# Tag Info

2

The surface charge compensates the outer electrical field. If you change that field by moving charges around in the surounding air then the surface charge of the conductor will change.

9

$\def\vE{{\vec{E}}}$ $\def\vD{{\vec{D}}}$ $\def\vB{{\vec{B}}}$ $\def\vJ{{\vec{J}}}$ $\def\vr{{\vec{r}}}$ $\def\vA{{\vec{A}}}$ $\def\vH{{\vec{H}}}$ $\def\ddt{\frac{d}{dt}}$ Faraday's law in integral form: $$\oint_{\partial A} \vE \cdot d\vr + \ddt\int_A \vB\cdot d\vA = 0$$ Thereby, $A$ is some surface and $\partial A$ its boundary. The boundary can be split ...

0

When an electric dipole is placed in a uniform electric field making an angle with the direction of the field as shown in the figure. Force on charge $-q=-q\overrightarrow{E}$ (opposite to $\overrightarrow{E}$) Force on charge $+q=q\overrightarrow{E}$ (along $\overrightarrow{E}$) Thus, electric dipole is under the action of two equal and unlike ...

1

$\def\vE{{\vec{E}}}$ $\def\vS{{\vec{S}}}$ $\def\vA{{\vec{A}}}$ $\def\rot{\operatorname{rot}}$ $\def\grad{\operatorname{grad}}$ $\def\div{\operatorname{div}}$ $\def\ph{{\varphi}}$ $\def\vn{{\vec{n}}}$ $\def\vr{{\vec{r}}}$ The charge floating through the wire causes a current density $\vec{S}$ in the wire (not only on the surface). This causes an electrical ...

3

An atom in isolation offers a potential well, and electrons form bound states in the well. The energy of those bound states can be calculated exactly in the case of a single-electron (hydrogen-like) atoms or by variational computational methods for more complicated cases. Now when you put several atoms together in a tight and regular array, they offer a ...

0

In theory, it should be (lumpy), however: Even if you could observe the peaks, or nodes and antinodes of the EM vectors of a single photon, its energy would be absorbed by the instrument you employed to observe it. Fast enough doesn't cut it. If the instrument absorbs no energy from the photon, it will also not be detected. Intensity for light is not ...

1

Although Chris White’s answer to the question “Why Moving Charges Produce a Magnetic Field?” posted by a High School teacher (Claws) last year, was selected as the best answer, I think it contains several pitfalls. Chris White imagines a stream of positive charges flowing in the $+z$ axis direction, while a test charge $+q$ initially located at $(1,0,0)$ is ...

2

In a single free atom, electrons have well defined energy levels and are somewhat bound to atom. Consider the following quantum mechanical model of atom to get an idea about an isolated atom. When all this isolated atoms come together to form the crystal, the atoms do not have well defined energy levels. There will be molecular orbitals. When the atoms ...

1

The intuition is that the valence electrons are so far away from their nucleus that when they combine to form metals, they feel the attraction of all the other nuclei as strongly as from theirs. In a more rigorous description, the orbitals for the valence electrons fully overlap with their neighbouring atoms, so their "play field" extends all over the ...

0

In general, when formulae call for "area" they are talking about the surface area on which the force or whatever is occurring. Note that this is not always the entire surface area, it might just be the curved part of a cylinder for example. This is one such example, because we actually aren't interested in the end effects at the bases of the cylinder.

0

If it were not, it would cause current to flow, and propagation of current involves the dissipation of energy, and this cannot occur without any external sources of energy. Hence, it follows that any charges in the conductor must be located on its surface.

2

As Mostafa says, it is macroscopically at equilibrium, not necessarily microscopically. There may be one misunderstanding you have, which is about "surface". I will talk about it later. In my opinion, equilibrium should be understood as no electron moving. It is easily to show that the electric field in conductor is zero. If the electric field is non-zero, ...

1

An inductor stores energy in a magnetic field. After current has been flowing in the inductor for a period of time, it has built up a magnetic field around the wire making up the inductor. In that state the inductor offers no opposition to current flow. If it were then disconnected from it's energy source (battery perhaps) then the the magnetic field will ...

1

Mostly, yes on both counts. It depends whether the solenoid also gets longer as it gets wider. The approximation that the field outside the solenoid vanishes is valid for points whose distance to the solenoid's centre is much smaller than the distance to both ends. This is impossible for a point outside a solenoid that's wider than it is tall. The field of ...

0

i think that both the bobs will reach the ground at same time because mass of both are nearly the same as they are of equal sizes and consideration of earth's magnetic field is not very effective in my point of view.

0

You have an issue in your argument. It's not right to describe the conductor as 100 electrons and 100 ions. These are not ions. Consider ion crystals such as NaCl: that's where the ions are, and this thing's an insulator. The fact that electrons are free doesn't mean that they left the ions in cold. It's like a cooperative, or a collective farm. All ...

0

Equilibrium means that there is no net change with time. A glass of water at room temperature is in equilibrium because, even though the molecules are fiddling around, their net movement is zero. Or, in another words, macroscopically, you can't see any overall change. In most practical situations, this means the state of the system after enough time has ...

-1

Listen guys..I don't think we need to just work out that much on the problem. This is CBSE and it doesn't require that much use of brain...and it also won't go much out of syllabus! I think that the question demands us to think of earth as a magnet...a bar magnet...and a magnet would at any case attract metallic bob...like a bar magnet attracts metallic ...

0

It may be pointed out that the word electromotive force is a misnomer. It does not represent force on the carriers of electricity. Instead, it represents the potential difference between the two poles in an open circuit (when no current is drawn from the cell).

0

There is no creation of anything, but it can be assumed that a circuit creates a voltage when the power, combination of that voltage and any current which would flow from it, has been gained from "outside the circuit" - e.g. through chemical processes (batteries), or electromagnetic processes (dynamo that converts mechanical power to electrical). This very ...

0

I dislike the term EMF (Electromotive force) as it is very confusing. Electromotive force, also called emf (denoted $\mathcal{E}$ and measured in volts), is the voltage developed by any source of electrical energy such as a battery or dynamo. Which means that all EMF are voltages but not all voltages are EMF. A voltage is only an EMF if it is a ...

0

Well, trying to get in the form of what Mark Wayne put down...you can start with the dfinition of the energy density "u" and find the partial wrt time. From there you can use Maxwell's Equations to make some substitutions. Finally in the end you will NEED a not so known/popular vector identity to give you the (ExB) term. Good luck...not a bad derivation at ...

0

1) With a constant and DC power source eventually the solenoid will become fully 'charged'. At that point its 'resistance' term vanishes because it no longer produces an emf against the battery. At this point, the $\frac{di}{dt}$ term will be zero, because the current isn't changing. 2) When you cut power, the magnetic flux is no longer maintained by the ...

2

Emf on a conducting object induces eddy currents. These in turn decay due to the electrical resistance of the object. What you end up with is energy in the form of heat. When you compare the two objects (essentially a conductor versus a non-conductor), a portion of the potential gravitational energy goes into generating eddy currents. That means the ...

-4

metal ball would hit the ground first. there wouldn't be any effect of induce emf or magnetic field. it only depends on gravitational force, as F=mg the mass of metal ball is greater, therefore, it will reach faster than glass ball, whose mass is less than metal ball.

0

Inside the solenoid there's no (component of the) magnetic field that isn't perpendicular to the loops, so treating a solenoid as if it has closed loops is valid approximation. Perhaps you can understand it better if you approach the solenoid as a single wire and see what the emf does from that point of view.

0

I suspect this is an approximation that works well. Each piece of the coils is almost exactly aligned with the induced electric field $\vec{E}$ at that position. (To be more accurate, some are exactly aligned, but most aren't. Try imagining a coil wrapped around a cylinder.) In this way, the EMF $\oint \vec{E}\cdot d\vec{l}$ for the closed loop is nearly ...

0

Do you mean a coil in which conductors are made of ferromagnetic material instead? If that's the case, you would probably not be better off: the field which is created is concentrated around the conductor (mostly at core of the solenoid) and in another direction, so it would not benefit from the increased magnetic permeability (which is what ferromagnetic ...

0

It results from the fact that every conductor in which flows a current generates a magnetic field which is concentric to it (iso-potential lines are concentric circles). The Biot&Savart law quantifies this behaviour under an integral form, and you can solve those equations in any point of space to get the distribution of the magnetic field (find it on ...

3

Maxwell is being misunderstood. First, Maxwell makes very clear that length, time and mass are the fundamental types of units. Then he discusses a totally different convention that isn't used today, saying "in the astronomical system, the unit of mass is defined with respect to its attractive power". In other words, Maxwell is talking about a concept of ...

-2

Think about the density of Lithium which is lower than the density of glass. The lithium ball many hit the ground later than the more dense glass due to air resistance. The experiment requires a procedure to put coating on the lithium ball to avoid it from burning in air.

1

Draw the graph of $\frac 1 x$ . You can see that it is a decreasing function for positive $x$. Hence conductance decreases as resistance increases. We could have defined conductance as any other decreasing function also but $\frac 1 R$ appears in many equations so we defined it that way. You might want to look at derivation of $J=\sigma E$ to get better ...

1

The electrical resistance of an electrical conductor is the opposition to the passage of an electric current through that conductor; the inverse quantity is electrical conductance, the ease at which an electric current passes. Consider resistance of $0.0001$ ohms, what is $\frac{1}{0.0001}ohms$? It is equal to $10000$ siemens. I hope this helped you in ...

0

Conductance is indeed defined as the inverse of resistance. $$G \equiv \frac{1}{R}$$ To see what this means physically, consider that when the resistance is large (i.e. it is "difficult" for current to get through), then conductance is low, and when resistance is small (i.e. it is "easy" for current to get through), then conductance is high. There's an ...

3

The explanation is Maxwell's text: If, as in the astronomical system, the unit of mass is defined with respect to its attractive power, the dimensions of $[M]$ are $[L^3T^{-2}]$. To motivate this, it is perhaps useful to be aware of some of the different systems of units for electromagnetism used historically. One of the units of charge commonly in use ...

0

Consider two bodies, $m_{1}$ and $m_{2}.$ From Newton laws, we have: $$F=G\frac{m_{1}m_{2}}{s^{2}}\tag1$$ but $$F =m_{1}a,\tag2$$ and combining $(1)$ and $(2)$ two we obtain $$m_{1}a=G\frac{m_{1}m_{2}}{s^{2}}.\tag3$$ We also know that $$a=\frac{s}{t^{2}}.$$ Then, $(3)$ becomes $$m_{1}\frac{s}{t^{2}}=G\frac{m_{1}m_{2}}{s^{2}},$$ which, after a little bit ...

0

Faraday's law could not be explained from basic principles at the time of discovery. In that sense, it doesnt have an explanation. It was incorporated as a new law of nature, and included in what today are maxwell's equations.

0

electrons will only distribute evenly if the conductor is spherical. if it is not, coronal discharge will occur at sharp points until the voltage is too low to sustain it. of course, breakdown will occur even with a spherical conductor, e.g., van de graff generator, but much more charge will accumulate before any one point has sufficient potential to cause ...

0

For inside (and outside) the sphere apply Gauss' Law, which is $$\oint \vec{E}\cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_{0}}$$ Where $\vec{E}$ is the electric field puncturing the Gaussian surface, $A$ is the area of the Gaussian surface, and $Q_{enclosed}$ is the total charge enclosed in the Gaussian surface. So main thing is figuring out the total ...

2

The definition of Ampere is obtained by the below equation of force between two infinitely long parallel current carrying conductors. Where $F$ is force, $\triangle{L}$ is small length element, $\mu_0$ is absolute permeability of vaccum or free space, $I_1, I_2$ are current flowing through two conductors. By calculation we can obtain that ...

0

Well, those electrons tend to run into things, both other electrons and the atoms, a process called scattering. Scattering off the ions can generate phonons (lattice vibrations), and that transfers energy from the electrons to the lattice. Similarly, phonons can scatter off electrons. In thermal equilibrium the rates are the same forward and backward. If ...

0

You are basically confining two negative charged small spheres in a bigger, neutral sphere (this image is better since the you don't need to worry about quantum effects, and there's no need to consider unphysical situations). Since both spheres would repel each other, they'd try to maximize the distance between them, so they would set up in a configuration ...

1

First, your force equation is wrong, as you're missing the electric field. Wait what electric field? That's the point! A changing magnetic field induces an electric field $\nabla\times E=-\frac{\partial B}{\partial t}$, and this "pushes" the current. Note that the applied magnetic field is perpendicular to the circuit/wire, so that at least part of the ...

3

You are right in that a magnetic field is build up, which generates a electric field opposing the given potential. But the consequence is not an oscillation of current, but only a damping of the increase of the current. Therefore, if you have a Heaviside step function for the voltage, it'll result in an "exponential" increase of your current ($I(t) = I_0 ... 0 I drew the magnetic field vector for each wire in red, and the resulting one with pink color. black lines are for a reference so you can see how magnetic vectors are at 90 degrees to the wires. if you draw something like this for every point inside the solenoid you'll see how the magnetic field is formed. here's how it looks in real world: 3 Here's the logic (well a particular rendition): Recall that$n$is defined as the ratio of the speed of light$c$in vacuum to the speed of light$vin the given medium; \begin{align} n = \frac{c}{v} \end{align} Note that in a linear medium, Maxwell's equations are exactly the same as in vacuum, except\mu_0$and$\epsilon_0$are replaced by$\mu$and ... 0 Power is, in words, the rate at which work is done and work done equals the amount of energy converted from one form to another. For electric circuits, the power associated with a circuit element is the product of the voltage across and the current through. We can verify this via dimensional analysis: $$v \cdot i = \frac{J}{C}\cdot \frac{C}{s} = ... 0 You will force a voltage across the conductor in series with the internal resistor of the battery and since it represents a very very low resistance, a huge current will flow (see Ohm's law), eventually heating up the wire by Joule's law, possibly burning/melting it. Though the battery will be the first one to suffer, and it can be dangerous for some types ... 0 The English Wikipedia article for reluctance uses the term ‘magnetomotive force’. I like Tobias's answer better, so I'm accepting it, but I'm also recording this one. If somebody else adds another better answer, then maybe I'll switch! 0 Starting from the definition of power$$P=\frac{dW}{dt}$$We can solve for the work (after integrating). The thing to know is what power is in terms of current and voltage or resistance$$P=IV=I^2R=V^2/R$$Clearly,$P=IV$is what we want to use. Last thing is what is$V$for a inductor? It is$V=L\frac{dI}{dt}\$ From here I will let you (and future ...

Top 50 recent answers are included