Lab 5 excel sheet

how do you calculate the e/m value in the excel sheet? the equation in the above post doesnt make sense to me..thanks

a =(a_v1+a_v2+a_h1+a_h2)/2/4
delta r=0.002 from the error lab

e/m= slope/constant
slope you get it from the graph, try putting in the number from the graph Y= Slopex+ number
constant is just (4*pi*(10^-7)*130)^2)/(2*(5/4)^3*(a^2))

I still cannot get any of these numbers to turn green!!! e/m, error of slope, error of e/m, error of constant!!! nothing is working!!!

So I tried everything, from the equation that a previous comment to plugging in about 50 different numbers, lets just say I can not get the Slope Error to save my life.

My slope is 63436, pretty huge to me... What can I do? I really need help with this one, because ABSOLUTELY NOTHING IS WORKING!

Does anyone know how to find this value?

someone posted this equation for error in slope:

for error in slope:

(delta(Ir)^2/(Ir)^2) * slope value

but which value of delta(Ir)^2 and which value of (Ir)^2 are you using?

I tried all of them, and tried the average of the max and minimum too with no luck =/

any thoughts, please? thanks

hey guys, were you able to get the 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?


- for the first one, is it an overestimate because it doesn't account for the doubling of a in the denominator?

- for the second one, the correct equation decreases the field by a factor of 1/(5/4)^(3/2)

I can't tell if my logic is correct o.o

thanks for your time

i used the first values off of my excel sheet for both delta(Ir)^2 and (Ir)^2 and multiplied that answer by the value i got for my slope which you get directly off your graph (the # before the "x") ...for me my #s were (.0022/.0121) x 9988 = 1816 --> that is my error in slope..it worked for me so maybe your #s for delta(Ir)^2 and (Ir)^2 aren't correct..check to make sure you did the math right for that before plugging those #s into the error equation.

delta(Ir)^2 was given to you in the lab prep assignment...

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

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