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}