# Tag Info

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The energy is located in the magnetic field $\vec{B}$ around the wire. The magnetic energy density is $$\frac{dU}{d^3r}=\frac{1}{2\mu_0}\vec{B}^2$$ with the magnetic field given by $$B(r)=\frac{\mu_0I}{2\pi r}.$$ Then you can calculate the total energy by integrating this energy density over the whole volume outside the wire. I recommend using cylindrical ...

3

Elementary particles are never at rest in Quantum Mechanics or QED. If you take a plane wave of a "free" electron $\text{e}^{\text{i}\bf{p}\bf{x}}$, the rest frame corresponds to $\bf{p}=0$ with arbitrary $\bf{x}$. So the dipole field given above corresponds to a highly localized calssical magnetic dipole, and not to a free elementary particle at ...

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Both electrons and protons generate an electric field, and electrons, protons and neutrons generate a magnetic field. The equivalent of this statement is the following. Electrons and protons have an electric field and electrons, protons and neutrons have a magnetic field. The second statement better points to the fact that both fields are intrinsic (existing ...

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Both electrons and neutrons at rest produce magnetic fields because they have non-zero magnetic moments. The electron moment is $${\boldsymbol \mu}= \frac{eg}{2m} {\bf S}$$ where the spin ${\bf S}$ has magnitude $|{\bf S}|=\hbar/2$ and $g\approx 2$. A point dipole with magnetic moment ${\boldsymbol \mu}$ produces a field with spherical polar ...

1

A current carrying solenoid produces a magnetic field. If a ferromagnetic material is placed in that field, the atomic dipole moments are subject to a torque and brought into some degree of alignment with that field. If the material is outside the end of the solenoid (where the field is spreading), the dipoles are also subject to a net attracting force ...

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I will give a more qualitative answer than just formulas. Considering the given proposal one has to be very careful since $\mathbf{H}$ not only has to fulfill Maxwell-equations and material equations, but also the boundary conditions which are not the same for $\mathbf{B}$ and $\mathbf{H}$. The Maxwell equations in case of a permanent magnet without any ...

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If you have a steel ball bearing with high permeability, then the magnetic potential energy of this configuration will be proportional to the square of the B-field at the position of the ball bearing. For a simple dipole magnetic field on-axis, this will be proportional to $r^{-6}$, where $r$ is the distance from the centre of the magnet. The force ...

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First off, if the wire is "ideal" the answer is undefined. The ideal models provide no solution if you try to solve them. What actually happens is defined by the non-ideal aspects of the system. The magnetic field collapsing causes back-EMF, which induces a voltage on the wire. In small voltage cases, this ends up turning the wire into an ...

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There will be a spark between the cut ends and the energy goes into the heat, light, and bang associated with the spark.

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There are two questions here: can the electric field of a static collection of charges have non-zero curl, and do any electric fields of the form $\mathbf{E}(r,\theta,\phi)$ describe the electric field of a static collection of charges? As for the first question, for a static collection of charges, the magnetic field does not change with time. Therefore, one ...

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You could, but unclear how this helps you. Ampere's law would, in this case, tell you that $$\oint \vec{B}\cdot d\vec{l} = 0\ ,$$ because you have no current flowing through the closed loop. I suppose it does tell you there isn't a toroidal field aligned with the Amperian loop.

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Suddenly the 3 charges enter a B field. The magnetic field points into the page, so there is a magnetic force pointing towards the right on the middle charge. Where isn‘t a magnetic force pointing towards the right. The deflection is the result of the interaction of the electrons magnetic dipole with the external magnetic field. By experiment it is so that ...

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At the moment the system enters the region with the magnetic field it experiences a time dependent magnetic field, which, because this is the same thing as a rotation of the electric field, can do work. This displaces the charge to the left until the magnetic field becomes static. The charges stays in its new position. Perhaps transient oscillations may ...

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In presence of a uniform magnetic field, the spin of a particle mostly enters the Hamiltonian, and therefore the energy, via a Zeeman term $\hat{H}_{Zeeman}=-\mu\vec{B}\cdot\hat{\vec{S}}$, characterized by the scalar product of the magnetic field vector $\vec{B}$ and the spin operator $\hat{\vec{S}}$. Given some spin state $|\chi\rangle$ consisting of a ...

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The wire is only shown to end at X. If it would indeed end there as shown then there would be no current in it in the first place. Currents through wires need a closed loop to flow. So this diagram probably means that the wire is going up and loops around at some far off place. And in that case the magnetic field will be the same as is shown in your XY plane ...

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https://ocw.mit.edu/courses/materials-science-and-engineering/3-a08-attraction-and-repulsion-the-magic-of-magnets-fall-2005/assignments/mag_relativity.pdf Have a look at this link - it will probably be better than any of our explanation.

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The rate of doing work is $\vec{F}\cdot \vec{v}$. Since the magnetic component of the Lorentz force is $q\vec{v}\times \vec{B}$ then this force is always perpendicular to the velocity and does no work. In the absence of electric field from the other charges, the middle charge would execute a circular path at constant speed and kinetic energy. No work would ...

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What happens to the middle charge? It will get deflected to the right as well if the magnetic force is greater than the electric force trapping it by the two charges as you stated. That is, if the net electric force manages to keep the middle charge restricted to move in the y-axis, and the magnetic force is greater than this force, there will be motion in ...

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what will happen if the insulating coat of some wires fell off and current can travel through the soft iron core? If the core is ungrounded then the core will just be energized to the voltage level of which ever winding (primary or secondary) became shorted to it (neglecting electrostatics etc.). If the core is grounded (likely) then current can flow and ...

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The derivation of the formula: $$F = \oint I \vec{dl} \times \vec{B} = - I \oint \vec{B} \times \vec{dl}$$ Now using identity in this MSE post $$\oint \vec{B} \times \vec{dl} = \int_S (\nabla \cdot \vec{B}) \hat{n} - (\nabla B_i) n_i dS$$ The first term vanishes due to maxwell's identity that $\nabla \cdot B=0$ and the second term is the directional ...

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First: the iron core of a transformer is not made of a solid iron , but is composed of thin sheets of iron, isolated against each other. Otherwise you would have circular currents in the core. But even so, if just e very short piece of your wire has no insulation, there would not be a current trough the core, since the second pole is mission, so just the ...

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For a spin in an not excessively strong external magnetic field we can ignore the spatial part of the wave function and use an effective spin Hamiltonian. For a single electron spin this is the $$H=g\mu_B \vec B \cdot \vec S$$, where g=-2. The solutions are simply $|+\rangle$ and $|-\rangle$, with eigenvalues $\pm \frac{1}{2} g\mu_B$. Whether these wave ...

1

The principled way to answer this question is to write down the Schrödinger equation for a particle in magnetic field and try to solve it. From the context of the question it seems that you are talking about an electron with spin-1/2 - the case that is treated extensively in textbooks, including its motion in magnetic field. However, it is not clear whether ...

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I think the author's main point is to introduce the concept of a magnetic moment (similar to a dipole moment) which can be thought of as a small current loop. The real magnetic moments (such as spin) are not really current loops... so it is just a pedagogical tool. Regarding evaluating the moment - the integral here is a contour integral, so you have to ...

3

In Figure-01 we have a solenoid of 1 turn. In case of uniform magnetic field $\mathbf{B}$ parallel to the axis of the helical wire the flux through the helical surface is equal to the flux through its projection on the $xy-$plane that is the circle shown. This equality, which is exact and not approximate, is proved in the Figure-03 below. In Figure-02 we ...

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You are right that z component of velocity remains unchanged. But this is taken care of automatically when you calculate the Lorentz force using vector notation: $$\vec F= Q\vec v \times \vec B = Q(a \hat i+b\hat j +c\hat k)\times B_0 \hat k=QB_0(-a \hat j+b\hat i)$$ We have used the fact that $\hat k \times \hat k =0$. So there is no force component on Q in ...

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The magnet generates a magnetic field $\mathbf{B}_{\text{magnet}}$. The electron has a dipole moment $\mathbf{m}_{\text{electron}}$ due to its spin, which will experience a torque $\boldsymbol{\tau}$: $$\boldsymbol{\tau}_{\text{on electron}} = \mathbf{m}_{\text{electron}} \times \mathbf{B}_{\text{magnet}}.$$ By Newton's third law, the electron itself will ...

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The changing magnetic field from a moving magnet can induce a pulse of current in the coil. The magnetic field from the current can act back on the magnet. For a magnet of given strength and size the interaction with the coil should decrease as the coil increases in diameter (while keeping the number of turns constant).

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I assume that you'd have no difficulty working out the flux linked with a closed circular loop if the flux density is uniform and normal to the plane containing the loop. With a long solenoid (a helix of small pitch) of $n$ turns we usually treat each turn as a circular loop having, in the case you present, a flux $\Phi=\vec B. \vec A =BA$ through it. The ...

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A compass needle is magnetized and will experience a force or torque when placed in an external magnetic field such as one produced by a nearby current carrying wire. The compass needle is said to have a "magnetic moment" which measures how magnetized it is, and therefore how strongly it responds to external magnetic fields. Sometimes, magnetic ...

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You can think of it like an S-shaped curve in a magnetic field line where the S is oriented orthogonal to the radial solar direction. That is, looking down on the ecliptic plane and imagine the azimuthal direction is such that the radial ray we examine is along the horizontal. In that case a magnetic field line could be traced as starting radial, then ...

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An atom with an unpaired electron has a magnetic dipole moment. In a ferromagnetic material a significant fraction of the dipoles can be brought into alignment and will maintain that alignment in spite of thermal agitation. In a bar magnet, the alignment of the dipoles produces a magnetic field which is very similar to the field that would be produced by an ...

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The definition for any steady current distribution is $${\bf m}= \frac 12 \int {\bf r}\times {\bf J}\, d^3x.$$ This quantity is independent of the origin chosen for ${\bf r}$ because $$\int {\bf J} \, d^3x={\bf 0}$$ for a steady current that obeys $\nabla \cdot {\bf J}=0$. I leave it up to you to specialize this general case to one of a current ...

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In short, the answer is that a current will be induced in the conductive object, as the external field is time-varying. As an example, consider that the conductive object is simply a wire loop. If we let the external field be a magnetic field, we may apply Faraday's law directly, such that the induced electro-motive force (emf) $\mathcal{E}$ is \mathcal{E}... 2 The situation resembles alot to a rotating loop. So I'm using this as a hint .The formula for calculating magnetic field at the centre of a current carrying loop is directly proportional to i/R. We see that when the ring rotates the current in a small area remains constant. This is because when the loop rotates the part of loop which crosses a region and ... 2 I'm thinking that with the wheel rotating, the motion of the positive charges will constitute an electric current that offsets the change in the motion of the free electrons. 0 If you understand the Wien's filter, you must know that it depends on setting an equal force in opposite directions using E and B fields perpendicular to the trajectory of the charged particles in the beam. As in your experiment there is no opposing electric field, the charged tracks will follow the right hand rule of interaction to the magnetic field : to ... 1 It is a second order linear differential equation of the type \frac{d^2f}{ dx^2} + cf = 0 \tag 1 where c is a constant. Differential equations of this type have periodic solutions of the kind f(x) = cos(kx) + sin(kx) \tag 2 Think about the first derivative \frac{df}{ dx} = kcos(kx) - ksin(kx) and then the second derivative \frac{d^2f}{ dx^2} = -k^2 ... 2 Eq. B1.1.30 is obtained from the previous one by taking an additional derivative, which permits us to eliminate the cross-couplings between the x- and y- components. Considering the form of Eq. B1.1.30, the solution of the generic differential equation \ddot{y}=-\omega^2y is found by substituting the general ansatz y=\exp(\lambda t), as we are ... 7 A magnetic field is a form of energy. Energy is required to form a permanent magnet by, e.g., aligning the spins of electrons in a material. That energy is stored in the resulting macroscopic magnetic field. The same is true of an electric field: energy is stored in the macroscopic electric field when positive and negative electric charges are separated, e.... 4 You cannot magnetize something without bringing it into the magnetic field but that takes work and the amount of work you have done while keeping the magnetic field unchanged is the energy acquired by the newly magnetized body. If the source of the magnetic field is a permanent magnet whose magnetization does not change while moving another magnetizable body ... 0 A magnet is created by aligning the magnetic fields of very small atomic level domains throughout the material. We know this because a sufficiently powerful external magnetic field can realign or even reverse the field of a magnet. You don’t need to have two different metals to create a magnet. 0 Outside the magnetic field, space is isotropic and angular momentum is conserved. Outside the field, there is no preferred axis. That would only appear at a measurement. Classically: without a torque there is no precession. 2 Yes, the spin precession happens because of the external magenetic field. In the z basis, it adds a phase to the states like: |\psi(t)\rangle=c_{\uparrow}e^{-i\omega t}|\uparrow\rangle_z+c_{\downarrow}e^{i\omega t}|\downarrow\rangle_z$$but it does not change the probabilities of outcome c_{\uparrow} and c_{\downarrow}, that are given by the initial ... 2 Hint : If \psi\left(\mathbf{r}\right), \boldsymbol{\alpha}\left(\mathbf{r}\right) scalar and vector functions of \mathbf{r} respectively then \begin{equation} \boldsymbol{\nabla}\boldsymbol{\times}\left(\psi\,\boldsymbol{\alpha}\right)\boldsymbol{=}\boldsymbol{\nabla}\psi\boldsymbol{\times}\boldsymbol{\alpha}\boldsymbol{+}\psi\boldsymbol{\nabla}\... 1 The result given by the OP is off by a sign (v1). It should read$$\vec{B}=-2{A_0}(\hat{k}\times\hat{\epsilon})\sin({\textbf{k.r}-\omega t+\delta_{\omega}}). Let me do one component. If $\hat{\epsilon}=(\epsilon_x,\epsilon_y,\epsilon_z)$, then \begin{align} (\nabla\times\vec{A})_x&=\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\\ &=...

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A measurement can be done with a dynamometer. For calculations one needs finite-element software and the magnetic properties (hysteresis loop) of the magnets. There are approximations. The ones that you quote for large distances are not often very relevant. There are formulas for holding power of a magnet against a weak magnet.

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There are answers to that here. What is conserved, is energy. Ultimately, what is happening above is Lenz's law which is a statement of the conservation of energy.

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Good question and good answers. Worth adding that also the electric field energy and the magnetic field energy are going to be exactly equal in such plane EM wave. It is just the traditional definitions of different units for E and B that make B seem thousands of times smaller. Since the energies are the same it is good they both get represented fairly in ...

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All these proofs pertain to an external field pointing in the same direction as the solenoid's central axis. This longitudinal field is indeed zero when the solenoid is infinite in length. But, no matter how tightly the solenoid is wound, there is a met current in the longitudinal direction. Thus, by Ampere's law, there must be an external, toroidal field ...

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