IMPORTANT ANNOUNCEMENTS

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        INDUCTION
Here is the new Capstone version of the manual for Induction. Here is a decent page about Faraday's Law:
https://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html

MOVIES

Please note: If the movie is too small,
please watch via youtube using the YOUTUBE button on bottom right of each video.
And, some videos have much higher quality. Find the HD version using the gear icon.
Induction: Part 1, Set Up 



Induction: Part 2, Data Analysis 



Induction: Part 3, History of the Tube: Dr. Sid Freudenstein, Gymnast for 1968 USA Olympics!


DMM-01- Inputs and Selector for Amprobe 37XR-A Digital Multimeter



DMM-04- Measure Inductance with DMM Amprobe 37XR-A







Begin Q&A Forum for "INDUCTION"

Douglas, I was simply wondering what is going to happen with the induction lab that is suppose to happen today 03-18-03 for lab 2040-005. Thank YouAnonymous
Tue. 18, Mar 2003, 09:49
This is what I sent to the other instructors today.
Thanks for asking.

Due to the snow-day today we will re-arrange the schedule to perform the Induction Lab.

The week of April 7th, we will combine Optics and Diffraction.
We will perform Induction during the week of April 21st.
I will reset the quiz deadlines as such.

If there is a snow-day tomorrow, those classes will be scheduled likewise.

Please let me know if you have any questions.

Thanks,
Douglas
Douglas
Tue. 18, Mar 2003, 12:21

My lab is on April 1st. Since the prelab for induction is closed, should I do e/m? Will the prelab settings stay as they are on the original syllabus?
Thanks,
Marjorie Eliason
Anonymous
Tue. 01, Apr 2003, 07:53
Yes, e/m is what we will be doing next week - April 7, 2003.
I have just reset its deadlines for all the classes.
I am working on the rest of the labs for the semester.
All the labs will shift down the schedule by one week.
:)
Douglas
Tue. 01, Apr 2003, 09:52

For #6 of the post lab, what is the "offset" they are asking about? Also, I usually just approximate the uncertainty, is this okay? I just look at how the data was collected and try to make a guess of how much it could vary due to error. Is there a more precise way I should be doing this?Anonymous
Wed. 25, Jun 2003, 09:04
M.M.,
The offset for induction comes from the fact that we did not calibrate the voltage sensor. (Think for a moment about a bathroom scale that reads 5[lb] when nobody is on it. That is a voltage offset.)

One way to discover the voltage offset is to select a region of the data (maybe 0.2 seconds worth) where the magnet was not near the coil. Find the statistical mean value. That is the offset. Or you could use the run when there was no magnet in the coil; its mean value is the offset.
Does this voltage offset correlate (numerically) with the sum of the two areas measured? (That sum should have been zero. But it probably wasn't. Does the voltage offset affect this?)

Uncertainty does matter. Your instructor should be discussing this with the class every class period.

For each lab it is different. For example: Induction has uncertainty in the voltage measurement as can be seen from the run with no magnet. The mean value is the offset; the chi^2 value is the uncertainty. Even though the voltage probe should be able to measure to within 0.001[V], if the chi^2 value is bigger than 0.001[V] than you have to use the chi^2 value.
Induction also uses time as a measurement. I believe the sample rate for the voltage sensor is 1000 samples per second. Which means each sample has an uncertainty of plus or minus 1/1000 [seconds/sample] or 0.001[s] or 1[ms].

For another example, think back to Resistivity. The meter stick could measure to within 1[mm]. But your eye saw that the wire was not straight. It could have been off by as much as 10[mm].

Often you have two choices for our uncertainty statement: resolution of measurment, or standard deviation from repeated measurements. The larger uncertainty is the one to use.

For another example. E/M has a voltage measurement. The digital readout counts by 2[V]. Since you aren't going to open the device and check that value, all you can assume is that you know the voltage to within 2[V].

So, yes it does matter how you state (or "derive") your uncertainties.
I hope this helps you in beginning to understand how to critically look at the instruments as well as the numerical results.
Douglas
Thu. 26, Jun 2003, 06:25

#4 of induction prelab...
I am not sure about finding the change in the flux for the second peak. I understand that the flux is the area of the peak and the area is supposed to be given by the computer, right? So how am I supposed to calculate this question?
Thanks
Anonymous
Fri. 03, Mar 2006, 21:08
To find (measure) the change in flux for the second peak, you would need to know the area using a stats tool. But the stats tool is not active for the second peak. We could assume from theory that the area is equal and opposite the first peak; but that is assumtion not measurment.Douglas
Thu. 09, Mar 2006, 08:54

Douglas,
I have a question about the the calculation for "Average Area Under Second Peak" on the Excel sheet. Are we supposed to include the value for the South Pole area under the second peak? Or is this value only supposed to be the average area under the second peak for the North Pole runs?
Thanks,
Mike
Burke, Mike
Mon. 17, Mar 2008, 19:10
mjburke17@gmail.com
Good question.
We expect the average to show us a representative value for Change in Flux for a consistent measurement.
Use the areas that are of a consistent nature, all North, or all South...
Douglas
Tue. 18, Mar 2008, 10:29

Douglas,
I was wondering if the post lab website recognized something like 4.45 e-3 as three sig figs or if it should be written as .00445? Thanks, Brandon
Moore, Brandon
Mon. 13, Oct 2008, 11:10
Brandon.Moore@uchsc.edu
4.45 e-3 is not acceptable.
4.45E-3 (no spaces) is acceptable.

0.00445 is also fine.
Douglas
Tue. 14, Oct 2008, 07:52

I would just like a clarification of the reason for different voltages of the first and second peak in one run. I had thought that they were different due to the fact that when the magnet is exactly midway through (between the north and south poles) the voltage is only theoretically 0. Instead it is changing directions opposite when the magnet entered. Therefore the outgoing voltage was more because it was that little bit plus the voltage of the south pole going through.
But now that I have looked around, it seems like the real reason is because the magnet is traveling faster outgoing than incoming...I just want to make sure I'm understanding the right concept.
Thanks
Emerson, Tracy
Mon. 13, Oct 2008, 18:01
tsubasa726@hotmail.com
Yes, the |second peak| is greater because the magnet is moving faster. The faster the magnet is moving the more B changes relative to the coil's static location thus Emf is greater since ΔΦ/Δt is greater.Douglas
Tue. 14, Oct 2008, 07:56

On question #4 of the pre-lab, isn't it supposed to be the same as #3 but opposite in sign? It tells me it is wrong. How else would you calculate it if it is not a "live" data studio graph, and it isn't just the opposite sign?Anonymous
Tue. 10, Mar 2009, 14:31
Okay, are you sitting down for this one?
This is a laboratory, not a lecture class. :)
In lab we take measurements. It is not a good idea in a lab to look at data and suppose the theory IS the measurement.
Douglas
Tue. 10, Mar 2009, 17:31

Hi Douglas,
Do you want us to state the absolute uncertainty for area on #7 on the postlab? I know we discussed relative uncertainties during lab, but relative uncertainty changes for each given value, doesn't it?
If we do need to find absolute uncertainty for area how do you calculate it? By adding the absolute uncertainties for voltage and time, maybe?
Thanks!
Anonymous
Thu. 10, Mar 2011, 22:04
One cannot add absolute uncertainties for Voltage [V] and Time [s] because of the units.

δΔΦ / ΔΦ is the relative uncertainty in ΔΦ
(ΔΦ is the Change in Flux according to the area under the curve of Voltage vs Time.)

To find the absolute uncertainty in ΔΦ, multiply your decimal value for δΔΦ / ΔΦ by your ΔΦ measurement for that run. That would give you δΔΦ [Vs], which is the absolute uncertainty in ΔΦ.

Yes, these values will be slightly different for each fun. We might do better by finding δΔΦ / ΔΦ for each run and then apply an average.
Instead, we found δΔΦ / ΔΦ for one trial.

δΔΦ / ΔΦ = δV / Voltagemean + δtime / timecoil.

Now, we also found the standard deviation of a series of ΔΦ measurements. This value should prove to be numerically similar to the relative uncertainty sum (above.)

You may use either value for the PostLab question.
And you may show the absolute uncertainty or the relative uncertainty (multiply the later by 100% to make it more digestible.)

I prefer percent uncertainties in many situations, including this one since the δΔΦ is such a small number.

I know this was much overload in δ's and Δ's. Let me know, if you need clarification.
Douglas
Fri. 11, Mar 2011, 04:53

For PostLab #7, re measurements, offset, uncertainty:
Was this the same as the total uncertainty for the area, based on summing voltage and time?
If the uncertainty varied per measurement/trial, please explain how I should be calculating this.
(Please limit the symbol verbage; thanks)
Anonymous
Mon. 14, Mar 2011, 22:42
One cannot sum voltage and time because the units of Volt and Second cannot be added.

The uncertainty in the Change in Flux can be found as a relative uncertainty by summing the relative uncertainties for voltage and time.

Take one trial and find the relative uncertainties for voltage and time as follows:
uncertainty in voltage divided by Mean Voltage for one peak
uncertainty in time divided by Time Span for one peak

The relative uncertainty will vary per measurement; the uncertainty itself will likely remain constant. You only need discuss uncertainty for one trial.

Douglas
See graphic below. Here is a pdf.

Douglas
Tue. 15, Mar 2011, 14:54

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