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2)

$ \begin{equation*} \text{\underline{Biot-Savart}:} \qquad \qquad d\bar{H}=\frac{I(\bar{R})d\bar{l}\times(\bar{R}-\bar{R}')}{4\pi\abs{\bar{R}-\bar{R}'}^3} \end{equation*} $

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$ \begin{align*} \bar{R}&= 0\hat{x}+0\hat{y}+0\hat{z}\\ \bar{R}'&=y\hat{y}\\ \abs{\bar{R}-\bar{R}'}&=y \\ d\bar{l}&= (dy)\hat{y} \end{align*} $

$ \begin{equation*} d\bar{H}=\frac{(Idy)\hat{y}\times(-y\hat{y})}{4\pi\abs{y}^3}=0 \longrightarrow \boxed{\bar{H}=0} \end{equation*} $

$ \begin{align*} \text{\underline{Ampere}:}& & \nabla\times\bar{H}&=\bar{J} & &\longrightarrow& & \oint \bar{H}\cdot d\bar{l}&=I_{enc}\\ \text{at the origin:}& & I_{enc}&=0 & &\longrightarrow& & \boxed{\bar{H}=0} \end{align*} $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood