Lab 5 excel sheet

does anyone know how to find the error of e/m and error of constant? Also the error of slope doesnt work for me =/

nvm i got error of constant, but i still cant get error of slope and i tried everything =/

you can't find this error of e/m unless you have the error of the slope first. I can't seem to get the value of my error for the slope either

hi guys

wat r the units for (Ir)^2 and Delta (Ir)^2
thanx

Y= Slopex+ number
when i plug the value bolded i don't get it right?
any help

my question is how do u find Δa so i can use in this equation
Error in the constant = 2*(Δa/a)*constant

I cant get delta IR^2!!!!!!!! I have tried everything, the equation in the literature, the equation on the board, I tried it calculating altogther and in pieces and in the excel sheet itself and NOTHING HAS WORKED!!

All the values I am using to calculate are correct according to the excel. This is such crap

alright for the slope first adjust all the voltage values so that the line is straight meaning the line of the slope runs thru all the points perfectly and then start with 300 and go up by intervals of 20 til you hit the right number should b between 300-700

fot the equation I have
Y= 1E+06x + ##

so shouldn't my slope be 1E+06

but i am getting this wrong

heeeeeeelp plz

still not getting the error in slope.can somebody please help

can someone please post how to find the error in slope I cant find it either by the method described!!!

Hey does anyone know how to calculate Delta a? i'm confused on the average of delta a and the error of a single delta a. if someone could just plug in numbers and show me how to do it that would be great!

Please someone help with the error in slope?? I have tried everyway I can think of its not working!!!!

Okay, the Equation to get Delta Ir^2 has already been posted. I used it and it works perfectly, so once again its:

Delta Ir^2 = sqrt((deltar/r)^2) * (2*(Ir)^2)

The Answer to find a is simply half of what your diameter measurements were (a is just the radius). I guess using the really long random equation would get you at the same answer but I figure why kill myself?

a = measurement of diameter/2 (ie: .403/2 = .2015) in meters of course


For right now, that's all I could come up with...

DOES ANYBODY KNOW WHAT THE SI UNITS ARE FOR (Ir)^2 and Delta(Ir)^2

units for (IR)^2 are (am)^2 duhh

(am)^2 for delta (IR)^2 you pitiful soul

anyone get tge questions for execution sheet 1?

Look at the sketch of the magnetic field lines due to a single current loop. (See Ch18 sheet 26)
Use that sketch and the sketch below to help explain whether simply doubling the magnetic field of one coil is an under estimate or an over estimate of the magnetic field of interest.


Notice that the correct equation (2) and the incorrect equation differ only by a correction factor of 1/(5/4)^(3/2). When this correction factor is applied to equation does it increase or decrease the field? Does this agree with your answer above?

I'm having a lot of trouble with finding the error of e/m. I've seen the appropriate equation and I've been using it, but its not working for me. The reason i think its not working is cuz i randomly chose my error of the slope as 360...

can some1 explain the proper way to obtain the error of the slope. i tried using the equation given but the person did not specify which values of Ir^2 or DeltaIr^2 to use.

Thanks