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This paper by Lyutikov and McKinney describes this precise scenario. I'll provide a brief summary of their results here. One of the crucial underlying assumptions made when deriving the no-hair theorem is that the region outside the event horizon is a vacuum. However, a rapidly spinning, magnetized neutron star can induce vacuum breakdown, leading to the ...


8

Magnetars are neutron stars where the extremely strong magnetic field originates from something called the magnetohydrodynamic process in the turbulent extremely dense superconducting fluid, that persist even after the neutron star settles into equilibrium. These fields then persist due to persistent currents in a proton-superconductor phase of matter that ...


4

Clearly $\exp(\hat A)\hat A\exp(-\hat A)=\hat A$ by Baker-Campbell-Hausdorff. Here, set $\hat A=\hat H=-it\mu\vec B\cdot \vec \sigma/\hbar$. Note that \begin{align} \exp(-i t\mu \vec B\cdot \vec \sigma/\hbar)\ne \exp(-it B_x\sigma_x)\exp(-it B_y\sigma_y)\exp(-it B_z\sigma_z) \end{align} or any other kind of factorization suggested by your snippet of Mma ...


3

Yes, electromagnetic energy is stored in both E and B fields. The EM energy density is $U=\frac{E^2+B^2}{8\pi}$ in Gaussian units. Integrating U gives the energy.


3

The heat comes from current flowing through the wire, since we are told the wire has resistance. So it's just like your usual DC circuit with heat being dissipated in a resistor caused by charge carrier collisions with the molecules of the resistor. The energy comes from whatever is causing the varying magnetic field. This causes an induced EMF that induces ...


3

Suppose you have higher $n$ (i.e. more free charges per volume). Then these many charges need to flow only slowly (i.e. with smaller velocity $\mathbf{v}$) in order to make the same current $I$. Now the Lorentz-force ($\mathbf{F}=q\mathbf{v}\times\mathbf{B}$) is smaller when you have low velocity $\mathbf{v}$. And therefore you get a smaller Hall-voltage ...


3

You are quite correct that a magnetic field cannot do work on charged particles, so I find that your question is a very good one. The answer is that anything that causes a magnetic field to change is liable to introduce a change in the electric field too. This is the content of Faraday's law of electromagnetic induction. In particular, when a magnetic field ...


2

As SuperfastJellyfish points out in the comment above, a moving charge creates both an Electric Field and a Magnetic Field (a very good reference for this is the Feynman Lectures on Physics). You can't always make an electric field "disappear". There are many ways to see it, but I feel the simplest is to realise that the quantity (call it $Q$) $$Q =...


2

Sometimes it's a good method to see things on the particle level. In case of capacitors the energy is stored in electric field,... In detail, electrons are separated and accumulated on one side of a capacitor. For the separation is needed energy and connecting the two sides of the capacitor the electrons flow back to the other side. Having some device (...


2

"magnetic field does not do work" But when the magnetic field changes there is an electric field around the changing magnetic field. The electric field does work on charges, for example those in a coil.


2

edited No, it seems that the law of action is true for the cherry-picked case of parallel wires as shown by @Sagigever's response. For instance, this is certainly not true for the case of magnetic Lorentz force exerted by two charges (yellow in figure) moving perpendicular to each other (along the blue and red lines). In this case, the $\vec{F_{12}}$ is ...


2

Electromagnetic waves, also known as light, can exist without any charged particles at all. You could imagine a universe where there is no matter at all, but there are still electromagnetic fields due to light waves bouncing around. In our universe, however, pretty much all electromagnetism around is due to charged particles. Sometimes the fields travel very ...


2

Okay let's start with the changing magnetic field. We know that in the region of time varying magnetic field , electric field is induced and ( they have some energy released by the source and it depends on the process involved in bringing the change) and when a coil is placed in this region then the electrons experience a net force as shown in the figure. ...


2

I would encourage anyone to draw any lines that help either them or others to get a good understanding. But the reason field lines are so much used in electromagnetism, and much less used in gravity, is because they nicely capture a mathematical property of a field whose divergence is zero (on a flat spacetime background). The zero divergence of the field ...


2

Linear means that if you consider a combination of sources such as $\alpha_1I_1+\alpha_1I_2$, you will obtain the corresponding combination of magnetic fields $\alpha_1B_1+\alpha_2B_2$.


2

$B_z$ decreases with $z$ because the field lines are getting farther apart. That’s the important consideration. The field is going from intense at the tip of the point to spread out over a large area.


2

According to the flux approach, Φ=𝑐𝑥2𝑙 This step is incorrect. If I take any dx element at a distance x from the AB, then area of element is $ldx$ and magnetic field $$B=cx\tag1$$. Then Flux $\phi$ is given by: $$d\phi = B dA = cx l dx$$ Integrating the expression: $$=>\phi = \int cl xdx$$from x=0 to x=x, we get: $$\phi = \frac12 clx^2$$ EMF $\...


2

My question is that if magnetic field cannot do work, then what does the energy signify? The energy stored in the magnetic field of an inductor can do work (deliver power). The energy stored in the magnetic field of the inductor is essentially kinetic energy (the energy stored in the electric field of a capacitor is potential energy). See the circuit ...


2

From Jackson's "Classical Electrodynamics" third edition in Chapter 5 section 16 "Energy in the Magnetic Field". He talks about how "the creation of a steady-state configuration of current involves an initial transient period during which the currents and fields are brought from zero to the final values. For such time-varying fields ...


1

ToAmpere's law can be written (for static fields) $$\oint \vec{H}\cdot d\vec{l} = I_{\rm enc}\ ,$$ where $\vec{H}$ is the "magnetic field", which is integrated around a closed path, and $I_{\rm enc}$ is the current enclosed by that path. From this equation it is clear that the H-field has units of A/m (in SI units). It seems quite likely that you ...


1

Flux is defined as field strength TIMES area.


1

Probably not just as you ask, but there are interesting ideas for visulazations in Visualizing spacetime curvature via frame-drag vortexes and tidal tendexes. II. Stationary black holes by David A. Nichols, Robert Owen, Fan Zhang, Aaron Zimmerman, Jeandrew Brink, Yanbei Chen, Jeffrey D. Kaplan, Geoffrey Lovelace, Keith D. Matthews, Mark A. Scheel, Kip S. ...


1

G. Smith gave the answer (brute force method) but there is a trick to do the problem other way around which is usually done on UG level. Pass current in bigger coil and find the mutual inductance ($M$) between the two coil. The flux linked due to current in smaller coil will simply the product of $M$ and current in smaller coil. I would give one simple ...


1

Is their any way to form a mathematical model of this situation which is easy to compute as well as to visualize? If $r \ll R$, you can model the smaller coil as a point magnetic dipole with magnetic dipole moment $IA$. ($I$ is the current in the loop and $A$ is the area of the loop.) The field of such a dipole is reasonably simple.


1

In this case we have to assume that the support for magnet keeps it horizontal while allowing it to rotate around a vertical axis. Since it is not in line with the horizontal component of the earths field, there will be a torque acting on it of: τ = MBsin(θ) which is trying to bring it into alignment. The θ is the angle between the true north and the ...


1

See the definition of flux is $$d(\phi) = B . d(S)$$ So wherever $$B = 0$$ $$\implies d(\phi) = 0$$ So at positions where Field is zero or where field lines do not pass, flux is zero.


1

assume that the fringing fields around the gap can be neglected then from the continuity of the flux you have $\Phi=BA$ through every cross-section $A$, be it in the gap or in the core. Since $B = \mu_g\mu_0 H_g = \mu_c\mu_0 H_c$ where $\mu_c$ is the core's permeability and for air gap $\mu_g=1$; also from Ampere's law we get $NI=H_c\ell_c + H_g \ell_g$. Now ...


1

The $\mathbf r$ vector points from your observation point to the point on the wire which is generating $dB$. If your observation point is $(x,y,-h)$ and $dB$ is being generated at $(x',y',0)$, then you would have $\mathbf r = (x-x')\hat x + (y-y') \hat y + (-h) \hat z$. It's true that at the single specific point along the wire which you mention, $x-x'=0$ - ...


1

I suspect you were either told that the wave was in a vacuum or another non-conducting medium, in which case it can be assumed that $\vec{J}=0$. If not, then I think you can still assume $\vec{J}=0$ if you are told $k$ is a real number, because solutions to Maxwell's equations in conductors are of the form $$ \vec{H}(\vec{r}) = H_0 e^{jkz} = H_0 e^{-k'z}e^{...


1

Yes this is true, for example Lets take 2 parallel infintite wires carrying a current $I_1$ and $I_2$ that located distance $d$ from each other We know from biot savart law that magnetic field of wire at distance $d$ is $$\frac{\mu_0 I}{2\pi d}\hat \phi$$ so now we can calculate to Force of wire 1 on wire 2 by your formula $$\vec{F} = I \int \vec{d}\ell \...


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