# Tag Info

5

The force on the conductor must be zero. We will solve the problem in two steps. First, we will write down the external force $d\mathbf{F}$ on each infinitessimal charge $dq$ in terms of the external field $\mathbf{E}_{ext}$ and then we will integrate $d\mathbf{F}$ to get the total force. Note we need only consider the external force (i.e., the force from ...

5

Yes. There is a standard way to generalize to field theory. Let a theory of $n\geq 1$ fields $\phi^i$ with a Lagrangian density $\mathcal L = \mathcal L(\phi^i, \partial_\mu\phi^i)$ be given. Here we use that standard abuse of notation in which $\phi^i$ denotes the vector whose components are the fields; $\phi^i = (\phi^1, \dots, \phi^n)$. To obtain the ...

5

The are various crystal forms that iron and steel can adopt, the common ones being ferritic, martensitic and austenitic. The ferritic and martensitic forms are ferromagnetic (or just magnetic in everyday terms) while the austenitic form is not. So it isn't simply that iron is magnetic and steel isn't, it is specifically austenitic iron and steel that isn't ...

3

The reason is not quite as intuitively put as for ropes, but it is essentially to make the fields consistent with the electromagnetic boundary conditions, which in turn can be traced to (1) Kirchoff's voltage law and (2) no conduction currents can flow in a dielectric. Consider a tiny, thin rectangular loop running parallel to the interface with one half ...

3

I don't think it is that tough to analyse. If a conductor is present in a uniform electric field then there will be redistribution of charges to counter Electric Field inside the conductor (so that the net field inside the conductor is zero). However in uniform electric field this redistribution of charges will not cause any net force on the conductor. Why? ...

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Is there an equivalent formulation of classical electrodynamics in terms of action at a distance that is completely equivalent to the formulation in terms of fields (Maxwell's equations)? Yes, but only if special boundary conditions on the fields are assumed. For example, if the fields are purely retarded (wiki: Retarded and Advanced Potentials), one ...

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I will set constants like $c$ equal to one. Then he starts with the normal relativisitc lagrangian, $\mathcal{L} = -\frac{1}{4}F^{\alpha \beta} F_{\alpha \beta} - A^\alpha J_\alpha$. Translating this into non-relativistic language, we get $\mathcal{L} = \frac{1}{2}(E^2 - B^2) - \phi \rho + \mathbf{A} \cdot \mathbf{J}$. Now at this point he seems to ...

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A couple problems with your development: 1) (minor) You lose a couple factors of $\mu_0$ when you substitute for $B$ in the expression for $u_B$. 2) (major) You lose a minus sign in calculating the radial force; by your expression, the force should incorrectly tend to compress the coil, not expand it. What's going on? When mechanical work is done, the ...

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Partly this answer is just gathering together the comments above, though there are a couple of points that haven't been mentioned. Firstly, as mentioned in the comments electromagnetic waves do gravitate and the links in the comments cover this well. In the early universe (for the first 47,000 years after the Big Bang) EM radiation was the dominant ...

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So, I assumed a pair of electron and proton to behave as a dipole Classically, this is correct. It would behave as a dipole. But in quantum mechanics we don't talk about a localized electron, but about orbitals. This is because particles have an associated wavefunction $\psi(t)$, which tells you the probability of finding your particle at a particular ...

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If the wire is flexible, you could change its bounded area $A$, thus changing the magnetic flux. I'm imagining a "closed" loop where the two ends of the wire meet up. In your case of case of a uniform field that's perpendicular to plane of the wire, $\Phi_B=\pm BA$, depending on your choice for the direction of the corresponding area vector. Then, if you ...

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I believe the term you're looking for is "Magnetostatic Energy". Magnetostatics is the field that studies static (constant) magnetic fields, much like electrostatics. For a uniform material the magnetostatic stored potential energy is: $$E_{\mathrm{ms}} = \frac{1}{2}\mu_0 \int_V \mathbf{M} \cdot \mathbf{H}_{\mathrm{ms}} d^3 r$$ You can find a full ...

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