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I just read that. The post where you answered what k value was
So, I would just attach the copper wire (the length being the amount used in the coil) and connect both sides to the ends of a 1.5 v battery, essentially making a circuit and then cutting in between that circuit and then taking an ammeter to measure the current yes?
@gneill said it well in post #30. Use the amount of wire that the current would be flowing thru when in actual use. That is, the Blue colored turns in the photo gneill posted.Another question, I am assuming that the circuit for the experiment would be a series circuit. Doesn't current stay the same throughout the circuit? Or am I wrong here and will have to measure it by using the amount of copper wire used?
@gneill said it well in post #30. Use the amount of wire that the current would be flowing thru when in actual use. That is, the Blue colored turns in the photo gneill posted.
You do not have to cut the wire for this measurement, just make the connections for current measurement across the same wire length.
And Yes, it is a series circuit as in your photo of measuring current. The length of wire replaces the light bulb and resistor, and your small battery with magnets replaces that big 6V battery.
Cheers,
Tom
It seems to me that the only part of the coil that is energized by the battery lies between the magnets at either end of the battery. You might consider improving the efficiency by insulating all but the end magnets from the coil so that the inward magnets aren't shorting out the coil turns that they span. This might involve making the end magnets slightly larger in diameter (maybe wrap them with a conductive sleeve of some kind.
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The length of wire you should consider for finding the resistance that the battery "sees" is the length of the coil comprising the turns that are contributing to the magnetic field. Measuring the resistance of such a short piece of copper wire is probably more trouble than it's worth. Look up the resistivity of gauge of wire you['re using and plug in the length of wire in that many turns of the coil.
Yes!Oh, so the amount of current measured would be in a circuit that had a copper wire as long as the number of coils in between the two magnets?
I just saw your response, and 8 amps for a 1.5v battery? Just double checking, thank you very much.Yes!
Expect around 3 to 8 Amps for a new battery. (that's just a guess )
For a fresh, high quality, brand name, size AA, Alkaline battery; yes, up to 8 Amps short circuit current. Don't carry them in a pocket with loose change or keys, they get HOT. (Those are surprising little things!)I just saw your response, and 8 amps for a 1.5v battery? Just double checking, thank you very much.
I have figured out which equation to use and finding an equation isn't the issue, I have another question about the force calculated.
Since you did not post your calculations we have no way of knowing what you calculated, therefore your question cannot be answered.Although, One major concern I still have is that I'm not sure whether the electromagnetic force would be doubled after the calculation?
That's incorrect.Snip
After that I subbed it into w = f d (d being 0.4m, length of my track) and solved for w.
After this I subbed it into ke = mv^2/2 and I solved for v.
So would I just use kinematics to solve for v2?That's incorrect.
W is the work done by the force over one revolution. That's not equal to the KE of the train.
W is equal to the work done against friction and the _change_ in KE of the train over one revolution. If it moves at constant velocity the change in KE is zero and w is just equal to the work done against friction.
Do you know another method to solve for v then? As far as I know, using kinematic and work equations are the only way to get the velocity.I know that's what you want. However Newtons law F=ma tells you that a force produces an acceleration not a velocity. So you cannot calculate a velocity from a force.
If you assume no friction the net force is non zero so it should keep accelerating indefinitely. Others have already mentioned this.
Ok thank you.Here's a thought, perhaps not practical. See if the projectile can hold itself when trying to move vertically. This should greatly reduce the friction. If not, find the angle at which it will hold itself. Friction will be reduced by the cosine of the elevation angle, and motive force needed to stay still will be proportional to the sine of the elevation, once you account for friction.
For an approximation of the friction, try it with a completely dead battery. Tilt the track up until the projectile just starts to move.The cosine of the elevation angle gives you the coefficient of friction, at least for that orientation.
These are only approximations and the coil will have to be wound carefully so that each turn is in line with the next. Perhaps glue the coil to a ruler to help with this alignment.
Cheers,
Tom