Phys Lab 3a

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Feb 20, 2024

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Lab 3a Methodology 1. Why do we treat the IOlab measurements in this experiment statistically? We will treat this data statistically (recall Appendix 1.2) because each plateau (data collected at a constant current) is made up of many repeated measurements of the magnetic field. By selecting an interval of data, we can directly obtain the average (u) and the standard deviation (o) of the data within that interval. 2. Why can we use mass as an equivalent variable for the magnetic force on the wire? Mass can be used as an equivalent variable for the magnetic force because we cannot measure force directly, instead, we must use an equivalent force (from the equation F = ma) Force and mass are related according to Newton's equation F = ma, where the downward acceleration on Earth is constant, a = g- 9.8 m/s If we apply an external vertical force to an object on a scale, the object experiences two forces: the force of gravity and the applied force. Therefore, the "mass" displayed on the scale readout is actually a convolution of the object's true mass and some additional "apparent" mass (due to the applied force). gravity + Fapplied - (Mobject + Mapparent) X a If we are (zero) the scale before applying the external force, this effectively sets F gravity = 0 and Mobiect=O. so the expression above simplifies to: Fapplied = Mapparent X a Example: We tare a scale and then apply an external vertical force. The scale readout is now m=0.33 g, which equals 0.00033 kg. We can then calculate external applied force: Fapplied = 0.00033kg x 9.8m/s? = 0.003234 N (Remember to find uncertainties and significant digits when you report your result.) Interpretation of results 3. Do your solenoid current vs magnetic field measurements validate B=unI? Use the χ 2 reduced value from your plot to support your answer. The χ2 reduced value indicates how well a sample of data represents or matches the larger population measured. The χ2 reduced values we achieved were 0.17 and 0.21. These values are relatively close to 1 meaning the fit is good. χ 2 reduced is a measure of the quality of a fit. χ 2 reduced << 1 means that you either have a good fit with possible inflated uncertainties that make it hard to tell, or are overestimating uncertainties. χ 2 reduced >> 1 means that you either have a poor fit or are underestimating uncertainties. A poor fit could be either because you have outliers or the linear function does not represent the data well. χ 2 reduced = 1 means that the fit is good and the functional form described the data efficiently

According to B=unI, magnetic field and current are directly proportional therefore as the current increases as should the magnetic field. When the leads are exchanged current is negative so magnetic field will decrease, 4. What force did you measure in step 6 of the protocol? (Report a force, not a mass.) 0.003234 N 5. Did the force on the wire change when the leads sending current to the solenoid were exchanged? (Step 6 of the protocol). Why or why not? The force on the wire would only change reverse in direction when the leads sending current to the solenoid were exchanged, as the direction of current will only dictate the direction of the magnetic field according to the right hand rule. The sole factors affecting the magnitude of force on the wire would be the magnitude of current, the length of the solenoid and the number of windings. Bonus Suppose that the core was only inserted halfway into the solenoid. Would the magnetic field change? Why or why not? If the core is inserted only halfway into the solenoid, the magnetic field inside the solenoid will change. The magnetic field strength will dampen compared to the case when the core is fully inserted. This is because the core is made of a material with high magnetic permeability, which helps to concentrate and increase the magnetic field lines within the solenoid. When the core is only halfway inserted, the magnetic field lines will not be as concentrated and will instead spread out into the air surrounding the solenoid. This will lead to a decrease in the magnetic field strength inside the solenoid. The magnitude of the decrease will depend on the specific dimensions of the solenoid and the core, as well as the magnetic properties of the materials involved. However, in general, the magnetic field strength inside the solenoid will be lower when the core is not fully inserted. I = 0.37 A Step 4 Current (A) B (T^-6) 0.01 ± 0.01 154.3 ± 4.4 0.06 ± 0.01 615.3 ± 4.6 0.15 ± 0.01 1560.0 ± 4.4 0.22 ± 0.01 2379.8 ± 4.3

0.27 ± 0.01 2854.5 ± 4.2 Step 5 Current (A) B (T^-6) 0.03 ± 0.01 -285.7 ± 4.7 0.09 ± 0.01 -986.1 ± 4.7 0.13 ± 0.01 -1442.1 ± 4.7 0.17 ± 0.01 -1793.7 ± 4.8 0.22 ± 0.01 -2301.8 ± 4.7 -0.18 0.62

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