Here is my experiment
Year 12 Physics
Design Investigation
Aim:
The pupose of the following experiment is to select a variable from the equation F=IΔLsinθ and test the effect of the change of the force with regards to the selected variable.
Materials:
1. Wire leads
2. 1 variable rheostat
3. 1 variable power source
4. 1 solenoid
5. 2 retort stands
6. 1 iron bar
7. 1 magnet
8. 1 ammeter
9. 1 electrical balance
Method:
1. The circuit was built as per the above circuit diagram.
2. After the circuit was constructed, an iron bar was hung from the retort stand such that the top of the bar was level with the top of the solenoid. The solenoid was held in the second.
3. The magnet was placed on the electric balance such that the magnet was in the exact middle of the balance.
4. The solenoid was temporarily disconnected from the circuit and the wires were connected to an ammeter as above.
5. The variable power source was set to 4V and was then turned on. The rheostat was adjusted such that the ammeter had a reading of 0.5A
6. The electrical balance was turned on, care was taken to ensure that the electrical balance was zeroed.
7. The power source was turned off and the solenoid was then reconnected to the circuit.
8. The power source was turned on, the figure that was displayed on the electrical balance was recorded.
9. The above procedure was repeated for ½ A steps.
10. After the 5A test the power source was switched to 6V output and the experiment was repeated.
Results:
Voltage Current(A) Force(N*10-3)
4V 0.5 4.59
1.0 6.81
1.5 8.18
2.0 9.23
2.5 9.77
3.0 10.36
3.5 10.77
4.0 11.02
4.5 11.62
5.0 11.76
Voltage Current(A) Force(N*10-3)
6V 0.5 5.07
1.0 8.10
1.5 9.84
2.0 11.46
2.5 12.78
3.0 13.21
3.5 13.48
4.0 14.56
4.5 15.58
5.0 15.99
My Experiment
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deleted-71552
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Hello, Question Man!Question Man wrote:come on guys answer this thread please!
I can't see the circuits you attemped to post to this thread. Also, going back and forth between this thread and the other you posted to has a me a bit confused because I'm not sure you are talking about exactly the same project - the formulas you posted are not identical, for example.
I'm flying a bit blind without being able to see the circuits. Can I safely assume that all the voltages and currents you measured are DC? When you adjusted the rheostat for the current you desired, did you make note of the voltage on the coil or rheostat?
Why did you disconnect the solenoid from the circuit when setting the current? Why not just leave the solenoid in the circuit with the ammeter in place? I can't see the circuit, but it seems to me that the resistance of the solenoid is probably quite a bit greater than the resistance of the ammeter - the ammeter should be very near zero.
I can't tell, but it could be that the current through the solenoid isn't what you have assumed it to be.
Can you post your circuit to the web somewhere that I can see it, or "redraw" it using text and posting it here?
You can use the Code text type to get monospace fonts for drawing things, like this:
Code: Select all
+---------RESIST--------+
| |
| |
b + R
a E
t S
t - I
| S
| T
| |
| |
+-----------------------+
Good luck!
Brian Castelli (OneBriiguy)
Engineering Specialist
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Question Man
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yes they are all DC.I'm flying a bit blind without being able to see the circuits. Can I safely assume that all the voltages and currents you measured are DC?
thats what i asked my science teacher, he said resistance in the solenoid was negligible, which would indicate that if the current was 1/2A (as measured with an ammeter which was put in place after the rheostat), the current in the solenoid should be also 1/2A.Why did you disconnect the solenoid from the circuit when setting the current? Why not just leave the solenoid in the circuit with the ammeter in place?
i am aware of that however, it sort of hit me that if for the same current with a different voltage produces a different force on a particle in a magnetic field it would bring into question the equation F=ILsin(theta) as well as Ohm's Law (V=IR).the formulas you posted are not identical, for example.
the circuit was sort of built like this:
+terminal of variable power source -->rheostat-->solenoid-->- terminal of variable power source
i'll post a picture here after the weekend when the sevrer that my work is on is reconnected to our network.
A MAN WHO NEEDS HELP!
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deleted-71552
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He's right, of course, that compared to the other elements in the circuit the DC resistance will be negligible. But the resistance may not be negligible when compared to that of the ammeter.Question Man wrote: thats what i asked my science teacher, he said resistance in the solenoid was negligible, which would indicate that if the current was 1/2A (as measured with an ammeter which was put in place after the rheostat), the current in the solenoid should be also 1/2A.
You have performed an experiment that yielded unexpected results. To isolate potential variations in your method, you need to take a look at what changed between the two experiments. Since you removed the ammeter in each case, you really don't *know* what the current was. I suggest that you repeat the experiment with the circuit connected as follows:
+terminal of variable power source-->ammeter-->rheostat-->solenoid-->- terminal of variable power source
Then you will know that the current through the coil is the same. Switching from the ammeter to the solenoid means you're blind during the test. You might see some fluctuation in the current during the test, too.
If you repeat the test and find that the results are the same with the ammeter in the circuit, then it is time to try to isolate some other cause of variation.
Note that if you can't repeat the experiment, you might be able to get by with figuring out how to measure the resistance of the meter and the coil and showing how there difference affected your results.
Brian Castelli (OneBriiguy)
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deleted-71447
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Just wanted to add this link, which might facilitate discussion:
http://hyperphysics.phy-astr.gsu.edu/hb ... r2.html#c1
http://hyperphysics.phy-astr.gsu.edu/hb ... r2.html#c1

