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5

I guess you mean a level set? No, they're not. Magnetic field lines (which aren't real, but which we use to describe the vector strength of the magnetic field in a region of space) don't cross because the region around the crossing would have nonzero divergence, forbidden by Maxwell's equations. In regions of space where the magnetic field has zero curl — ...


4

Or at least "functions" as a monopole? I'm afraid not. A magnet is a magnetic dipole and there is, as far as I know, no configuration of magnetic dipoles that can give a monopole field.


3

As you may know, the friction is proportional to the normal force of an object or in this case the force of attraction between magnet and refrigerator. If your force is strong enough then the friction will be sufficient and the magnet will not slip (on earth the force of friction must exceed the mass of your magnet multiplied by 9.81 m/s). If we assume ...


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

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 ...


2

My question is do the byproducts(elementary particles such as higgs boson) emit photons as they are produced to be filmed on tape by the sensitive detectors? If you're imagining that the particles are emitting photons as the leave the collision center and that the tracks of the particles on the computer screen were recorded by collecting these ...


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 ...


2

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.


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

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 ...


2

No (or at least, not in three or more spatial dimensions). The reason is simple: in $d$ dimensions, level sets are $d-1$ dimensional surfaces. In particular, in two dimensions, level sets are lines, but in three dimensions, they're planes. But in any dimension, magnetic field lines are just lines, and therefore can't be level sets in three or more spatial ...


2

Q1: For photons of energies much less gamma rays, the quantum mechanical photon-photon interaction is negligible. This is consistent with the classical electrodynamic description where the principle of superposition holds (electromagnetic waves pass through each other unchanged, as well as through electric/magnetic fields). Q2: in reality, charge is defined ...


2

There are two main types of magnets to consider, permanent magnets, and electromagnets, when considering 'maximum' strength to weight ratio. Permanent magnets are exactly what they sound like, permanent, made from magnetic materials that work pretty much because at the atomic level, certain atoms and molecules can possess a Magnetic moment. Think of it as ...


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 ...


1

The energy difference along with the larger thermodynamical likelihood for occupation of the lower level is real. There is an application, nuclear magnetic resonance (NMR) spectroscopy/imagining/quantum computing. But due to the very small energy difference for technically achievable magnetic fields, the effect is usually negligible at roomtemperature. NMR ...


1

If the particle source is "unpolarized", that literally means it is equally likely to find particles from this source in either energy eigenstate - that's the definition of "unpolarized", so you shouldn't be surprised about that. When the spin of a particle is "perpendicular to the magnetic field", that's another way of saying that the particle is in an ...


1

The $H$ field is, by definition, (upto the constant $\mu_0$) essentially the contribution to the magnetic field if there were no material. We then relate that to the measurable magnetic field in the material through its permeability: $$B = \mu H$$ or $$B = \mu_0H + (\mu -\mu_0)H$$ $$ = B(\text{from current}) + B(\text{due to magnetization})$$ This is the ...


1

It's the inertial frame in which you wish to find the magnetic field. update It might help to make the definition more general. Consider an arbitrary origin in an inertial reference frame, the frame you wish to find the magnetic field. Let $\vec{r_0}$ be the position vector to the field point (referenced from the origin), and $\vec{r_1}$ the position ...


1

I can think of two ways to get the answer you looking for. 1) The easy way is to obtain the magnetization specifications from the Mu metal manufacturer. 2) You can make a magnetic amplifier and run some tests. As you magnetically saturate the Mu metal, the amplification decreases. Note: use Wikipedia or Google "magnetic amplifiers" for additional ...


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

The square bracket transformation This is just the application of chain rule. The LHS means a derivative over the primed spacial coordinates while keeping unprimed spacial and time coordinates fixed. $$\nabla'[ \rho(\mathbf{x'},t')]_{ret} = \left(\sum_i \frac{\partial }{\partial x_i'} \hat{i}\right)[\rho(x_i',x_j',x_k',t')]_{ret}\\$$ But the $\rho$ is a ...


1

If we put a magnetic dipole, $\mu$, in a uniform electric field of strength $B$ then the torque on the dipole is given by: $$ T = \mu \times B \tag{1} $$ where the $\times$ symbol indicates the cross product. If the angle between the dipole and the field is $\theta$ then equation (1) can be written as: $$ T = \mu B \sin\theta \tag{2} $$ For small angles ...


1

It sounds as though you're talking about the Gilbert Model and how to apply it. This is simply a phenomenological model that works out magnetostatic fields by putting sheets of magnetic monopoles near the ends of bar magnets: the field at a distance from the magnet is calculated by summing up all the forces owing to the monopoles worked out using an inverse ...


1

The resolution is that currents can change instantaneously. Remember that the reason that the current in a single solenoid cannot change instantaneously is that a change in current causes a change in flux, which causes an electromotive force (emf) to oppose the original change in current. However, suppose now there are two currents, as in your example. ...


1

I agree with Alfred Centauri that there is no configuration of a monopole, but on the second question, depending on how big the surface of the magnetic floor is, the strength of the magnets, and some other conditions I would imagine that it would levitate over the floor. On the third question, I believe that if it were a perfect sphere, the center ball would ...


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

$$\frac{\nabla'\times \vec{J}}{R}=\nabla'\times(\frac{\vec{J}}{R})-\frac{\vec{J}\times(\nabla'R)}{R^2}$$ 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: $$\int\nabla'\times(\frac{\vec{J}}{R})d^3x^{'}=\oint(\vec{n}\times\frac{\vec{J}}{R})d^2x^{'}$$ ...


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

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. ...



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