Question

5. Consider the flux across the line segment from (1,2) to (3,2) (a) Use vectors and following the process in the previous page to compute the flux across the line segment Include umi Segment Length = Total Flux ix- b) Explain why your answer to part (a) makes sense intuitively 6. Now consider the flux across the line segment from (1,2) to (2,1) (a) Uac vortors to comp the ux ars the l Include mits.

Answer 1

here flux is how much pass through that line element.

"Here flux is how much A^vector .dl^vector = A^vector.ndl they surely the unit vector so K^vetor it is also very clear from the diagram that there is no F ^vector pasing through the line segment so only fair result is that flux shouid zero

flux $\phi=\vec{A}.\vec{dl}=\vec{A}.\hat{n}dl$ where $\vec{dl}$ is the line and $\hat{n}$ is the unit vector showing the direction which is $\perp$ to the line $\vec{dl}$ .

(a) let $\vec{F}=F_0\hat{i}$

line element

$\left | \vec{dl} \right |=\sqrt{2^2}=2$

lets consider any vector $\vec{K}=a\hat{i}+b\hat{j}$ (dont need z-component as we are in x-y plane) which is $\perp$ to $\vec{dl}$ . then surely the unit vector $\hat{n}=\frac{\vec{K}}{\left | \vec{K} \right |}$

since $\vec{K}$ is $\perp$ to $\vec{dl}$ , the scaler product $\vec{K}.\vec{dl}=0$

$\vec{K}.\vec{dl}=0$

$\left ( a\hat{i}+b\hat{j}\right ).2\hat{i}=0$

$2a=0$

$\therefore a=0$

let the other component be $b=\lambda_1$ where $\lambda_1$ are some constant.

so, $\vec{K}=\lambda_1\hat{j}$ and

hence $\hat{n}=\frac{\vec{K}}{\left | \vec{K} \right |}=\frac{\lambda_1\hat{j}}{\lambda_1}=\hat{j}$

total flux= $\vec{F}.\hat{n}\left | \vec{dl} \right |=F_0(\hat{i}.\hat{j})\times2=0$

It is also very clear from the diagram that there is no $\vec{F}$ pasing through the line segment. so only fair result is that flux should zero.

(b) $\vec{dl}=(2-1)\hat{i}+(1-2)\hat{j}=\hat{i}-\hat{j}$

$\left | \vec{dl} \right |=\sqrt{2}$

again let any vector $\vec{K}=a\hat{i}+b\hat{j}$ which is $\perp$ to $\vec{dl}$ . then we have

$\vec{K}.\vec{dl}=0$

$\left ( a\hat{i}+b\hat{j} \right ).(\hat{i}-\hat{j})=0$

$a-b=0$

$\therefore a=b$

let $a=b=\lambda_2$ where $\lambda_2$ is a constant

then $\vec{K}=\lambda_2(\hat{i}+\hat{j})$ and $\left | \vec{K} \right |=\sqrt{2}\lambda_2$

unit vector $\hat{n}=\frac{\vec{K}}{\left | \vec{K} \right |}=\frac{\hat{i}+\hat{j}}{\sqrt2}$

then total flux across the line, $F_0\hat{i}.\hat{n}dl=F_0\hat{i}\times\frac{\hat{i}+\hat{j}}{\sqrt2}\times\sqrt2=F_0$