Recent questions in Electric current

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photo

What is the strength of the electric field at the position indicated by the dot in the figure?

$E=\overline{\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}}\frac{N}{C}$

What is the direction of the electric field at the position indicated by the dot in the figure? Specify the direction as an angle above the horizontal line.

$\theta ={\overline{\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}}}^{\circ}$ .

Resistivity
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Waylon Mcbride
2022-05-10

First, I tried to solve this like this: $dR=\zeta \frac{l}{S}$

In our case the length is dr, and therefore I suppose that the area of this ring is 2πrdr: $dR=\zeta \frac{dr}{2\pi rdr}$ The solution sheet says: $dR=\zeta \frac{dr}{2\pi rL}$

I know that something is wrong with my equation, because dr goes away and then I cannot integrate from a to b. But why is in the solution L instead of dr? As I understand problem instruction L= b-a. And therefore L=dr which doesn't make sense to me.

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lifretatox8n
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Aedan Gonzales
2022-05-10

$\begin{array}{r}({u}_{1}^{T}x){u}_{1}\\ ({u}_{2}^{T}x){u}_{2}\end{array}$

Certainly we have

$\begin{array}{r}x=({u}_{1}^{T}x){u}_{1}+({u}_{2}^{T}x){u}_{2}\end{array}$

Looks like a simple problem, but how can I prove this?

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London Ware
2022-05-09

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Annabel Sullivan
2022-05-09

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velinariojepvg
2022-05-09

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fetsBedscurce4why1
2022-05-08

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hyprkathknmk
2022-05-08

The equation is in one dimension and P and O are points on the x axis.

The Electric field is oriented in the $+x$ direction ans has the formula $c\frac{1}{r}$ where c is a coefficient and r is the distance form the origin of the coordinate system.

The vector ${\overrightarrow{r}}_{P}-{\overrightarrow{r}}_{O}$ has the same direction as that of the electric field.

Now, Why do we put P below the integral and O above it? Can someone explain the equation for me?

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velinariojepvg
2022-05-08

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Peia6tvsr
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Micah Haynes
2022-05-08

${\int}_{0}^{1}\frac{{\mathrm{log}}^{2}(x+1)\mathrm{log}({x}^{2}+1)}{{x}^{2}+1}dx$

Any kind of help is appreciated.

Resistivity
Answered

deformere692qr
2022-05-08

$\rho =\frac{RA}{L}$

where $\rho $ is resistivity, $R$ is resistance, $A$ is cross-sectional area, and $L$ is the length of the conductor.

We can see from the formula that $A$ and $L$ are involved, why then does resistivity not depend on dimensions?

Resistivity
Answered

ga2t1a2dan1oj
2022-05-08

${\rho}_{a}(s)={s}^{2}{\int}_{0}^{\mathrm{\infty}}T(\lambda ){J}_{1}(\lambda s)ds\phantom{\rule{2em}{0ex}}(4)$

Where ${\rho}_{a}$ is the apparent resistivity and $T(\lambda )$ is a resistivity transform function. Using a Hankel Transform, we get:

$T(\lambda )={\int}_{0}^{\mathrm{\infty}}{\rho}_{a}(s){J}_{1}(\lambda s)\left[\frac{1}{s}\right]ds\phantom{\rule{2em}{0ex}}(7)$

New variables are then defined by

$x=ln(s)\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}y=ln\left(\frac{1}{\lambda}\right)\phantom{\rule{2em}{0ex}}(8)$

Substituting (8) in (7), we get

$T(y)={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{\rho}_{a}(x){J}_{1}\left(\frac{1}{{e}^{y-x}}\right)dx\phantom{\rule{2em}{0ex}}(9)$

Please note that any missing equations were for another resistivity equation.

The practical use of the electric current problems is met basically everywhere even if you are not majoring in Engineering. Take an example of data programming and the use of safety (data transfer) questions when the data is being transferred from place to place via cloud storage solutions. The same is related to automation and logistics where the electric current equation will help to calculate the limitations and the available resources. See some answers to the questions dealing with the electricity principles to get a better idea of how it works and compare the existing solutions to your task objectives!