# Tag Info

-1

I am looking from the unified theory. I feel that all basic forces of Nature should have same origin. Gravitational , Electrostatic , Magnetic, Nuclear . Let us put them on a force spectrum, one end may be Gravitational and other end may be nuclear forces. I am trying to understand the basic unifying link between all these forces. Like moving electron ...

1

Yes, it comes from perturbation theory. The Hamiltonian is $H'=-\boldsymbol{\mu} \cdot \mathbf{B}$ and the assumption is that you align your system so that the magnetic field points in the $z$ axis: $\mathbf{B} = B\hat{\mathbf{z}}$. The magnetic momentum operator $\boldsymbol{\mu}$, for your basis, is usually expressed as: $$\boldsymbol{\mu} = -\frac{\mu_B ... 2 The fact is that the 'R' you are using in the first simplified equation is the perpendicular distance of the point from the wire. Whereas the R you are using in the integration is not the same , the R used here is the distance of the point from the end point of the infinite wire , which is infinite.Here, see the image below ... 3 First case: Second case: You seem to be confused to between r and R. r, used in the Biot-savart law, is the distance of all the elements along the wire from point, whereas R is the distance of the midpoint of the wire from the point. 0 In the integration, r is the distance of the elementary part from the concerned dl part. So you cammot directly put that 2R, as this r is something that depends on theta, and is 2R only for theta= 90 degress. So it is recommended to use the first one, as in that r is the perpendicular distance from the line to the concerned point. 1 The so-called Faraday rotation, or magneto-optic effect, is an interaction between a magnetic material and linearly polarized EM wave, see https://en.wikipedia.org/wiki/Faraday_effect. The rotation in question is the turning of the linear polarization of the EM wave (light) that is moving within a homogeneous magnetic material and in a parallel direction ... 1 Update - The repulsion definitely increases with faster spinning as wikipedia says: Faraday's law states that the EMF is also given by the rate of change of the magnetic flux where epsilon is the electromotive force (EMF) and \phi_B is the magnetic flux.$$\varepsilon = - \frac{-d \phi_B}{dt}$$I think repulsion will also be greater with stronger magnets.... 0 In the case of static electricity, the voltages may be significant but the currents generated are extremely small. Therefore any magnetic fields generated will be extremely small as well. 2 Any flow of electron signifies current flow. And any flow of current generates magnetic field. 9 You have the definition of the vector potential.$$\mathbf{B}=\nabla \times \mathbf{A}$$According to Stokes' theorem this is equivalent to$$\iint_S \mathbf{B}\ d\mathbf{S} = \oint_{\partial S} \mathbf{A}\ d\mathbf{r}$$where S is any surface area and \partial S is its boundary line. Now choose for S a circle around the z-axis. Then the integral on ... 8 One way to do this (though I wouldn't blame you for accusing me of cheating) is to "remember" the identity (in Cylindrical Coordinates):$$\nabla \times \left(\frac{\hat{\mathbf{\varphi}}}{\rho}\right) = \hat{\mathbf{z}}\,\, 2 \pi \,\delta^2(\rho).\tag{1}\label{1}$$This is reminiscent of the divergence identity (in Spherical Polar Coordinates):$$\...

2

As suggested by @hyportnex, and correctly understood by the OP, there is an essential difference between Griffith's example in the warning and the Problem 6.12. In the former case, the cylindric magnet is finite, while in the latter is infinite. The two cases differ for a region of non-zero divergence of magnetization, precluding the possibility of getting ...

4

A quick google search pulled up this article, which reports on this paper in Nature Physics. The essence of it is that pure graphite is (at least mostly) diamagnetic, but the presence of defects - seven or eight membered rings, vacancies, and nontrivial domain edges - have ferromagnetic behaviour. Pencil lead is an amalgam of small, crude graphite grains and ...

1

The thing to remember is that there is no such thing as a tiny "north" magnet since every magnet contains a north and south pole no matter how small it is. Magnetic field lines always form closed loops and this is a statement of the fact that the are no magnetic monopoles (that have been observed) no matter how tiny the magnet is, or in other words,...

1

First, there is no such thing in real life as an isolated North or South pole. This implies that magnetic field lines always form closed loops, as you see in the image. This is codified in Gauss's Law of Magnetism. Your textbook, while I can sort of see the logic, gives a very inelegant definition for field lines. Don't think of the field lines as a force ...

1

Yes, you can shape field lines much like how you can shape light rays. But it's important to remember that a static magnetic field is not a wave, and therefore it does not reflect or refract. A magnetic field in motion propagates as an electromagnetic wave, that is to say, light. So, the propagation front of a moving magnetic field, which is the part with ...

1

To some extent, you can- but not in the same sense that light can be focused. Here's how: If you give a magnetic field the choice between propagating through empty space or through a piece of iron, it will choose the iron. This means that if you have a magnet sitting alone by itself, the magnetic field lines emanate from one end, loop around, and re-enter ...

0

As a first approximation for the field outside you can treat the toroid as a single loop of current (as in the sketch). Look up the field of a magnetic dipole. Other than that, you can calculate the field at points of interest by numeric integration.

2

This is, it turns out, a deceptively hard question. The equation you've found is actually only applicable for points on the axis of symmetry of a circular loop. You could, if you wanted, use this formula to find the magnetic field along the axis of a solenoid. The basic technique is described in the answers to this question. The integral you get for points ...

1

Not if you're observing them from their own rest frame - that is, you're near them and also comoving with Earth's surface. An observer in the charge's rest frame measures an electric field and no magnetic field, and measures the charges moving as predicted by Coulomb's law. An observer moving relative to the charges measures an electric field and a magnetic ...

1

(a) "when you use current to magnetize something the north end of the magnet is the end that was negative" I'm afraid that this suggests a misunderstanding. The direction of magnetisation is determined by the direction of current around the coil, not along its length. You need to consult a textbook and learn about the right hand grip rule. (b) If, ...

3

You cannot magnetize a human, or just about any object that is not ferromagnetic. Ferromagnetism basically refers to the mechanism by which certain materials can become magnetic. Human tissue is not ferromagnetic, and so you could not make a person permanently magnetic by the use of magnets. Certain substances, like iron, copper, sodium etc., can become ...

1

You might consider the case of a circular path around the middle of a long solenoid, and co-axial with a it. Let us call four equally spaced points around the circular path: N, E, S, W. The solenoid has a current through it that is rising at a constant rate, so the emf around the circular path is constant (with value $\mathscr E$, in the sense NESW, say.) We ...

1

In an orthogonal coordinate system like the spherical coordinates system, $dr,d\theta,d\phi$ are always orthogonal to each other, therefor the $\alpha"$ you mentioned acutally is always $\pi/2$, however, $r\times d l$ is a vector, during the integration, only its $ \sin \theta"$ componet is reserved, the $\cos \theta"$ componet will cancel out during ...

2

For a particle on confined to a 2D plane, a magnetic field in that plane does not make sense. Such a field would try to force the particle off the plane. The only part of the magnetic field that can meaningfully interact with the particle is the vector component perpendicular to the plane. That way, the Lorentz force remains in the plane. In 2 dimensions, ...

0

The interior of the earth is too hot to support ferromagnetism but some articles I've read suggest that as the liquid iron core at the center of the earth cools, some parts solidify. The heat of fusion from this process heats the surrounding liquid, causing it to circulate. The loops of circulating liquid metal interact with any existing magnetic field, ...

1

No it doesn't prove that, but it does prove that any such field is nothing to do with the current. It could be the Earth's magnetic field or some other stray field, but it's not generated by the current.

1

(a) What you are forgetting is that the field beyond the North Pole of a bar magnet is not parallel to the magnet's longitudinal axis of symmetry, but has a component directed outwards from this axis. The magnetic field lines splay out! [The exception is along the axis of symmetry itself.] So the motor effect force on your loop element, $dl$, will have a ...

0

The meter seems to show the magnetic field intensity at the point of interest. The magnetic field intensity, in this case, has been measured in units of gauss ($1 G = 10^{-4} T$). The angle $\theta$ is the angle between the $x$-axis and the magnetic field. $B_x$ and $B_y$ are the $x$ and $y$ components of the magnetic field. Since the angle $\theta$ is zero, ...

2

I imagine that they are the magnitude and $x$ and $y$ components of the ${\bf B}$ field measured in gauss. If the device is to measure the Earths magnetc field (which is a few gauss) the $\theta$ is probably the dip angle. The $x$ axis would thebe oriented to true north.

1

Sure. The easiest way is to make the one you want to move out of iron and make the other two out of copper. If the material of the masses is fixed beforehand then you may not be able to move any of them. If all are made of the same material and all are magnetic then you still may be able to be selective by using the shape of the magnet’s field. At the ...

2

You are correct. The emf is present in the left hand loop and the outer loop, because the changing flux is linked with both these loops. There is no net emf in the right hand loop, because no (changing) flux is linked with it. It is therefore fine, for purposes of circuit theory, to represent the emf as if it were concentrated in the place you have shown (or ...

3

Why does the a light wave crest, rather than continuing off in the direction it's heading prior to the moment it begins cresting? A light wave does not involve any particle moving with an oscillatory trajectory. What's "waving" in a light wave is the magnitude and direction of the magnetic (B) and electric (E) fields. One reason neither the B nor ...

0

Assuming the free electrons actually shift outward (away from the axle) as a result of their circular motion, then the separation of charge produces an electric field pointing outward in the rod. This field exerts a centripetal force on the electrons preventing any further shift. For the given directions of motion and applied magnetic field, the magnetic ...

0

The electron charge is negative, so the magnetic moment is anti-aligned with the spin.

1

There’s an azimuthal electric field outside of the solenoid. This electric field is what causes current to flow through the outer coils (e.g. as given by Ohm’s law $\bf{J} = \sigma \bf{E}$). There's no single physical explanation for the origin of this external electric field; "why" questions in physics rarely have a single answer if there are ...

1

Is it possible to create this environment that the magnets are in perfect balance in every direction, both pushing and pulling? If by magnets you mean fixed magnetic moments in a static configuration, then the answer is no per Eanrshaw's theorem. If the magnets are electromagnets or at least can change their magnetic fields dynamically, then maybe something ...

1

You can create a spatial array of magnets which generates a specific spatial force profile against a corresponding magnet array. Check out polymagnets: https://vimeo.com/107166551 Spring Polymagnets attract until they pass through a defined transition point, passed which they will repel. These Polymagnets will come to rest at an equilibrium distance. I don'...

1

I believe I have heard that there is a train in Japan which is supported by magnetic levitation, but that would be for linear motion, and might require a significant amount of power.

0

Current creates a magnetic field through one of the four “fundamental interactions of the universe”, namely the electromagnetic force. This manifests as electrical attraction/repulsion between two or more charges, and as magnetic attraction/repulsion between two or more currents. Even though they appear to be two distinct types of forces, they both arise ...

1

Actually the statement and converse are both true. If by statement we mean: $\oint \vec{B} \cdot d\vec{l}=0 \implies$ enclosed current is zero And converse: Enclosed current zero $\implies \oint \vec{B} \cdot d\vec{l}=0$ This does not imply no field through the surface defined by the loop, just that it integrates to zero around the path of the loop (dot-...

1

In order to go from $\oint \vec B\cdot d\vec \ell= \vert \vec B\vert L=B L$ where $L$ is the length of your contour, you must have that $\vert \vec B\vert$ is the same on all points of the contour, and $\vec B$ is parallel to $d\vec \ell$ everywhere. Here, this is not the case. If you consider an Amperian loop of infinitesimal radius $\delta r$ in the plane ...

3

Ampere's law: $$\oint_C \vec{B} \cdot d\vec{l}=\mu_0I_e$$ The integral tells us that if the magnetic field is zero at every point along the closed curve C, then there must be no net current enclosed in the loop. However, the converse is not true -- there are many ways for the magnetic field to be non-zero at most or every point along that curve, and yet ...

0

First, solve for the B field in the wire (if current is constant, the current density will be uniform, and most wire is cylindrical). Second, note that each element of the cross section is a wire, carrying some current, in a magnetic field. So, there's a Lorentz force on that wire element, current crossproduct with B, toward the center of the entire wire. ...

Top 50 recent answers are included