Physics lab 3b

Lab Report 3b Methodology 1. Why can you use current through a solenoid as an equivalent variable for the magnetic field strength of the solenoid? You can use current through a solenoid as an equivalent variable for the magnetic field because they are related through the equation B=μnI, Ampere’s Law where magnetic permeability (μ) and the number of turns per length (n) will be constant in the experiment. Therefore, because the other variables remain unchanged, the current would be proportional to the magnetic field strength and would act as an equivalent variable that can be used alternatively as we cannot measure the magnetic field directly using the equipment in our lab. Data and analysis - Table(s) of all measured & calculated data used in your analysis, with uncertainties. Show sample calculations for your calculated values and calculated uncertainties. (Remember significant digits!) - A curve.fit plot showing how the force on a wire depends on current through the wire. (You must have force (N) and current (A or mA) as your axes: That is, where needed, convert equivalent variables. (See Appendix 3.2.) Any conversions you do should be included in a table and sample calculations must be provided.] - A curve.fit plot showing how the force on a wire depends on the external magnetic field. (You must have force (N) and magnetic field (T or mT) as your axes: That is, where needed, convert equivalent variables. (See Appendix 3.2.) Any conversions you do should be included in a table, and sample calculations must be provided.

Sample calculations for force changing B ext (T): B=μnI = (5E-05)(100)(0.52) =0.0026 T 𝛅 B = B ( δ𝐼 𝐼 ) 2 + ( 0 100 ) 2 + ( 0 5𝐸−05 ) 2 = (0.0026) ( (0.52) (0.01) ) 2 + ( 0 100 ) 2 + ( 0 5𝐸−05 ) 2 = 0.01 F=ma = (0.00005)(9.8) = 0.0004905 N 𝛅 F = F ( δ𝑚 𝑚 ) 2 + ( 0 9.8 ) 2 = (0.0004905) ( (0.00001) (0.00005) ) 2 + ( 0 9.8 ) 2 =0.0001 Table 1. B ext (T) vs. Force (N) Trial # I solenoid (A) Uncertainty I solenoid (A) B ext (T) Uncertainty B ext (T) Mass (kg) Uncertainty Mass (kg) Force (N) Uncertainty Force (N) 1 0.52 0.01 0.00260 0.00005 0.00005 0.00001 0.0005 0.0001 2 0.70 0.01 0.00350 0.00005 0.00010 0.00001 0.0010 0.0001 3 0.90 0.01 0.00450 0.00005 0.00012 0.00001 0.0012 0.0001 4 1.10 0.01 0.00550 0.00005 0.00014 0.00001 0.0014 0.0001 5 1.31 0.01 0.00655 0.00005 0.00018 0.00001 0.0018 0.0001 Table 2. B ext (T) vs. Force (N) with leads in solenoid reversed Trial # I solenoid (A) Uncertainty I solenoid (A) B ext (T) Uncertainty B ext (T) Mass (kg) Uncertainty Mass (kg) Force (N) Uncertainty Force (N) 1 0.52 0.01 0.00260 0.00005 -0.00004 0.00001 -0.0004 0.0001 2 0.70 0.01 0.00350 0.00005 -0.00006 0.00001 -0.0006 0.0001 3 0.90 0.01 0.00450 0.00005 -0.00013 0.00001 -0.0013 0.0001 4 1.10 0.01 0.00550 0.00005 -0.00014 0.00001 -0.0014 0.0001 5 1.31 0.01 0.00655 0.00005 -0.00018 0.00001 -0.0018 0.0001

Interpretation of results 2. Do your wire current dependence findings validate F = I wire L x B ext ? Use the X^2 reduced value from your plot to support your answer. According to the equation F=ILB, the wire current and the magnetic force on the wire are directly proportional. Therefore, as the current increases and all other variables remain constant as should the force. The data in the plot shows a linear and increasing relationship between the force and the current indicating the two are directly proportional. The χ2 reduced value indicates how well a sample of data represents or matches the larger population measured. The χ2 reduced values achieved with the plot is 0.83. This value is extremely close to 1 meaning the fit is good and the plot represents the data efficiently. Furthermore, as seen by the residual plots, almost all of the trial’s error bars crossed the residual = 0 line, indicating a close relationship between our fit and our values. As trial 3 where the current I = 0.9 A, we can see that it does not cross the residual line, showing that this trial can potentially be seen as an outlier. 3. Do your magnetic field dependence findings validate F = I wire L x B ext ? Use the X^2 reduced value from your plot to support your answer. According to F=ILB, the external magnetic field strength and the magnetic force on the wire are directly proportional. The magnetic field (B ext ) exerts a force on the current-carrying wire, perpendicular to both current and field. Therefore, as the magnetic field increases and all other variables remain constant, the force on the current should increase proportionally. The χ2 reduced value indicates how well a sample of data represents or matches the larger population measured. The χ2 reduced value achieved was 1.07, very close to 1 meaning the fit is good and the uncertainties were appropriately taken into account through the trials, confirming the validity of the hypothesis. 4. If you reverse the direction of the current in the wire, what happens to the force on the wire? If you reverse the direction of current in the solenoid, what happens to the force on the wire? Explain. If you reverse the direction of the current in the wire the force on the wire would change in direction, according to the right hand rule. As if the magnetic field of the solenoid remained in the same direction, the force would reverse with the inversion of current flowing through the wire. If you reverse the direction of the current in the solenoid, by the right hand rule, the magnetic field produced by the solenoid would change in direction, as the direction in which your fingers would coil would force your thumb in the opposite direction. As the direction of magnetic field strength reverses, the force on the wire would also reverse, according to the right hand rule, so long as the direction of the current running through the wire remained constant. However in both scenarios, only the direction of force would differ, not the magnitude, as the factors affecting the magnitude of force on the wire would include the magnitude of current, the length of the solenoid and the number of windings.

5. If you turned off the current to either the wire or the solenoid, would there be a magnetic force on the wire? Explain. Solenoids can act as on/off switches because they have a unique ability to create magnetic fields only when current is flowing through them. If current through the solenoid is not flowing, there would be no magnetic field produced, and therefore no magnetic force resulting from a magnetic field. From the equation F = I wire L x B ext where F is the magnetic force, I is the current in the wire, L is the length and B is the external magnetic field, we can see that if there is no current supplied to the wire (I = 0A), the magnetic force will subsequently also be equal to 0. Similarly, if we turned off the current in the solenoid, we would get a magnetic field inside the solenoid as being 0, as described by Ampere’s law or by considering the sum of the contributions of all current loops that form it; B = unI. Bonus What would happen if we used a non-magnetic material as the solenoid core? Explain. When we have a magnetic material in the solenoids core, we increase the overall magnetic field as the solenoid’s magnetic field will align with the miniscule magnetic dipoles inside the material such that the core’s magnetic field will add up with the solenoid’s magnetic field. On the contrary, if a non-magnetic material was used it would decrease the induced magnetic field, and so it would be weaker, and therefore will have dampened. This is essentially because the absence of a magnetic material (which would have increased the strength of the magnetic field), will thus be weaker than a field induced by one with a magnetic material within the solenoids core. Overall, using a non magnetic material as the solenoid core will result in a less efficient solenoid (as it will use up more heat), consume more power, and would be weaker than one with a magnetic material.

Plot 2: