# Tag Info

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The battery terminal appears "small" when you hold it in your hand. But from the perspective within the wire, close to the terminal, the battery terminal appears large. The electric field close to a large object like this is roughly constant (pitcure the constant field between capacitor plates). Further away from the terminal, inside the wire, the charges ...

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Length contraction by itself doesn't give the result. But let's step back a tiny bit further and notice that Maxwell and Lorentz do not together predict that two wires feel an acceleration in any particular direction just because they have current flowing between them. And that stems from the fact that Maxwell is a Partial Differential Equation (PDE) and so ...

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No, it will rotate at a speed determined by the load. Witness that the current in, and thus the magnetic field produced by the stator coils is either in-phase with or anti-phase with the rotor current, with the $\pi$-phase change triggered by the split ring commutator. So the torque in each half of the rotor's rotation will throb at twice the AC line ...

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The glass sphere will reach the ground earlier, because as the metallic sphere is freely falling within the magnetic field of the earth, a large amount of current will be induced on the metallic sphere called eddy current. The eddy current will oppose its cause and the cause is the acceleration of the metallic sphere. Now it will stop the metallic sphere to ...

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Special relativity is indeed the origin of magnetism.* As you correctly note, the actual speeds involved in conduction currents in real-life conductors is at crawling pace and it is much smaller than the speed of light. However, most objects tend to be neutral and therefore exert no electrostatic forces on each other, in which case the magnetic contribution, ...

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I had read here that magnetism arises from a current because of the special relativistic effect associated with the speed of the moving charges in that current. Yes, you can read about that in apparently authoritative sources, but it just isn't true. It doesn't make sense either, because charged particles move rotationally in a magnetic field, and the ...

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It's because of Lorentz contraction. From the reference frame of the wire, the charge appears to be moving. From the reference frame of the charge, however, it's the wire that's moving. If the wire is moving at all, then Lorentz contraction will shrink the wire in that reference frame a minuscule amount. This results in an unbalanced number of charges in ...

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In case of varying magnetic fields the electric field is considered as nonconservative. Thus, it is simply not possible to define a potential for induced electric field (See also this post with answers). Alternatively, one can think of the induced current as charges moving in the circle of radius $r$. Through every part $r \,d\phi$ flows the same amount of ...

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The EMF created by a changing magnetic field is not considered to arise from a potential. This can easily be seen because when there is an emf, a charge can move around in a complete circle and dissipate energy the whole way around, but a potential cannot drive a charge around in a circle, because potentials are conservative. The two pieces of the electric ...

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This is a very simple problem to tackle. The points A and B are connected by a conducting wire and nothing else. Therefore, there is no potential difference from A to B. The magnetic field is a red herring. We are not told otherwise, so we assume the wire is an ideal conductor. That means there is never a potential difference across it. Remember Ohm's Law ...

1

Here, there is a time varying magnetic field at work, and it's flux through the given loop changes. Thus there is a non-Electrostatic Field induced along the wire, proportional to the rate of change of the flux. However, it is a non-electrostatic field", for example the closed integral over a path for this field isn't zero. Also, the line integral of this ...

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Perhaps the easiest way to see that there can't be a potential difference between $A$ & $B$ is a symmetry argument. You're tempted to say that $A$ is at a higher potential than $B$ so that current will flow from $A$ to $B$. But continuing along the loop, I find that current must also flow from $B$ to $A$, which would lead me to conclude that $B$ is at a ...

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Your question is unclear. If you add transverse dimensions, your problem is no longer 1-dimensional. If you mean adding transverse components of your interested quantities, then there is no change needed to the usual FDTD method. For example, if you are analysing the propagation of an electromagnetic plane wave in one-dimension, say the positive $z$ ...

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This is assuming the earths magnetosphere is the minimum required to shield an object from cosmic rays which is, in fact, incorrect. Here is the what you need to figure out to calculate a satellite system to do what your proposing .. 1-what is the required minimum strength of a magnetic field so that it deflects virtually all cosmic rays and ...

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After some more thoughts, I will attempt to "answer my own question." Here is what I think is going on: Since the 40 game solutions all have the same length, I am guessing their total areas are close enough so they all form inductances within some specified range. For the two-loop example (which is not one of the intended game solutions): I suspect ...

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Yes, in electrical circuits of only passive elements (Resistors, Capacitors, and Inductors) only the Resistors dissipate power (as heat). Active elements like transistors can also dissipate power, and if the currents in the circuit are changing with time, then power can be radiated away in electromagnetic waves. Therefore, the electrical resistance of the ...

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Yes it is true. The only DC magnets that use "no" power are superconducting magnets (like in MRI systems). Of course for those, there is significant power needed to keep the windings at superconducting temperature... and the cooling system will typically use several kW. "How much power does a junkyard magnet use" is not an easy thing to answer: but ...

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The angle of rotation is proportional to the length of the path the light ray spends inside the active material. This needs to happen because each bit of the path only 'knows' what's happening there and it does not interact with the rest of the material's optical activity. This means that the rate of rotation of polarization must be constant, i.e. that the ...

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Another important issue is impedance matching between the output of the antenna and the receiver input. The length being a full, half or quarter wavelength creates a resonance situation that helps to increase the sensitivity. But then the situation is less favorable if the received frequency deviates from a resonance frequency. For that reason designers of ...

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Antenna sensitivity for a fixed length depends on the wavelength of receiving radiation. Antennas are usually designed having a length of half or full wavelength or even quarter.

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I'm going to agree with you and guess that it involves magnetic induction, too. As for ensuring some selectivity and not lighting up whenever someone puts just any closed loop down, or puts down two closed loops, I think that there are ways of conveniently doing that. For example, there could be simple wire coil loops around each of the 25 individual dot ...

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having more metal bars wont really effect the magnetic flux as magnetic flux is related to the magnetic field and the area of the loop.As long as the current through the bars is constant the magnetic field will be constant and so will be the flux. one thing you could possibly do to vary the magnetic flux would be to use use a rotating device to change the ...

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I would use something like this - a stack of bars in parallel to the direction you want for the field, above and below the box: Since you have limited wire, you want to wrap the largest possible amount of metal bars: the ratio circumference / area gets better as the thing you wrap is larger. But there is no point to make it larger than the area of the ...

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The Maxwell equations do not remain invariant in form when changing to a rotating coordinate system and therefore predictions made from them, like the Larmor radiation formula, cannot be held to be true anymore. While that sentence is sufficient to answer your question, if you want to dive deeper, here are some quick resources: ...

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it is because line integrals of magnetic field through these portions is zero.Yes, the vertical paths are perpendicular to magnetic field

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Yes; as the integral has a dot product between magnetic field and length element. $$B \hat{x} . dl \hat{y}=0$$

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The definition of magnitude of induced emf(which produces induced current) is, $|e| = \frac{d\phi}{dt}\space$ that is the induced emf in magnitude is equal to change in magnetic flux (say of a coil), also, $\phi = BAcos\theta\space$ Hence the flux through the coil is changed when there is change in either Magnetic Field, $B$, The area of the coil inside ...

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You're really asking about "engineering" issues rather than "physics" issues. The equation you presented gives the B-field in a solenoid. From a physics perspective, that's it. Physicists don't have to "worry" about anything else but engineers - the guys responsible for actually getting things to work in the real world - do. Now as for the wire material, ...

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What's wrong is that you started with the waves $\pi/2$ out-of-phase. They should be in-phase, as this description shows, otherwise they will not obey the Maxwell Equations and you cannot use Poynting's Theorem (which itself is derived from the Maxwell Equations): Image credit: nde-ed.org. Doing out the expression with $E \propto \cos(\omega t - k x)$, ...

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Are you talking about a spinning particle traveling along the axis of two coaxial current rings? In that case, the force is dependent upon the quantity $\frac{\partial B}{\partial z}$ and losing all the kinetic energy doesn't have any effect on the force. Consider a simple scenario: if you throw a ball upwards, it loses all its kinetic energy when it ...

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Well in my book it says that " An EMF is induced in a circuit to oppose the flux change, while closed circuit is special case in which since the circuit is complete, a current flows"

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Clearly, there is a solution for q1 and q2 in terms of q. I think you just got a little careless with rushing through the algebra. What is the distance from the charge q at point C to the measurement point A? It's not 2d (as implied by your first equation) nor d (as implied by your second equation). Hint: Use the Pythagorean theorem.

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Charge $q$ creates electric field at $A$ that is directed along $AC$. You need to calculate this electric field using the Coulomb law, but you need to take into account its direction. Remember that electric field is a vector value.

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It is true that given an electric field, then you can define uniquely the charge density that created it, by Gauss' law, as you have done. But the converse is not true: given a charge density you cannot define uniquely the electric field that it will create since you have to solve a differential equation (again Gauss' law) to do that and each differential ...

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Roughly speaking, the conductivity is inversely proportional to the scattering rate $1/\tau$ (in the Drude model, the longitudinal conductivity is $\sigma_0=\frac{n e^2 \tau}{m}$). From Fermi's golden rule, $1/\tau$ is proportional to the number of density of states a quasiparticle can scatter into, which is proportional to the density of states near the ...

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when a straight conductor is moved in and out of magnetic field,a potential difference is setup because when conductor is in magnetic field,force is exerted on free electrons and work is done by them (elctric potential) and when conductor is out of magnetic field,no force is acting on charges and so work done is zero.In this way,a potential difference is ...

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Yes, inductance exist in a DC circuit. The problem here is the similarity between the words inductance and induction. Inductance is not about change. In fact, inductance is measured in Henrys, which is a Weber per Ampere. Hence, there is no change. In contrast, induction is about change and does not exist is a DC circuit. You would be amazed how many ...

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EM waves don't "stop" they just slowly become weaker as $r^{-2}$, so one could conceivably answer "forever." On the other hand, the wave will quickly become so dissipated/spread out that there isn't much to measure, so you have a practical limit where it won't be detectable. However if this is your intent, you haven't given us enough information to answer ...

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I don't think your argument works, for a few reasons. All you've done is shown that interference terms arise with a frequency given by the difference between the energy levels. The problem is that there's no connection to a photon here, which is what I assume you mean by "Bohr's equation." You are correct that you're seeing kind of the same effect, but the ...

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An EMF from a source is defined as a force per unit charge line integrated about the instantaneous position of a thin wire so for an electromagnetic source: $$\mathscr E=\oint_{\partial S(t_0)} \left(\vec E + \vec v \times \vec B\right)\cdot d \vec l.$$ Where $S(t_0)$ is a surface enclosed by the wire at time $t=t_0$ and the partial means the boundary, so ...

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This depends on exactly what you mean by non-uniform, or (equivalently) on how big the loop is. In particular, the important criterion is whether the field changes appreciably over distances that are about the same size as the loop. If the field changes throughout space, but the loop is small enough that the field doesn't change much from point to point on ...

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Broadband Emission from 400 to 2200 nm"...which means it will transmit all mixed wavelength of infrared light between 400 to 2200? Yes, this is correct. But the light intensity will likely not be equal at all wavelengths, but will have some dependence, which may be in your spec sheet. Note that 400 - 700 nm is the visible range. I don't know that ...

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Let's restate Faraday's Law of Induction carefully in integral form: $\varepsilon = -\frac{d}{dt} \int_S B \bullet da = - \frac{d \Phi}{dt}$ Where C is a closed curve, and S is any smooth surface whose boundary is C. So as you can see, no matter how the generation of electric field is interpreted (take the two scenarios in your question), the motional ...

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When you have a non-uniform magnetic field, the magnetic force felt by the charges in the loop $\vec{F} = q(\vec{v} \times \vec{B})$ will vary from point to point, and therefore so will the resultant torque $\vec{\tau} = \vec{r} \times \vec{F}$ in each point vary. The net torque on the entire loop can be found by calculating the torque in each point of the ...

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In principle yes, though in practice it's far beyond our current capabilities. The problem is that a charged black hole has to have a Schwarzschild radius greater than $2r_Q$, where $r_Q$ is given by: $$r_Q = \sqrt{\frac{Q^2 G}{4\pi\varepsilon_0 c^4}}$$ If this isn't true the black hole will be superextremal and this (probably) means it's unstable. For ...

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Yes, it does, A dynamo diriven in reverse simply changes the sign of the generated voltage. As a dynamo produces AC and changes the sign of voltage a few time each second, you wont notice that. So even alterating the rotation every minute has close to no effect (except from having to slow down and reaccelerate a potentialy heavy axis)

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Consider a current carrying rectangular coil with its length $l$ and breadth $b$ place in a non uniform magnetic field $\vec{B(x)}$ . Here I assume that magnetic field variying along x axis. Let the current through the loop be $I$ . Let the normal of rectangular loop make an angle $\theta^0$ with direction of $\vec{B(x)}$ . In such field loop experiences ...

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It has to do with the frequency response of the materials in the wall. Different molecules absorb different frequencies (or wavelengths) producing an absorption curve called the materials spectral response. Lots of materials are very absorptive in the frequencies typical in visible light but start to open up (get clearer) in longer wavelengths. Generally the ...

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In electricity and magnetism, we use the scalar potential to derive the electric field and the vector potential to derive the magnetic field because ∇⋅B=0 and ∇×E=0. IMHO there's a deeper reason than the mathematical expressions: the field concerned is the electromagnetic field. See section 11.10 of Jackson's Classical Electrodynamics where he says ...

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One of the defining characteristics of an electromagnetic wave is its wavelength (which is related to its frequency). Radio waves have wavelengths ranging from 1 millimetre to 100 kilometres, while light has wavelength on the order of hundreds of nanometres. Interaction between electromagnetic waves and objects can be roughly predicted with the relationship ...

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