / not converted m

and / or an incorrectly

calculated F

allow 0.96 or 0.956 (m/s)

Total 13

How to answer it

The Motor Effect and Magnetic Forces

What this question tests

This question tests your understanding of the magnetic effect of a current (the motor effect), your ability to apply Fleming's Left-Hand Rule to predict motion, and your skill in combining multiple physics equations to solve multi-step calculations involving force, mass, acceleration, and velocity.

Question 09.1

Explaining Fleming's Left-Hand Rule [5 marks]

✅ Model Answer

  • Hold the thumb, first finger, and second finger of your left hand at right angles to each other.
  • Point the second finger representing current in the direction of the current (out of the paper).
  • Point the first finger representing the magnetic field from North to South (downwards).
  • The thumb will then point in the direction of the force / thrust / acceleration.
  • Therefore, the copper rod moves from left to right.

💡 Key Knowledge

  • First finger = Field (N to S).
  • seCond finger = Current (+ to -).
  • ThumB = MoBion / Force.
  • Always state clearly that it is the left hand being used.

🧠 Exam Technique

To secure all 5 marks, you must methodically link every finger of the left hand to its specific orientation based on Figure 16, ending with a definitive statement about the direction of motion.

❌ Common Errors

  • Using the right hand instead of the left hand.
  • Confusing the field direction with the current direction.
  • Omitting the final conclusion of which way the rod actually moves.
Mark breakdown: 1 mark for hand configuration, 1 mark for second finger (current), 1 mark for first finger (field), 1 mark for thumb (force), 1 mark for concluding the direction of motion (dependent on scoring field or current marks).
Question 09.2

Increasing the Magnetic Force [2 marks]

✅ Correct Answers (Any two)

  • Decrease the resistance of the variable resistor (which increases the current).
  • Use a stronger magnet / a magnet with a greater magnetic flux density.
  • Increase the potential difference of the power supply.

💡 Key Knowledge

The magnitude of the force on a conductor in a magnetic field is given by the equation:
F = B × I × l

To increase force F , you must increase magnetic flux density ( B ), current ( I ), or length ( l ).

Mark breakdown: 1 mark for each valid suggestion (max 2 marks).
Question 09.3

Multi-Step Calculation of Velocity [6 marks]

📐 Step-by-Step Calculation

  1. Calculate the magnetic force (F):
    F = B × I × l
    F = 0.30 T × 1.7 A × 0.050 m = 0.0255 N
  2. Convert mass into kilograms (kg):
    m = 4.0 g = 0.0040 kg
  3. Calculate acceleration (a):
    F = m × a ⇒ a = F / m
    a = 0.0255 / 0.0040 = 6.375 m/s²
  4. Calculate maximum velocity (v):
    v = a × t
    v = 6.375 × 0.15 = 0.95625 m/s
    (Accept rounding to 0.96 or 0.956 m/s)

❌ Common Calculation Traps

  • Unit conversion trap: Forgetting to convert grams into kilograms ( 4.0 g = 0.004 kg ). If left as 4.0, acceleration and final velocity will be out by a factor of 1000!
  • Significant figures: Keep full calculator display values through intermediate steps and round only at the final answer stage.
Mark breakdown: 1 mark for calculating F (0.0255), 1 mark for converting mass to 0.0040 kg, 1 mark for substituting into F = ma, 1 mark for calculating acceleration (6.375), 1 mark for calculating velocity change, and 1 mark for final correct unit/value (0.956 m/s).

Topics

Physics · P7: Magnetism and Electromagnetism · P5: Forces

Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 2 (Higher), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.