The slopes of these graphs represent what? (Magnetic Fields)

1. Mar 23, 2013

RockThis52

For a magnetic fields lab I am asked to graph the data and from there, use the slope to find a certain value.

For one of them, I am asked to plot the current (x) vs the magnetic field (y).
The slope is supposed to give me a value, I have the slope, no clue what the value would represent. After some research I found that people are saying it represents the loops?

No clue what that is.

Anyway I also have to plot the distance (1/r) (x) vs the magnetic field strength (y) and the slope is supposed to be the permeability of a vacuum. I get 2E-7. Using 4piE-7, that's a deviation of 85%. I do have a feeling something is not right here.

Units for 1/r, are 1/m and units for magnetic field are T, while units for current are A.

Relevant equations:

B=(µ0I)/2πr
B=(µ0NI)/2R

This is no homework question, in fact I don't need a numerical answer. I just need to know if I'm on the right track.

Thanks.

2. Mar 23, 2013

Staff: Mentor

The slope is "magnetic field per current", and this is proportional to physical constants and the number of windings in your loop.
That cannot work, the units (and physics) do not match. Can you post the full problem statement?

3. Mar 23, 2013

RockThis52

Yup.

5. Plot a graph of magnetic field vs. the reciprocal of the distance from the center of the
conductor. Use Logger Pro or another graphing tool. Page 2 of the experiment file is set
up for this graph.
6. Calculate the permeability of the air from the slope of this graph. How does it compare
with the permeability of the vacuum?

4. Mar 23, 2013

Staff: Mentor

"Calculate" is better. You can rearrange your equations for B to have the form B*r = ...

This is equivalent to B/(1/r)= and the slope of your graph. You have to know I (and probably N), too.

5. Mar 23, 2013

RockThis52

Oh, that makes total sense. I do have I, but N why do I need? Can't I use the first equation I posted?

6. Mar 24, 2013

Staff: Mentor

If your coil has multiple windings, you have to take that into account.
If you have a straight wire, things are different.

Both formulas apply to different situations, I don't know which setup you have.

7. Mar 24, 2013

RockThis52

Well my set up was just a straight wire, so I'm assuming I use the first equation since there are no visible windings? Either way I could rearrange in both equations, correct?

8. Mar 25, 2013

Staff: Mentor

This is possible, but you still have to use the correct one - the first one, in this case.