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Ok I'll give this one a go - the reason is that the effect is just not strong enough. Andre Geim showed that in a strong enough magnetic field you can levitate a frog. Frogs are mostly water and water is diamagnetic. You can also buy a kit from your local electronics store that has a piece of (strongly diamagnetic) pyrolytic carbon and a neodymium magnet ...

3

What do you consider to be Ampere's law? I think the RHS is not the current that cuts some plane, it is actually a surface integral of the current density, over any surface that is bounded by the loop. In other words it does not matter what surface you define ; so long as it is bounded by the loop, Ampere's law works just fine. So usually, you choose the ...

2

No, your understanding is wrong: $B$ isn't proportional to $H$, the relationship is $\vec{B}=\mu_0 (\vec{H} + \vec{M}(\vec{H}))$ where $\vec{M}$ is the magnetization (see Wiki page of this name). And $\vec{M}$ saturates for precisely the reason you state: its maximum value is reached when all the magnetic dipoles in a medium are perfectly aligned with the ...

2

Using a process called interference, we can find wavelength, because the way that waves interfere is reliant of wavelength. Interference is based off of two key principles of waves: they are made up of peaks and troughs. When troughs overlap, they go lower. When peaks overlap, thy go higher. When a peak meets a trough, they cancel. Of course, the positions ...

2

Yes ... but let's be careful to understand that the sensation of touch is a psychophysical phenomenon. The electrons at the surface of an object "push" against the electrons at the surface of your fingers. The electrons never touch each other. Your skin deforms a bit, and the nerves in your fingers detect this deformation, and send a signal to your brain. ...

1

In the case of a planar loop, the surface that it bounds does not have to be a plane. It could bubble out. Ampere's Law is still valid (assuming the other conditions for validity are met). Ampere's Law applies to any loop, and any surface bounded by the loop. I guess I'd better add that I don't know what happens in pathological cases where, for example, ...

1

While your assumption of $J$s cancelling out in the intersection is valid, your conclusion that $B_1$ and $B_2$ are along the same line only holds along the line joining the centers of the circles. More generally, that is not the case; we can get a more complete solution (which also correctly takes care of the directions) by using the vector quantities ...

1

This answer contains some additional resources that may be useful. Please note that answers which simply list resources but provide no details are strongly discouraged by the site's policy on resource recommendation questions. This answer is left here to contain additional links that provide little or no commentary. Staelin's Electromagnetic Waves. This is ...

1

The electric and magnetic fields transform like a second rank tensor not a four vector. I would suggest you look in Jackson classical Electrodynamics for the transformations. One observer will see a moving magnetic field which will have an electric field component. At low velocity this will just be interpreted as magnetic induction.

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I'm not sure what you mean by macroscopic? The experiment was first done with silver atoms which, due to having 1 unpaired electron of spin 1/2, split into 2 distinct beams. This was a quantum mechanical event that was visible macroscopically. If by macroscopic you mean on a large object instead of a beam of atoms then I don't think it would be feasible or ...

1

Pay close attention to how you defined $x$: it's not the coordinate of the desired point on the $x$ axis, it's how far left of the positive charge that point is. So when your equation tells you $x = 4.83$, that means the desired point is $4.83$ units left of the positive charge. But, presumably the question you've been asked wants the coordinate of the ...

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In my opinion, your solution is correct. The book's solution is correct as well, but you would need to draw a different diagram for it to make sense. Their assumption is to take by default, r as lying on the positive x axis. In this way, the distance between the $Q_2$ and the required point is the mod of $r-2$. Since this expression is being squared, it ...

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For the moving wire situation in which $d\Phi/dt=Blv$, the wire is typically part of a larger closed loop of conducting material (a circuit). The change in flux is due to the changing area of this loop, which is immersed in a uniform magnetic field. Here's a plagiarized image illustrating this:

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$$\frac{\nabla'\times \vec{J}}{R}=\nabla'\times(\frac{\vec{J}}{R})-\frac{\vec{J}\times(\nabla'R)}{R^2}$$ Thanks to Prof. Y. F. Chen I was able to figure it out. While in the integral the first term on the RHS can be converted into a surface integral as below: $$\int\nabla'\times(\frac{\vec{J}}{R})d^3x^{'}=\oint(\vec{n}\times\frac{\vec{J}}{R})d^2x^{'}$$ ...

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Maxwell's equations of electromagnetism don't allow an electromagnetic wave to just sit there without moving in empty space (in the absence of electric charges). An electric field is created by a magnetic field changing in time, and vice versa, so the coupled fields have to be changing in time (i.e. propagating) in order to exist and sustain themselves.

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The Gravitoelectromagnetic equations are exactly the same as Maxwell's equations with $\epsilon_0$ replaced by $(-4\,\pi\,G)^{-1}$, so, to the extent that the GEM equations approximate the Einstein field equations, the behavior of mass is very much like that of electric charge. Here are the differences: The minus sign in the "gravitoelectric constant" ...

1

We have, for a point charge $q$ at position $\vec r(t)$: $$\rho(\vec x, t) = q\delta^3(\vec x - \vec r(t))$$ $$\vec J(\vec x, t) = q \frac{d\vec r}{dt}\delta^3(\vec x - \vec r(t))$$ Let us for now work without worrying about what the derivative (more precisely, gradient) of the delta function actually is. We will also enforce the convention that $\vec\nabla$ ...

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