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Nathanael A. Fortune added subsection_Slope_Method_Initially_we__.tex
over 8 years ago
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\subsection{Slope Method}
%Initially, we fit our data to the form $V=V_{0}(sin(\theta_{1}+\theta_{2}))^2$, which using trigonometry, is $V=V_{0}\frac{(1-cos(\theta_{1}+\theta_{2})^2)}{2}$ (the actual equation that we fit to). However, we should have been fitting to the form $V=V_{0}cos(\theta_{1}-\theta_{2})^2$, which is $V=V_{0}\frac{(1+cos(\theta_{1}-\theta_{2})^2)}{2}$. Now,
%$$sin(\theta_{1}+\theta_{2})^2=sin2(θ1-(-θ2))$$
%2π-θ2=θ2’ to account for the incorrect sign in our original equation.
%θ2’-π=θ0 to change from sine to cosine, so we do end up with the form
%V=V0(½)(1+cos(2(θ1-θ0)))
%which is V=V0cos2(θ1-θ2), giving us the form we should have initially fit to.
As shown in Figure 2:
$$V=0.000443B+0.127$$
Therefore:
$$\frac{\Delta V}{\Delta B}=0.000443\frac{V}{mT}=0.443\frac{V}{T}$$
As shown in Figure 1:
$$V=\frac{0.021\times (1+cos(\frac{2\pi\times(\theta-162)}{180}))}{2}$$
Taking the derivative at $\theta=117$ degrees:
$$\frac{\Delta V}{\Delta \theta}=\frac{0.21}{2}\times(-sin(2\frac{\pi}{180}(-45)))=0.21V$$
$$\frac{\Delta\theta}{\Delta B}=\frac{\Delta V}{\Delta B} \times \frac{\Delta \theta}{\Delta V}=0.443\frac{V}{T}\times\frac{1 radian}{0.21V}=2.1095\frac{radians}{T}$$
$$V_{c}=\frac{1}{L}\times \frac{\Delta\theta}{\Delta B}=\frac{1}{0.1}\times2.1095\frac{radians}{T}=21.095\frac{radians}{T*m}$$
