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In refraction and reflection the incoming electromagnetic wave causes the electron density of the refracting material to oscillate. This happens because at any point in space the wave produces an oscillating electric field (and magnetic field, though that isn't relevant here) so any material that has a non-zero polarisability will respond by developing an ...


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No, a magnetic monopole a la the Dirac string does not "violate" gauge symmetry. Rather, the statement "we have a magnetic monopole" means only that we are forced to consider the gauge theory not on the whole spacetime, but on the spacetime with the location of the magnetic monopole removed. Why? Because, at the location of the magnetic monopole, the curl of ...


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The force on a charged particle is called the Lorentz force, and it is give by: $$ {\bf F} = q({\bf v} \times {\bf B}) $$ where the $\times$ symbols means a cross product. This means the force ${\bf F}$ is always at right angles to the direction of motion ${\bf v}$, and therefore the work done on the charged particle is zero. The Lorentz force can ...


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A battery is no capacitor, and the actual charge stored in the battery terminals is very low. When you connect the anode of one battery to the cathode of another, that charge is transferred very quickly, and the voltage drops to zero. When you connect anode and cathode of the same battery, a chemical reaction takes place, and charges flow inside the battery ...


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Well you want to go from QFT to Classical mechanics. Let's do this in three steps 1. QED to Dirac Equation QED lagrangian with electric dipole is $\mathcal{L} = \bar{\psi}\left(\gamma\cdot\Pi - \frac{\mathrm{i}d}{2}\sigma^{\mu\nu}\gamma^5 F_{\mu\nu}\right)\psi\\$ Where $\Pi\equiv \partial - \mathrm{i}eA$. This implies the hamiltonian $$\mathcal{H} = ...


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There are two questions.. 1. What is the concept of magnetic monopole here? 2. Why Maxwell's eqns remain unmodified? They are not elementary particles as anticipated by P.M.Dirac. But the concept comes from the non-zero divergence of magnetization field. In electrostatics when you have $\nabla\cdot \mathbf{P}$ (polarization vector) not equal to zero, you ...


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Witricity uses a rapidly changing magnetic field to induce a current in a receiver (coil). This is based on Faraday's law of induction: $$U = - \frac{d}{dt} \int \vec B \cdot d\vec A$$ From this you see that a changing magnetic field is necessary. The Earth's magnetic field, however, is largely static (compass!) and certainly not fast enough to induce a ...


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The energy of a photon doesn't change when moving from one medium to another as pointed out by Andrew in a comment. Considering that $E = h\nu$, $\nu$ being the frequency of the photon and $h$ Planck's constant, we see that the frequency has to stay the same when going from one medium to another. Since the frequency is the same, then the wavelength of the ...


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If there are no charges inside the cylinder, then the potential obeys Laplace's Equation: $$\nabla^2V = 0$$ In cylindrical coordinates, that's $$\left(\frac{\partial^2}{\partial r^2} + \frac 1 r \frac{\partial}{\partial r} + \frac{1}{r^2} \frac{\partial^2}{\partial \phi^2} + \frac{\partial^2}{\partial z^2}\right) V = 0$$ Based on your notation, it seems ...


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The function $A_z$ is only a function of $r=\sqrt{x^2+y^2+z^2}$ and $\theta$ (the integral over $z'$ is not a function of $z$, but does depend on $\theta$). I.e., $$ A_z=\frac{\mu e^{-iBr}}{4\pi r}{\tt C(\theta)} $$ Since only (apparently) $A_z$ is non-zero we only need to be able to take the derivative w.r.t. $z$ to get $\nabla \cdot A_z$. $$ \nabla \cdot ...


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To answer your question think of how we get light from the far away galaxies. It is the same problem as you posed, though inversing the path of the light. Each photon that we get from a galaxy, travelled for billions of years (depending how far is the galaxy). However, take in consideration that the further we look into the sky, we have knowledge about the ...


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High-energy radiation tends to get randomly deflected rather than slowed down by a medium, making lenses impossible. It also will bounce off of individual atoms of a mirror unless it hits at a very steep angle, so mirrors are difficult. Diffraction gratings won't work because you can't make slits smaller than the atomic scale. In the worst cases, even a ...


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No, it does not imply that the surface is spherical and charged uniformly. Imagine a charged conducting shell of arbitrary shape. (An ellipsoid is a simple example.) Gauss' Law tells us that the charges in the conductor fly to the outside surface of the conductor, and the distribution of charges is such that the E-field inside is zero. But for a ...


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A partial answer could show you that the Earth's atmosphere is actually a big obstacle for a large range of radiation, starting from UV light to shorter wavelengths. X-ray astronomy is largely impossible on Earth, which is why such telescopes are always located in space. Long wave radio signals however can be picked up also through the atmosphere, air is ...


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Correlation between two variables (or objects) is, very simply put, how much a change in one variable affects or determines a change in the other. Replacing variables with spins, highly correlated spins would mean that, due to some interactions between them, a change in the direction of one spin will cause a change in the direction of the spin it is ...


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According to Faraday’s Second Law, the magnitude of emf induced in the coil is equal to the rate of change of flux that linkages with the coil. if the flux change in both coils are same then emf created will be same.(but current can be different). http://electrical4u.com/faraday-law-of-electromagnetic-induction/


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When they say "Do not ignore electric force", they mean that there is both a magnetic and an electric force on the electron/positron, and you should not forget the electric force. In other words, you are asked to compute, for the $\vec v_+$, $q_+$ of the positron, the effect on the electron of its $\vec E$ and $\vec B$ field. Fortunately, 5 keV (kinetic ...


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According to classical mechanics the electrons moving outside an infinite solenoid do not feel the magnetic field. This is because the force they experience, according to the Lorentz law, depends only on the fields and not on the potentials. Thus according to classical mechanics the electrons beams passing from the different sides of the solenoid will move ...



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