Purpose
To understand the magnetic field generated by a long, straight, current-carrying conductor, the behavior of a conductor carrying a constant current in a magnetic field, and to verify the relationship associated with the force between two current-carrying conductors and the current in the wires.
Hypothesis
According to the theory that F = mg = μoLI2 / 2πdo, as the weight on the upper wire the increased, the current needed to return it to its original position will also increase as long as the length of the upper conductor and center-to-separation between the wires are kept constant. In addition, with the theory that d = rs / 2D and that do = d + rupper + rlower, iff the distance between the mirror and whiteboard were to increase, then less current would be needed to return the upper wire to its original position when a weight force is applied to it. Furthermore, the larger the value of so, the more current will be needed to for the upper wire to reach its original position when a weight force is applied to it. Finally, because the wires run in opposite directions, they will experience a force of repulsion when a current is run through them.
Labeled Diagrams
See attached sheet.
Data
First Separation
| Radius of lower conductor, rlower | 0.00164 m |
| Radius of upper conductor, rupper | 0.00179 m |
| Length of upper conductor, L | 0.272 m |
| Distance from mirror to end of upper conductor, r | 0.208 m |
| Distance to whiteboard, D | 2.21 m |
| s = s1 – s0 | 0.0120 m |
| Surface-to-surface separation of conductor, d | 0.000565 m |
| Center-to-center separation of conductors, d0 | 0.00908 m |
| Mass in the pan, m (kg) | Weight in the pan (N) | Current I (A) |
| 20.0 x 10-6 | 196 x 10-6 | 3.50 |
| 40.0 x 10-6 | 392 x 10-6 | 5.50 |
| 50.0 x 10-6 | 490 x 10-6 | 6.00 |
| 70.0 x 10-6 | 686 x 10-6 | 7.10 |
| 90.0 x 10-6 | 882 x 10-6 | 8.50 |
| 100 x 10-6 | 980 x 10-6 | 8.80 |
Second Separation
| Radius of lower conductor, rlower | 0.00164 m |
| Radius of upper conductor, rupper | 0.00179 m |
| Length of upper conductor, L | 0.272 m |
| Distance from mirror to end of upper conductor, r | 0.208 m |
| Distance to whiteboard, D | 2.21 m |
| s = s1 – s0 | 0.0470 m |
| Surface-to-surface separation of conductor, d | 0.00221 m |
| Center-to-center separation of conductors, d0 | 0.00564 m |
| Mass in the pan, m (kg) | Weight in the pan (N) | Current I (A) |
| 20.0 x 10-6 | 196 x 10-6 | 4.70 |
| 40.0 x 10-6 | 392 x 10-6 | 6.70 |
| 50.0 x 10-6 | 490 x 10-6 | 7.40 |
| 70.0 x 10-6 | 686 x 10-6 | 8.70 |
| 90.0 x 10-6 | 882 x 10-6 | 10.50 |
| 100 x 10-6 | 980 x 10-6 | 11.10 |
Graphs
Part 1
Part 2
Questions
Part 1
1. Plot a graph in Graphical Analysis of F vs. I (F is the weight force of the mass in the pan). What type of graph is it? What appears to be the relationships between force and current? Does your graph verify what is expected from theory?
It is a quadratic graph. The relationship between force and current appears to be that F is proportional to I2 multiplied by some constants (μoL / 2πdo). Yes, the graph verifies what is expected from theory.
2. Linearize your graph. To do this, you will need to perform some operation on the x-axis variable. Include both your graph of F vs. I and your linearized graph in your lab report.
See graphs section above.
3. What is the slope of your linearized graph?
For part one, the slope of the linearized graph is 1.19 x 10-5 N/I2. For part two, the slope of the linearized graph is 7.65 x 10-6 N/I2.
4. How does the slope of your linearized graph compare to μoL / 2πdo? What is the percent difference?
For part one, μoL / 2πdo = 5.99 x 10-6 Tm / A, so the percent difference is 66.1%. For part two, μoL / 2πdo = 9.65 x 10-6 Tm / A, so the percent difference is 23.1%.
5. Assuming your linearized graph is a fairly good straight line and the slope is about right (see previous question), what physical relationships have you confirmed in this experiment?
The physical relationships that should have been confirmed are that F = mg = μoLI2 / 2πdo.
Conclusion
To begin the experiment, a current was connected to two parallel wires in a manner so that it would run in opposite directions. No current was actually run through the wires at this point. The length of the upper wire was measured, along with the radius of both the upper and lower wires, the distance from the mirror to the end of the upper conductor, and the distance from the mirror to the whiteboard. A He/Ne laser was turned on and reflected off the mirror onto a whiteboard, making sure that the upper and lower wires were touching. The initial position of the laser was marked on the board, and then the wires were allowed to repel to a distance of about 2 mm apart, using a counterweight to achieve this separation. The new position of the laser on the whiteboard was marked, and the distance between the two noted points was measured and recorded.
Masses were then added in increments of about 20 mg to the pan on the upper wire. At each weight, current was run through the wires to return the laser to its secondary position. After reaching the secondary position, the current was recorded and the next trial was performed, until six trials were completed. This whole process was repeated after using the counterweight to achieve an initial separation between the wires of about 4 to 5 mm.
Talk about results calculation number from questions in comparison to slope (percent difference).
Sources of error (elaborate on these a little bit and add any more you can think of): Inaccurate so measurement (dots marked were not extremely precise). The laser was pointed at an angle at the mirror, not head on. This would cause the calculation of d to be slightly thrown off. Hard to match laser perfection with marking because fluctuation in reaching equilibrium. Any other inaccuracies in measurements. Overall the perfect differences are probably not too bad considering all the possible sources of error.
Equations
B = μoI’ / 2πdo
F = ILB = IL μoI’ / 2πdo = μoLI2 / 2πdo
F = mg = μoLI2 / 2πdo
d = rs / 2D
do = d + rupper + rlower
Percent Difference = |x1 – x2| / (x1 + x2)/2 x 100%
