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The teacher closes the switch and the copper rod accelerates - AQA - GCSE Physics - Question 9 - 2021 - Paper 1

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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 wil... show full transcript

Worked Solution & Example Answer:The teacher closes the switch and the copper rod accelerates - AQA - GCSE Physics - Question 9 - 2021 - Paper 1

Step 1

Explain how Fleming's left hand rule can be used to predict the direction in which the copper rod will move.

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Answer

Fleming's left hand rule provides a method to determine the direction of force acting on a current-carrying conductor within a magnetic field. To apply this rule, one holds the left hand such that:

  1. The thumb, first finger, and second finger are all at right angles to each other.
  2. The first finger points in the direction of the magnetic field (from North to South).
  3. The second finger indicates the direction of conventional current (from positive to negative).
  4. The thumb then points in the direction of the force experienced by the conductor, which indicates the motion of the copper rod. Thus, if the current flows in a specific direction, the rod will move perpendicularly in the direction indicated by the thumb.

Step 2

Suggest two changes to the equipment that would increase the force on the copper rod.

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Answer

  1. Decrease the resistance of the variable resistor, allowing a higher current to flow through the copper rod.
  2. Use a stronger magnet or increase the magnetic flux density, which would enhance the force acting on the copper rod.

Step 3

Calculate the maximum possible velocity of the copper rod when it left the magnetic field.

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Answer

To find the maximum velocity, we first need to calculate the force acting on the copper rod using the formula: F=BILF = BIL Where:

  • ( B = 0.30 , T ) (magnetic flux density)
  • ( I = 1.7 , A ) (current)
  • ( L = 0.050 , m ) (length of rod in the magnetic field)

Calculating the force: F=0.30imes1.7imes0.050=0.0255NF = 0.30 imes 1.7 imes 0.050 = 0.0255 \, N

Next, we convert the mass of the copper rod from grams to kilograms: ( m = 4.0 , g = 0.004 , kg )

Using Newton's second law to find acceleration: F=maa=Fm=0.02550.004=6.375m/s2F = ma \Rightarrow a = \frac{F}{m} = \frac{0.0255}{0.004} = 6.375 \, m/s^2

To find the velocity after 0.15 seconds of acceleration: Δv=at=6.375×0.15=0.95625m/s\Delta v = a \cdot t = 6.375 \times 0.15 = 0.95625 \, m/s

Thus, the maximum possible velocity of the copper rod when it left the magnetic field is approximately:

Maximum velocity = 0.95625 m/s

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