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7

That looks like a Moire pattern to me. You have a camera with a grid of pixels on the imaging element and a screen with a grid of (colored) pixels. These elements don't line up exactly, so you get the odd patterns. Try taking another image with the camera slightly twisted along the lens axis or slightly angle the lens axis to the laptop. If it's a ...


4

The top magnet pulls on the bottom magnet with the same force that that magnet pushes up and vice-versa. Therefore, the pushing forces (going both ways) cancel. So yes, it does have have to do with equal and opposite reactions.


3

One of Maxwell's equations is $$ \nabla \times \vec{E} = - \frac{d\vec{B}}{dt} \, .$$ Consider an imaginary disk whose normal vector is parallel to the axis of the coil and which is inside the coil. If you integrate this equation over the area of that disk you get $^{[a]}$ $$\mathcal{E} = - \dot{\Phi}$$ where $\Phi$ is the flux threading the disk and ...


3

The poles of the magnetized bar (I assume it is magnetized along its main axis for simplicity), will attract the other bar independently of the contact point (with only minor difference in strength), just as a paper clip is attracted to the poles of a permanent magnet regardless of the paper clip's orientation. But the ends of the unmagnetized bar will ...


3

The basic electric motor you described (see here and here for more complete descriptions than that provided in the question) is interesting and quite subtle to treat quantitatively, and I have not been able to find a good explanation of it on the Internet. I will provide one below. Even if the OP has moved beyond this problem, I hope others might find my ...


3

Besides the electric field $\vec E$ and the $\vec B$ field there are two other macroscopic fields, the displacement field $\vec D$ and the magnetic field $\vec H$. In a vaccum, $\vec D= \vec E$ (up to a scaling constant) and $\vec H = \vec B$ (up to a scaling constant). The magnetic field $\vec{H}$ is often what you make with a permanent magnetic, and it ...


3

Have a look at the Wikipedia article on the left hand rule. It says: The direction of the electric current is that of conventional current: from positive to negative.


3

The one tells you only the magnitude of the induced field. The minus sign associated with Lens' Law is a reminder of a procedure for determining the directional sense of the induced EMF (tending to opposed the applied change in flux). In a differential vector-calculus form the law can be written $$ \nabla \times \mathbf{E} = -\frac{\partial ...


3

Here's some pictures to explain: For the case of constant velocity: And for the case of accelerating (the case you are actually interested in): And, one final note to sum it all up, the induced EMF is the coil is $\left(-\frac{d\Phi}{dt}\right)$. I chose my $\vec a$ (the area vector) pointing "to the right" which means that the positive sense of ...


3

One way to look at this is to look at the stored energy in a magnetic field. Stored energy goes with the square of the field. And the fields of multiple magnets are just the sum of the individual fields. This means that if you take two magnets and look at their individual field maps, then they will be trying to move to the situation where there is less ...


2

There's no real symmetry argument that can explain why the $z$ component is zero. This is because it doesn't have to be: there can always be a uniform magnetic field added to the existing field, without affecting Maxwell's equations. They are all differential equations in the fields, like the following: $$\nabla\times\mathbf{B} = 0$$ which is Ampere's ...


2

I know that Wikipedia is not the best source for reference, but according to this page the superconducting parts that are cooled by liquid nitrogen in most cases. Most of the superconductors are High TC ones, which still needs to be cooled. Here are some links related to this topic: Toy train Video Paper on HTS (High Temperature Superconducting) Maglev ...


2

E.M.F(induced)= -(rate of change of magnetic flux). Which implies that the direction of E.M.F depends upon the sign of the Rate of Change of magnetic flux. When the magnet is moving into the coil, magnetic flux is increasing(Because magnetic feild(and thus flux) are greater nearer we get to the poles), while when it is coming out magnetic flux is ...


2

In general, Ampere's law does not necessarily give the value of the magnetic field. It only gives the integral of the field along a closed path. That integral can sometimes be used to deduce the magnetic field at any given point, but only if you know something about the magnetic field from symmetry or other considerations. For example, along a circular ...


2

The mathematical model for classical electromagnetism just doesn't forbid magnetic monopoles by construction. Consider an arbitrary vector field $X$ in 3d. Such a vector field is totally characterized by its divergence and curl. Suppose the following is true: $$\nabla \cdot X = \sigma, \quad \nabla \times X = Y$$ Then knowing $\sigma$ and $Y$ ...


2

So many ways to be confused... I will try to tell you how I think about these things. When I have a coil (one turn, N turns...), and I try to change the magnetic field through the area that the coil surrounds, then I have to have lines of B field "cross" the wires into the area. This may not be a scientifically accurate way to think of it, but it's very ...


2

It is not that there is no "interaction" - at any point in space, the two magnetic fields will add up with the resultant pointing in another direction. In other words, the magnetic field caused by the wire will appear to distort the externally imposed field. However, this distortion is radially symmetrical: you can think of the magnetic field lines as ...


1

Let us imagine that the charge carriers in the rod are electrons (negatively charged). An electron moving to the right is equivalent to a (conventional) current to the left. Alternatively, you can use a "left hand rule" for electrons (since the current is to the left when the motion is to the right, you can represent electron motion with the thumb of your ...


1

$ \renewcommand{\div}{\vec{\nabla} \cdot} \renewcommand{\curl}{\vec{\nabla} \times} \newcommand{\e}{\vec{E}} \newcommand{\b}{\vec{B}} $The answer is yes. We know that $\div \e=\rho$ and $\curl \e = -\dot{\b}$. The the helmholtz decomposition theorem tells us $\e$ can be written as the sum of a irrotational piece $\e_\mathrm{ir}$ and a solenoidal piece ...


1

In analogy with Gauss's law for the electric field $\nabla\cdot \vec{E}=\rho/\epsilon_0$, the flux of the gravitational field through a closed surface is proportional to the mass contained inside the surface. There is an approximation to General Relativity called Gravitoelectromagnetism (see Wikipedia page of this name. It's relationship with Newton's law ...


1

Actually you can formulate Gauss's law for the gravitational field as well: $$\oint_S \vec g \cdot d\vec A=-4\pi G M, $$ where on the left you have the gravitational flux through a closed surface and $M$ is the mass inside the volume. $G$ is the gravitational constant. When you call this quantity on the left $\Phi_G$ and write the mass as an integral of ...


1

Yes, of course that if a field - magnetic field - is able to make a bar magnet rotate or move, it is doing work. The statement that magnetic fields don't do any work only applies to point-like pure electric charges. - Lubos Motl


1

Suppose we hold two bar magnets with opposite poles facing each other a certain distance away. We let go and the magnets fly towards each other. We observe that the magnets gain kinetic energy. Then work must have been done on their masses. I'm afraid not. You do work when you pull the magnets apart. That's when you add energy to the system. When you let go ...


1

To use rules without knowing what is the reason is boring. See my paper about vector product for Lorentz force, for generators and for electric drives, in a reduced form for perpendicular vectors only. If one isn't sure that this equations could be derived see this answer from mathematicans. See my answer Why does one call $B$ the magnetic induction? too. ...


1

Outside of the wire where $r>R$, the curl of the B-field is zero (no current density) and inside, it is non-zero. Using symmetry, we know that if $\vec{j}$ only depends on r, which is often the case, then the B-field must also only depend on r. $\vec{B}=\vec{B}(r)$. If we look at the curl in cylindrical coordinates, we find that $\nabla \times ...


1

According to Lenz's Law, an EMF generated by changing magnetic flux in a coil creates an electric current that flows in the direction which generates an induced magnetic field opposed to the changing flux that produces it. So, for example, as the north pole of a magnet passes through a coil, current travels counter-clockwise in the coil (looking in the ...


1

What you seem to be referring to (a guess from your mention of quantum tunneling) is perhaps near-field magnetic induction. Here's a summary - magnetic coils have a near-field region around them, the primary contriubtion to which involves minimal radiation of power. This region is dominated by the response of the coils (for example) to the fields, and is ...


1

Layman's Explanation: Long ago the magnet was far far away and the magnetic flux through the coil was essentially zero. If you measure the induced emf as a function of time with time as the x axis and the induced emf as the y axis then the area under the curve an above the x-axis between the vertical lines $t=t_1$ and the vertical line $t=t_2$ is the ...


1

It is said to be perpendicular to the surface in the outward direction.


1

Why does the s-pole induce an emf in the opposite direction to the one induced by the n-pole when a bar magnet falls through a coil? I'm sure that a complete description in 'layman terms' is possible, for this we haqve to talk about basics of EM induction. You use a metallic wire which is bended to a coil. Inside the wire are available valence ...



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