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Ning Zhu edited subsection_Slope_Fit_Calculation_begin__.tex
over 8 years ago
Commit id: 02c55046a57ad2b4dfdb43ad9ee462927c6ca2eb
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\subsection{Slope Fit Calculation}
\begin{eqnarray}
From %From Figure \ref{fig:VB}
V&=&0.000443B+0.127\\
Therefore: %Therefore:
\frac{\Delta V}{\Delta B}&=&0.000443\frac{V}{mT}=0.443\frac{V}{T}\\
From %From Figure \ref{fig:directmethoddata}:
V&=&\frac{0.021\cdot (1+cos(\frac{2\pi\cdot(\theta-162)}{180}))}{2}\\
Taking %Taking the derivative at $\theta=117$ degrees:
\frac{\Delta V}{\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 \frac{\Delta\theta}{\Delta B}=\frac{1}{0.1}\cdot2.1095\frac{radians}{T}=21.095\frac{radians}{T \cdot m}