| Blank data sheet here | |
| current I | B (mT) |
| 0.5 | |
| 1.0 | |
| 1.5 | |
| 2.0 | |
| length of coil | |
| number of turns N | |
| n = N/L | |
| Note the rounding to the 0.001 place in the I column; most of your recorded uncertainties were rounded to that place. For example you could have recorded 0.1320 +/1 (plus or minus) 0.0011453 for the current. | |
| Recall the graph paper has a least count of 0.002 mT. Thus, the above numbers can easily be plotted. If the last digit is even, the point lies exactly on a small division. If the last digit is odd, the point lies at the midpoint between two divisions. | |
| mbest
=
|
|
| mmax =
|
|
| Thus mmin
=
|
|
| (mmax
- mmin ) =
|
|
| Compute LCy/(delta
I) =
|
|
| Choose delta m as the
largest of the previous two: delta m =
|
|
| Now the slope should be equal to uo*n, where uo is the magnetic permeability of free space and n is the number of turns per meter of the slinky. Note n = N/L, where N is the number of loops and L is the length. | |
| Compare your
results with the actual, accepted measured uo*nacc
.Compute the percent difference which is the absolute value of the
difference divided by the average of the two and multiplied
by 100 to achieve a percentage.
|
|
| Also check if the
accepted value falls within the range in a way defined by
this inequality: | uo*nbest - uo*nacc | < delta m + (uo*N/L2)*delta L, where delta L = 0.003 m. Watch your sig. figs. !
|
|