(Created page with "<math> \begin{equation*} \boxed{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*} </math> <math> \begin{align*} \tex...")
 
Line 1: Line 1:
 +
2)
 +
 
<math>
 
<math>
 
\begin{equation*}
 
\begin{equation*}
\boxed{d\bar{H}=\frac{I(\bar{R}')d\bar{l}'\times(\bar{R}-\bar{R}')}{4\pi\abs{\bar{R}-\bar{R}'}^3}}
+
\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*}
 
\end{equation*}
 
</math>
 
</math>
 +
 +
[[Image:biot.jpg|Alt text|270x222px]]
  
 
<math>
 
<math>
 
\begin{align*}
 
\begin{align*}
\text{\underline{along x}:}& & \bar{R}&=(0,0,0) \longrightarrow &&\bar{H}&=\int_1^{\infty}\frac{Idx\hat{x}\times(-x\hat{x})}{4\pi x^3}=0\\
+
\bar{R}&= 0\hat{x}+0\hat{y}+0\hat{z}\\
& &\bar{R}'&=x\hat{x} &\\
+
\bar{R}'&=y\hat{y}\\
& &\abs{\bar{R}-\bar{R}'}&=x && &\\
+
\abs{\bar{R}-\bar{R}'}&=y \\
\text{\underline{along y}:}& & \bar{R}&=(0,0,0) \longrightarrow &&\bar{H}&=\int_1^{\infty}\frac{Idy\hat{y}\times(-y\hat{y})}{4\pi y^3}=0\\
+
d\bar{l}&= (dy)\hat{y}
& &\bar{R}'&=y\hat{y} &\\
+
& &\abs{\bar{R}-\bar{R}'}&=y && &
+
 
\end{align*}
 
\end{align*}
 
</math>
 
</math>
Line 18: Line 20:
 
<math>
 
<math>
 
\begin{equation*}
 
\begin{equation*}
\text{\underline{Superposition}: Total field} \qquad \boxed{\bar{H}=0}
+
d\bar{H}=\frac{(Idy)\hat{y}\times(-y\hat{y})}{4\pi\abs{y}^3}=0 \longrightarrow \boxed{\bar{H}=0}
 
\end{equation*}
 
\end{equation*}
 +
</math>
 +
 +
<math>
 +
\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*}
 
</math>
 
</math>

Revision as of 19:43, 18 June 2017

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*} $

Alt text

$ \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

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010