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Ning Zhu edited subsection_Slope_Fit_Calculation_begin__.tex
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\subsection{Slope Fit Calculation}
\begin{eqnarray}
From Figure \ref{fig:VB}
$$V&=&0.000443B+0.127$$\\ V&=&0.000443B+0.127\\
Therefore:
$$\frac{\Delta \frac{\Delta V}{\Delta
B}&=&0.000443\frac{V}{mT}=0.443\frac{V}{T}$$\\ B}&=&0.000443\frac{V}{mT}=0.443\frac{V}{T}\\
From Figure \ref{fig:directmethoddata}:
$$V&=&\frac{0.021\cdot (1+cos(\frac{2\pi\cdot(\theta-162)}{180}))}{2}$$\\ V&=&\frac{0.021\cdot (1+cos(\frac{2\pi\cdot(\theta-162)}{180}))}{2}\\
Taking the derivative at $\theta=117$ degrees:
$$\frac{\Delta \frac{\Delta V}{\Delta
\theta}&=&\frac{0.21}{2}\cdot(-sin(2\frac{\pi}{180}(-45)))=0.21V$$\\
$$\frac{\Delta\theta}{\Delta \theta}&=&\frac{0.21}{2}\cdot(-sin(2\frac{\pi}{180}(-45)))=0.21V\\
\frac{\Delta\theta}{\Delta B}&=&\frac{\Delta V}{\Delta B} \cdot \frac{\Delta \theta}{\Delta V}=0.443\frac{V}{T}\cdot\frac{1
radian}{0.21V}=2.1095\frac{radians}{T}$$\\
$$V_{c}&=&\frac{1}{L}\cdot radian}{0.21V}=2.1095\frac{radians}{T}\\
V_{c}&=&\frac{1}{L}\cdot \frac{\Delta\theta}{\Delta B}=\frac{1}{0.1}\cdot2.1095\frac{radians}{T}=21.095\frac{radians}{T \cdot
m}$$ m}
\end{eqnarray}