# Fucntional Analysis

1. Mar 5, 2006

### Nothing000

Will someone please help me? I am having trouble with a problem. It asks me to find the relationship between the magnitude F of a magnetic force that one wire exerts on another wire and the amount of current flowing in the wires. I am supposed to use the fact that F is proportional to I^2/y. y is the distance between the two wires.

Last edited: Mar 5, 2006
2. Mar 5, 2006

### Nothing000

I am pretty sure that the relationship is F=kI^n. But it then asks me to find n, and I can not figure out how to do that. I have recorded data for the values of F and I for 5 different measurements. I have plotted the data, and I get a parabolic curve. But I can not figure out how to find the value of the exponent in the function of the graph.

3. Mar 5, 2006

### Nothing000

I know how to solve for exponential variables using logarithms, but what values do I put in for F and I?
I know this is super simple, but for some reason I just don't get it.
I am really stuck and any assistance is greatly appreciated.

Last edited: Mar 5, 2006
4. Mar 5, 2006

### Nothing000

Can anyone help?

5. Mar 5, 2006

### vaishakh

What you know and don't know is not much clear. Also How you know the things known by you and what is the difficulty in knowing the needed.
Have you learnt vector cross products and magnetic induction.

6. Mar 5, 2006

### Nothing000

I have read my text book. I am reading the page about functional analysis over and over again. I just don't understand how to find the answer to this question.
And yes, I do know cross products, but we have not covered magnetic induction yet.

I will try to make my question more clear.

If I have a graph that is in the shape of a typical parabolic curve, then I know that the type of equation that would describe the graph would be of the form y = mx^n .
I get that. Now it asks me what is the value of n. My question is how I find the value of n. I know how to find n with logarithms, but I don't know what values to plug in for y and x to make the equation a single vairable one.
Do I just pick any point on the graph and plug those values in for x and y?
I have changed all the variables to common mathematical ones, you can ignor what I wrote earlier. The question I just asked is all that I really need help with. Thank you.

7. Mar 5, 2006

### arunbg

For a wire, the force exerted by a magnetic field is i*L*B (B perp)
Also for an infinite conducting wire ,
magnetic field B=constant*i/a

so for two long conductors carrying same current i,
putting both eqns together,

Force on one wire = constant*i*L*i/a

a is the distance bw wires and l is length

So force per unit length is prop to i^2/a

Hope this helps

8. Mar 5, 2006

### Nothing000

Thanks for the input. But that is not exactly the information I was looking for. I want to know what value I plug into the variable F and I in the equation F = kI^n in order to solve for n. I have graphed the equation with experimental data, so do I just pick any point on the graph and plug in that x and y coordinate (or in this case I and F coordinate since I have graphed I vs F with F as the ordinate and I as the abscissa.)

9. Mar 5, 2006

### topsquark

You mentioned graphing it as a logarithm. That's a good step. So we have:
$$log \, F=n \, log \,(kI)$$

What I don't know is what kind of graph you have. Is this from an experiment that you did? If so I would imagine you have calculated a best-fit for the slope and intercept? Again, if so I would re-do the calculation for the log-log plot equation above. In this case the best-fit slope would be n and the intercept around zero. If you are sketching the best-fit line, then get the approximate slope of that line (but from the line directly, NOT the data points.)

-Dan

10. Mar 5, 2006

### Nothing000

The graph is a parabolic curve not a line. The equation of the parabolic curve is of the form F = kI^n. I need to solve for n. I think all I should do is plug in values for F and I (assume k to be 1) then solve for n.
My question is what values do I plug in for F and I? Do I just pick any point on the graph and plug in the corresponding coordianates for F and I (the graph is F vs I)? That is my question.

11. Mar 5, 2006

### arunbg

just plot F against i^n for different values of n(integer)
see which one gives you a straight line

12. Mar 5, 2006

### Nothing000

Isn't there a way to simply PLUG IN VALUES FOR F AND I into the equation and solve for n? If so, what values do I plug in for F and I? Do I just pick a point on the graph and plug in those coordinates for the values of F and I?

13. Mar 5, 2006

### topsquark

If you have 10 points on your curve, then you have 10 equations of the form:
$$F_x=kI_x^n$$
where x indicates your data points. It is, in general, a monster to solve. Basically we are saying, NO!

However, if you suspect a value for n is true ahead of time, such as n=2, you can plug it into your formula and check to see if they work out okay. As you have been told that your equation is supposed to have n=2, I would do it that way.

In general you would need to do something like the log-log plot if you don't know n.

-Dan

14. Mar 5, 2006

### Integral

Staff Emeritus
If you plot the data in the form given by TopQuark in post number 9 you should get a straight line with a slope corresponding to the n you are looking for. The idea is to create linear function of your data. Topquark has provided that function.

This is use of something called http://www.csun.edu/~vceed002/ref/measurement/data/graphpaper/4-cylce_log-log.pdf" [Broken]. If you were to plot your data on a log log graph you would get your straight line.

http://www.physics.uoguelph.ca/tutorials/GLP/" [Broken] are instructions on how to use it.

Last edited by a moderator: May 2, 2017
15. Mar 5, 2006

### Nothing000

Excellent. I got it. Thanks guys. I am stuck again though. It now asks what is the signifigance of the slope of the graph of the F vs I^2 graph. I have.
Now I know that the slope should be a straight line. Therefore I can simply use the rise over run method of finding the slope. Using units with dummy values of 1, I am finding that $$N/A^2\Rightarrow(kgm/s^2)/(A^2)$$
Shouldn't I end up with $$(kgm/s^2)/(Am)\Rightarrow(N/Am)\Rightarrow(Tesla)$$?

Last edited: Mar 5, 2006