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Therefore:  $$\frac{\Delta V}{\Delta B}=0.000443\frac{V}{mT}=0.443\frac{V}{T}$$  As shown in Figure 1:   $$V=\frac{0.021\times (1+cos(\frac{2\pi\times(\theta-162)}{180}))}{2}$$$$\frac{\Delta V}{\Delta\theta}=\frac{0.21}{2}\times(-sin(2\frac{\pi}{180}(-45)))$$