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1

Note that a current carrying wire produces a circular magnetic field that's why it doesn't matter how you hold your hand ie how you rotate your hand around your arm as long as your thumb shows the direction of the current. Edit after comments: See the illustration I've added below. Now use your hand in the way that you've learned and convince yourself ...


2

I'll answer the concrete question, because it's one of those fun ones where the units are all wrong and the scales are just absurd. Does this also mean that if I release a million amperes of current into the earth, every living entity walking barefooted should immediately die? It depends on how long you do it and with how much power. And ...


1

Trying to address this misconception: I start of with a resistance of 1 ohm by the wire and 6 amperes which result in 6 volts. When I meet the resistor however the resistance increases to let's say 3. Does the current decrease at the same rate the resistance increases? So if the resistance goes down to 3 will the current be 2 so that in the end I have a ...


0

There's an energy transfer whenever there is a change in potential, not potential difference. The (electric) potential, measured in volts, is the electric potential energy (EPE) of a unit charge at a particular point in the circuit. So, imagine a particle of unit charge travelling in the direction of current. It starts off with a higher potential (therefore ...


0

You must put a battery with the positive terminal on (a) that has such potential difference that gives you a high enough current thay in its turn gives you a force equal but opposite of the gravitational force of the mass.In that way, you can determine the mass of the object. The magnetic force will pull upwards while the gravitational one will pull ...


8

Firstly we are not the best conductors, so current might be having a relatively hard time getting through us. But I believe the real reason is that you also need a high potential difference in order to get current flowing through you. Like lightning which needs a huge potential difference between the clouds and earth (so big that most of times a neutral ...


26

Electricity isn't a gas that expands out to shock anything in contact with it. Electricity is a flow from high voltage to low voltage. Touching a charged object is only dangerous if you become a current path--if it uses you to get somewhere. Even if the earth had a net charge, you aren't providing it anywhere to go, so you will not be shocked. It's somewhat ...


1

The minus sign is wrong.The reason for this is the x which you have chosen to be positive but is in fact negative. x points positively to the right and negatively to the left,and the horizontal vector that you are using in your picture is opposite to the direction of x.So,its x=-acotĪø. Cheers!


0

As you already mentioned $$ I = \int \vec{J} \cdot d\vec{A} $$ current is the charge-flow through a given surface. So If one talks about currents, the surface (and therefore its normal direction) is to be understood beforehand. You could assign a direction like this: $$ \frac{\vec{I}}{I} = \frac{\int d\vec{A}}{A} $$ which is the mean normal direction of the ...


0

What I think is the best way to look at it is that current is completely local. What I mean is that actually current is defined a distinct region of space. We care about how much charge there is in a region of space and then how much charge is in that same region without worrying to what happened to the first amount of charge. On the other hand, the current ...


6

Typically this is explained by the saying, "current kills." It's not the charge (or potential above ground) that a body attains that hurts biological systems, it's the current that flows through them and either 1) heats them or 2) disrupts important electrical signals in the body. Heating damage occurs and can "cook" (cause 1st, 2nd, or 3rd degree burns ...


3

Take a capacitor and put it across a battery. There will be a transient current as the electrons go towards the anode . This happens very fast and the current is small. If you short the capacitor with a wire, the battery will empty all its charge on the short, which, depending on the battery can really be damaging. Your body accumulates some charge which ...


8

If you have an excess of electron in your body, your hair might stand on end and you might feel a bit negative (I couldn't help that pun), and you should probably avoid touching people or metal object if you don't want a static shock, but other than that, it's mostly harmless. The real danger comes from flowing electrons. Because the body basically runs on ...


2

The whole electrical power grid is connected to ground. I don't know the details of other regions, but if you are in North America, the two current carrying conductors in a residential electrical outlet are called "hot" and "neutral". The "neutral" conductor is connected to the Earth at many places. If your bare feet touch wet Earth, and your hand touches ...


1

If you are not in a complete electrical circuit, any electric shock caused by touching a charged object or wire is brief. These "static shocks" are slightly painful, but they are (rarely) dangerous or fatal. I'm sure you've experienced a minor static shock. By wearing insulating footwear, you break a complete circuit and forbid a flow of electricity from ...


-1

We have to add both of them. It is no more a choice between two of them.


3

The Hall voltage can indeed be equal to zero if the electrons and holes balance out. You can find the formula in these Hall Effect lab notes by Pengra, Stoltenberg, Dyck, Vilches, eq. 16: $$R_H = \frac{1}{|q|}\frac{n_h \mu_h^2 - n_e \mu_e^2}{(n_h \mu_h + n_e \mu_e)^2}$$ where $\mu$ is mobility and $n$ is density and $e$ means electron and $h$ means hole ...


3

In real-life conductors you always have some of both kind of charge carriers (electrons and holes), so, the main question is, which of those is more abundant? In fact materials with positive and negative hall coefficients can be found.


1

The full Maxwell equation $j=\nabla\times H+\partial_{t}D$ ($j$ current, $H$ magnetic field and $D$ electric induction) is recover from the action $$S=\int dx\left[L\left(A_{\mu}\right)\right]=\int dx\left[-\dfrac{F_{\mu\nu}F^{\mu\nu}}{4}-j_{\mu}A^{\mu}\right]$$ in a standard way, provided we define $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$ ...


1

From the Maxwell-Dirac Lagrangian $$ \mathcal L = -\frac{1}{2}F^2 + \overline{\psi}(i\gamma^\mu D_\mu +m) \psi $$ where $D_\mu$ is the gauge covariant derivative it is clear that the 4-current that acts as the source term in Maxwell's equations is $$ j^\mu_D = q\overline{\psi} \gamma^\mu \psi. $$ Using the Dirac equation it can be shown that (see, e.g., ...


2

I think you bring up an interesting question, I'm not sure why there were some derogatory comments to this... First off, the fermionic current doesn't couple to the gauge field due to its dimension. The Dirac field is dimension $ 3/2 $ and the current is dimension $ 3 $. Therefore, the coupling of the fermion to the gauge fields is of higher order. On the ...


-1

Nichrome is used to connect the two terminals of the battery or any source that is connected to the circuit. It is not used for making filaments. Therefore, it strong text offers low resistance to the flow of the electric current.


0

I don't have access to the Sampayan paper, but I think you're conflating two different things: The total anode-to-cathode accelerating potential $V_{AK}$. (Note it's the difference between the two potentials that matters.) The cathode is assumed to be capable of emitting unlimited numbers of electrons, so it is solely the repelling electric field of said ...


1

Only a few electrons in the outer shell of a copper atom are ever displaced (removed) due to charge forces. The rest all stay in place. So even if copper atoms are mediating a current, the integrity of the atoms themselves is never in doubt.


1

No electrons are being lost; for a current to flow there needs to be a circuit, ie some kind of loop. Also, when current flows, it is not necessarily the electrons themselves that are moving as fast as the current, but rather the signal to move. The copper atom also consists of protons and neutrons in its nucleus, as well as most of its electrons which stay ...


4

Electrons in a metal are delocalized, which means they don't know to what atom they belong. The cores are bound in a crystal lattice and the electrons flow around it like a gas, similar to this. When you attach a voltage to both ends of such a material, the electrons bounce into each other and push each other a tiny bit down the wire. Electrons are flowing ...


0

The $\frac{1}{I}$ term is the analogue of $y$ in the equation for a straight line, $y=mx+b$. I'll guess that $R$ is a variable resistance you control and $r$ is some internal or fixed, possibly unknown, resistance in a circuit. When you plot the controlled resistance $R$ on the horizontal axis, that is the analogue of $x$ in the linear equation pattern. ...


1

What I don't understand is the fact that more energy is dissipated within a resistor of length l than would be dissipated in a wire of length l. It just doesn't make sense. You mean, why is it not like this: Electron goes through a resistor while there's a 1 V voltage between the resistor ends -> resistor is heated by energy of 1 V * 1 elementary ...


3

The dissipation in a resistor or wire of the same length $\ell$ doesn't have very much to do with the length - it has to do with the resistance. For a given current (number of electrons per second crossing a particular point), there will be a voltage drop associated with a given resistance - this is Ohm's law. A nice intuitive way to think about this is ...


2

For your circuit, $V = I\cdot R$. You are plotting (unusually) R along the X axis and $\frac{1}{I}$ along the Y axis, so the slope is $\frac{1}{V}$. Now the fact that this slope is a straight line tells you that the voltage is constant. This means that (over the range of your experiment) your voltage source has a low internal resistance. Imagine for a ...


1

If it's a simple circuit where Ohm's law applies, then we should get $$V=IR$$ so we see that $$V/I = R$$ $$1/I=R/V$$ $$1/I = (1/V) \times R$$ The gradient should then be $1/V$. Seems like a slightly bizarre plot but if you got a straight line then that makes the maths simple at least!


0

It is easy to show that the differential and integral forms of Maxwell's equations are equivalent using Gauss's and Stokes's theorems. Correct, they are equivalent (assume no GR, and no QM) in the sense that if the integral versions hold for any surface/loop then the differential versions hold for any point, and if the differential versions hold for every ...


0

Yes, in general it affects the current, in particular if that current was going to change, the self inductance $L$ makes it change less quickly than it otherwise would change. Here are some examples: You could put a resistive loop inside a solenoid as a good example. The total $\vec{B}$ field is the $\vec{B}$ field due to the solenoid $\vec{B}_1(t)$, and ...


2

If you have current flowing one way through a resistor, then the electrons flow through the other way. Since current flows from the high voltage end of a resistor to the low voltage end, then the electrons come in at the low voltage end and come out at the high voltage end. When electrons (which are negatively charged) go from low voltage to high voltage, ...


0

There may be a confusion between enery and power. While the current, which is the number of charges per second, and the energy of the ,electron do'nt change at the output of the resistor, the power does. Power is the amount of energy per unit time, and that does not affect the current which, again, is the amount of charge per unit time.



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