# Tag Info

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Read the blue box in NCERT on page no 219 with the heading "ELECTROMAGNETIC DAMPING". If we do the experiment in a aluminium pipe, then due to eddy curretns the glass bob will reach first as the motion of spherical bob is opposed. BUT if we do the experiment in a PVC/Plastic pipe or in air...both will reach at same time.

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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.

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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 ...

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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 ...

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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.

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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.

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I assume that by zero frequency, you mean zero momentum transfer. Zero momentum transfer corresponds to the $k=0$ value of the Fourier transform. The value of this part of the fourier transform is the integral of the scattering strength over all space. So you can think of this value as being the total amount of stuff that is there. Another thing to keep in ...

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In fact a sharp boundary is not required. It is sufficient that the charge density $$ρ(\bf{x})=∂_{x}⋅E(x)=[∂_{x}ɛ⁻¹(x)]⋅D(x)$$ is not constant. The wikipedia article is somewhat outdated. A fairly recent discussion (of which I was one of the authors) can be found in B. Lastdrager, A. Tip and J. Verhoeven: Phys. Rev. E 61, 2000, p 5767. It presents a detailed ...

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I am not sure if my answer is correct, from what I understood: (i) At the relativistic limit, $m<<E$, so the second and third terms in (6.15) will be negligible, just as you said. (ii) P&S is aiming at $\hat{k}$ parallel to $\mathbf{v}$ or $\mathbf{v}'$ and integrating around $\theta=0$, since (6.15) peaks there (also ref my comment below ...

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1) one way to find out is to measure motion of charged particles in their own fields on the microscale and infer which EM fields give best agreement with the observed motion. This was never done, because there are always EM fields of other sources and the motion is hard to measure accurately enough. 2) it depends on what you mean by "force is time-reversal ...

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A "point charge" does not exist (it is only a mathematical abstract). A charged particle exists, but it has a finite radius ( even though the radius may be very small) therefore it is NOT a point charge. If a charge exists, it must have a radius. The solution to the self-force is to treat the charge as a sphere, and use its radius to do the required ...

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The system solves the Euler-Lagrange equations of the Lagrangian: $$L\left(\phi, \frac{d\phi}{dz}\right) = \frac{1}{2}\left(\frac{d\phi}{dz}\right)^2 - 4\pi \int_0^{\phi(z)} \rho(s) ds$$ since the corresponding Lagrange equations are: $$\frac{d^2\phi}{dz^2}= -4\pi \rho(\phi)\:.$$ As the Lagrangian does not explicitly depend on "time" $z$, "energy" is ...

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The expressions above deal with macroscopic electrodynamics, the theory of electromagnetism in some matter. It stands as convenient extention of Maxwell-Lorentz vacuum elecrodynamics, for studying fields in both conducting and insulating media, but it bases on same picture of charges in a free space. It would be strange magnetic monopoles to occur. It is ...

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Here, it states that for a tightly wound coil of wire, composed of $N$ identical turns, each with the same $\Phi_B$, Faraday's law of induction states that: $$\epsilon = - N \frac{d \Phi_B}{dt}$$ So the Electro Motive Force (voltage) $\epsilon$ induced will be proportional to the number of turns $N$. And be proportional to the rate of change of magnetic ...

3

I'm not sure if this problem was ever solved in classical electrodynamics. However, it is (somewhat) solved in quantum field theory electrodynamics (QED). In QED, self-interaction has noticeable effects on quantities such as the observed mass of a particle. Furthermore, the self-interaction effects create infinities in the theoretical predictions for ...

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I suppose my first major question is simply, has this problem been solved yet? After a bit of research I came across the Abraham-Lorentz force which appears to refer exactly to this "problem of self-force". As the article states the formula is entirely in the domain of classical physics and a quick Google search indicates it was derived by Abraham and ...

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There is a confusion in this question between classical electrodynamics and quantum electrodynamics. The multipole expansions of the vector potential are classical. A photon is a single elementary particle, and elementary particles are described with quantum mechanics. Physics is continuous and quantum mechanical ensembles of photons do build up classical ...

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I think you are confused about what the Hall effect does. You ask: "Why does the magnetic field stop them from continuing to flow?" The answer is: it doesn't. Take this setup: Here, the magnetic field is pointing up, in the +z direction. The conventional current in the purple conductor is flowing towards us (electrons going in the opposite direction). ...

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$\newcommand{\B}{\vec{B}^\times} \newcommand{\e}{\vec{E}} \renewcommand{\b}{\vec{\beta}} \newcommand{\bv}{\vec{B}}$ The field tensor can be written $\begin{pmatrix} 0 & -\e \\ \e & \B \end{pmatrix}$, Where $\B$ is the dual tensor to $\vec{B}$ defined by $\B \vec{v} = \vec{B} \times \vec{v}$. Equivalently, $(\B)_{ik} = \epsilon_{ijk} B_j$. Note ...

2

We start with a collision between two particles. Particle 1 has mass $m_1$, momentum $\mathbf{p}_{\rm lab}$, and total energy $E_{\rm lab}$ while Particle 2 (mass $m_2$) is stationary. Particle 1 then scatters off at an angle $\psi$ while particle 2 scatters off in angle $\zeta$. By using a transformation of coordinates, we can jump into a frame in which it ...

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As Jan noted, the Hamiltonian should have a minus sign: $H=\frac{(p-qA)^2}{2m}$ where $p$ is the canonical momentum, and the expression $p-qA$ is the kinetic momentum $P$. A homogenous magnetic field is an interesting case, because the vector potential in a given gauge does not exhibit translation invariance, but the physical system clearly does. The ...

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Maybe an example helps. Let $B$ be a constant magnetic field. Then we can take $A=\frac12B×x$. Now $$\frac{(p+qA)^2}{2m}=\frac{p^2}{2m}+\frac{q}{2m}(p⋅A+A⋅p)+\frac{q^2A^2}{2m},$$ and $$p⋅A+A⋅p=l⋅B$$ where $l=x×p$. Thus $$\frac{(p+qA)^2}{2m}=\frac{p^2}{2m}+\frac{q}{2m}l⋅B+\frac{q^2A^2}{2m}.$$ Here we recognise the $l⋅B$-term as the Zeeman term. If we now ...

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Hamilton's equations state $\dot{P_i} = -\frac{\partial H}{\partial q^i}.$In this case, this is $\dot{P_i} = -\frac{\partial H}{\partial q^i} = -\frac{\vec{P}}{m} \cdot \frac{\partial \vec{A}}{\partial q^i}.$ So the canonical momentum is not conserved.

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Hint: $P$ is conserved if it is not explicitly time-dependent and if its Poisson bracket with the Hamiltonian is zero. So you just neet to check that: $$\{P,H \}=0$$

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