Resistivity Q&A: Expert Assistance

The Hall coefficient is defined as this:
$R_{H} = \frac{E_{y}}{j_{x} B_{z}} .$
Always as $R_{H}$ . I am currious as to how to use this coefficient? Is it the y-direction resistance/resistivity (it is very close to the general resistivity from Ohm's law $ρ = E / j$ )? Then I should be able to find it by $R_{H} = V_{y} / I_{y}$ (or $R_{H} = E_{y} / j_{y}$ ).
or is it simply defined as this as a proportionality constant (it does indeed look good with the three factors combined around one constant) and not an actual physical quantity?

agrejas0hxpx 2022-05-15

Derivative of $e^{- x}$

Daphne Haney 2022-05-15

A Binomial Coefficient Sum: $\sum_{m = 0}^{n} (- 1)^{n - m} (\binom{n}{m}) (\binom{m - 1}{l})$

Jazlyn Raymond 2022-05-15

What would happen with the drift velocity of a cylindrical resistor's diameter increases, with a given voltage between its terminals? According to the expression:
$\begin{aligned} R & = ρ \frac{L}{A} \\ I & = n e A v_{d} \\ Δ V & = I R \\ v_{d} & = \frac{Δ V}{ρ L n e} \end{aligned}$
The resistivity does not change, neither does the length of the resistor nor the term ne but the resistance does change as well as the current, so the area is eliminated from the expression. I wonder if the drift velocity would be the same after increasing the diameter or if my derivation is wrong.

Nubydayclellaumvcd 2022-05-14

Velocity of projectile with both quadratic and constant resistive force
Suppose we have a projectile of mass m that impacts a material with velocity $v_{0}$ . As it travels through the material it is subject to the resistive force $F = α v^{2} + β$ . How can I determine $v (t)$ , the velocity of the projectile at time t where t=0 is the impact time, and $v (p)$ , the velocity of the projectile when it reaches a penetration depth of p?
This question addresses v(t) for an exclusively quadratic resistive force ( $β = 0$ ), but is there a more general formula for when there is a constant component?

Yasmine Larson 2022-05-14

Resistivity in terms of temperature coefficient of resistance
$R = R_{0} * (1 + α (T - T_{0}))$
Where $R_{0}$ and $T_{0}$ are the reference resistance and temperature.
Is it then safe to say that:
$ρ = ρ_{0} * (1 + α (T - T_{0}))$
(where $ρ_{0}$ is the reference resistivity)?

Blaine Stein 2022-05-14

How to prove $\int_{0}^{1} \frac{1 + x^{30}}{1 + x^{60}} d x = 1 + \frac{c}{31}$ , where 0

syaoronsangelhwc17 2022-05-14

What is the minimal polynomial of $α = \frac{3^{1 / 2}}{1 + 2^{1 / 3}}$ over $Q$ ?

vilitatelp014 2022-05-14

Is the resistivity of all (conducting) alloys more than that of all (conducting) metals?

Osmarq5ltp 2022-05-14

The wires in (Figure 1) all have the same resistance; the length and radius of each wire are noted.Rank in order, from largest to smallest, the resistivities $ρ_{1}$ to $ρ_{5}$ of these wires.

Logan Lamb 2022-05-13

I know what resistivity of materials are like, and is given by the formula:
$R = \frac{ρ L}{A}$
However what is the volume resistivity of materials like copper and aluminium for instance? How do we calculate the volume resistivity of materials, because this was confusing when I read the site:
Method of calculating eddy currents of a conductor, is this correct?
Where there is the answer given by Floris states the equation:
$F = \frac{v \cdot B^{2} \cdot A}{ρ / t}$
Where $ρ$ is given to be the volume resisitivity.
My question:
how do we measure the volume resistivity of a material for example aluminium?
what happens if the thickness of the material changes?

Waylon Mcbride 2022-05-10

Coaxial cable has radius a of copper core and radius b of copper shield. Between there is an insulator with specific resistivity $ζ$ . What is the resistance of this cable with length L between the core and the shield?
First, I tried to solve this like this: $d R = ζ \frac{l}{S}$
In our case the length is dr, and therefore I suppose that the area of this ring is 2πrdr: $d R = ζ \frac{d r}{2 π r d r}$ The solution sheet says: $d R = ζ \frac{d r}{2 π r L}$
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.

Nubydayclellaumvcd 2022-05-10

What is the strength of the electric field at the position indicated by the dot in the figure?
photo
What is the strength of the electric field at the position indicated by the dot in the figure?
$E = \bar{} \frac{N}{C}$
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.
$θ = {\bar{}}^{\circ}$ .

Aedan Gonzales 2022-05-10

This is kind of a stupid question and I am taking some risk of getting some down-votes here, but, I can't resist posting it. Suppose $(u_{1}, u_{2})$ is an orthonormal basis for $R^{2}$ , and let x be an arbitrary vector in $R^{2}$ , then we can decompose x by projecting on $u_{1}, u_{2}$ , i.e.
$\begin{aligned} (u_{1}^{T} x) u_{1} \\ (u_{2}^{T} x) u_{2} \end{aligned}$
Certainly we have
$\begin{aligned} x = (u_{1}^{T} x) u_{1} + (u_{2}^{T} x) u_{2} \end{aligned}$
Looks like a simple problem, but how can I prove this?

Annabel Sullivan 2022-05-09

I was looking at the data sheet for a copper cable and noticed the conductor resistance to be specified at around 5 mOhm/m. This is magnitudes higher than pure copper, which has a resistance in the order of 10 nOhm*m. Why is the resistivity so much higher?

hyprkathknmk 2022-05-08

$V_{P} - V_{O} = \int_{P}^{O} E . d r$
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 ${\vec{r}}_{P} - {\vec{r}}_{O}$ 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?

Peia6tvsr 2022-05-08

In the grating equation $n λ = d \sin θ$ , the quantity $θ$ will be determined in the lab using the angular scale on the experimental apparatus. using meter sticks and geometry/trigonometry. using a protractor.

Micah Haynes 2022-05-08

How can we evaluate
$\int_{0}^{1} \frac{\log^{2} (x + 1) \log (x^{2} + 1)}{x^{2} + 1} d x$
Any kind of help is appreciated.

deformere692qr 2022-05-08

The formula for resistivity is:
$ρ = \frac{R A}{L}$
where $ρ$ 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?

ga2t1a2dan1oj 2022-05-08

The equations are as follows:
$ρ_{a} (s) = s^{2} \int_{0}^{\infty} T (λ) J_{1} (λ s) d s (4)$
Where $ρ_{a}$ is the apparent resistivity and $T (λ)$ is a resistivity transform function. Using a Hankel Transform, we get:
$T (λ) = \int_{0}^{\infty} ρ_{a} (s) J_{1} (λ s) [\frac{1}{s}] d s (7)$
New variables are then defined by
$x = l n (s) a n d y = l n (\frac{1}{λ}) (8)$
Substituting (8) in (7), we get
$T (y) = \int_{- \infty}^{\infty} ρ_{a} (x) J_{1} (\frac{1}{e^{y - x}}) d x (9)$
Please note that any missing equations were for another resistivity equation.

Electric Engineering and robotics will require the studies of the electric current and resistivity. These are meant for the selection of the materials that are used in electrical and electronic components. Therefore, the resistivity problems with solutions will revolve around conductors and the resistant wires as you calculate the resistivity. The answers will include formulas that must be used first, so you can see and understand how things work. The resistivity examples can also be related to statistical data, which means that you can use the solutions as the templates with the different numbers where you simply replace the data.

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