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Question 3
3 (a) A student has a bar magnet, a piece of iron the same size as the magnet, and some paper clips. Describe how the student could use these items to demonstrate t... show full transcript
Step 1
Answer
To demonstrate temporary induced magnetism, the student can perform the following steps:
Interaction with the Magnet: Place the piece of iron close to or in contact with the bar magnet. The iron will become magnetized due to its proximity to the magnetic field generated by the bar magnet.
Attraction of Paper Clips: Use the magnet to pick up one of the paper clips. This shows that the magnet can attract ferromagnetic materials. After contacting the magnet, the iron can also pick up additional paper clips.
Temporary Effect Observation: Remove the magnet and observe that the iron no longer attracts the paper clips. This demonstrates that the magnetism is temporary, as the iron loses its magnetization when the external magnetic field is removed.
Step 2
Answer
To investigate the effect of the current size in the solenoid on the force of attraction, the student should follow these steps:
Set Up the Apparatus: Connect the solenoid to a power source and place the iron nail close to the solenoid without touching it.
Initial Measurement: Measure the initial position of the iron nail when the current is off. This creates a baseline for measurements.
Current Variation: Turn on the current and gradually increase it while noting the current values with an ammeter. Measure how far the nail moves towards the solenoid with each increment of current.
Force Calculation: Use the force of attraction formula to compute the force based on measurements taken at each current level, relating extension of the nail's movement to the varying current sizes.
Repeat for Accuracy: Repeat the experiment with different current levels to obtain reliable data, ensuring consistency in measuring the movement of the iron nail.
Step 3
Answer
To calculate the energy transferred in extending the spring, we will use the following equation:
Where:
Substituting these values into the equation:
Calculating this gives:
Rounding off, the energy transferred is approximately 0.17 J.
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