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5

No, one cannot explain the "cause" any deeper than the explanation that Lorentz force and Maxwell equations are postulated as a description and experimentally are found to foretell correct results. One can give certain motivations why these might be the correct equations: for example, the Lorentz force law is one of the simplest Lorenz covariant laws one can ...


4

You have a problem in the conceptual understanding here. The magnetic force is $\vec{F} = q\vec{v} \times \vec{B}$ If the field is constant if you double the velocity the force would be doubled as well. But your confusion is that you are not considering the radius in the equation bellow. It cannot be fixed. $F = qvB = \frac{mv^2}{r}$ If you double ...


3

I like the answer from @AndreMaizel and I will add another conceptual idea you need to remember. The form $m\frac{v^2}{r}$ is not a real force. It is a form of $ma$, where the $a$ is a centripetal acceleration. That centripetal acceleration is a "must behave this way" statement, not a "source of force" statement. As Andre says, the "real force" is the ...


3

According to Faraday's law of induction, $$\mathcal{E} = -N {{d\Phi} \over dt}$$ you will need a change in the magnetic flux $\Phi$ in order to get an EMF or an electric field. So if you just put your coil in the magnetic field of the permanent magnet, you will not measure a current. There will only be a current, if you move the coil around so that the flux ...


3

The field around the wire isn't uniform. When you calculate the force on a charge in a magnetic field, you use the value of the field at the point where the particle is. So, $$F_B = qvB\sin\theta$$ is not just for a constant field. If the field varies from position to position in space, then the force the particle feels will also vary. This equation is ...


2

Yes. The quantum-mechanical origin of the magnetization is largely irrelevant: all you need to know is the material's magnetization field $\mathbf M(\mathbf r)$, which is defined to be the (locally averaged) total magnetic dipole moment per unit volume at and near position $\mathbf r$. The magnetization then determines the magnet's magnetic field outside it ...


2

The claim, "the magnetic force does no work," while technically true, is misleading in so many situations and actually helpful in so few that we probably shouldn't put as much emphasis on it as many textbooks do. The magnetic field stores energy, and a change of the configuration of a system (like moving a current) can convert that stored energy into other ...


2

Two big factors: ferromagnetic objects (steel beams in buildings, structural elements in cars, etc) can bend the local magnetic field; and the stators of DC motors/alternators (where a significant current oases through many windings, deliberately increasing the magnetic field far above what you get from a single cable). When you have a DC magnetic field ...


2

Although Ampere's law and the Biot-Savart law are magnetostatic approximations, they will still approximately hold when the charge is moving slowly enough. The problem is you're just naively using an Ampere's law result derived for a steady current to a situation with a moving charge. Ampere's circuit law comes from $$ \nabla \times \vec{B} = \mu_0 ...


1

Not entirely, but a disk electromagnet works by routing a magnetic field though a ferromagnetic core. It's not a 100% efficient process and you do have a portion that is not directed through the core. The escaped portion is known as flux leakage, and is usually a fairly small percentage of the total field. From Wikipedia: The green lines represent ...


1

How does an inductor store [electro]magnetic energy? Rather surprisingly, it's something like a flywheel. You can see a mention of that here in Daniel Reynolds' electronics course: . It really is like this, check out the pictures of inductors on Wikipedia, and you'll notice they're rather like a solenoid. And there's the flywheel again: "As a result, ...


1

The energy is stored in the magnetic field. I usually think of it as "magnetic field lines repel" but that is not very precise (useful for intuition though). But along the same lines as your capacitor example (moving the plates to infinity takes work), if you look at a simple current loop there is a force on the wires from the magnetic field generated. This ...


1

Your conclusions are pretty much correct. The simplest way to see that magnetic fields don't do work is to consider the $\mathbf F_B = q\ \mathbf v \times \mathbf B$ : that is, the force due to the magnetic field is always perpendicular to motion, so no work is done directly by magnetic field. It's maybe more accurate to say the electromagnetic field can ...


1

Apart from philosophical debates: What is the cause of any force? A gradient in the energy. I'm not in the mood to do any actual calculation, but the energy density of a magnetic field is $\sim \vec{H} \cdot \vec B \sim B^2$ (here at least). Now, we are looking at a field that is created by superposition $\vec B = \vec B_1 + \vec B_2$ with $\vec{B}_i$ being ...


1

When electrons moving through a wire experience a magnetic force they change direction right? And when they push in that direction they don't gain any energy. So it must be the battery which gave the electrons their kinetic energy that is doing the work. As long as the current is flowing, the magnetic force is acting, and it is the battery that is sustaining ...


1

Therefore, in this case the magnetic field does positive work. How is this contradiction possible? The force given by the formula $$ \vec F= i \vec{l} \times \vec B $$ is magnetic force acting on the current forming particles - mobile electrons. The formula is accurate provided the wire does not move so the current forming particles follow the wire ...


1

To answer your question in one word, "Yes" Now, onto the explanation:- According to Faraday's Law, you will get a current in a conductor when the amount of magnetic flux linked with the conductor changes. Note that it is immaterial whether the source of the magnetic field is a permanent magnet or an electromagnet. All that needs to happen for you to ...


1

Only in a perfect diamagnetic. In a real conductor the induced magnetic field is limited by the resistance of the material, so it will always be smaller than the inducing field.


1

Clarification from another source: Source: Physics For Scientists And Engineers, Paul A. Tipler and Gene Mosca, Sixth Edition, W. H. Freeman and Company, New York, 2008, p. 971, Fig. 28-20. I maintain that the loop will act the same as the bar. In other words, if you cut a thin slit down the center of the bar and less than the length of the bar (you ...


1

It seems like the key problem is that you are conducting your integration over the wrong surface. The problem is, you have an infinitely extending sheet of charge on the surface $\rho=1.2$. Based on this, as you seem to have done, we work in cylindrical coordinates and consider a plane of constant $z$. Taking your expression: and converting it into ...


1

There is no asymmetry, for any observable quantity, because you'll always use the right hand rule twice to get an observable answer. For example, we might want to compute the acceleration of a charged particle near this wire. The force on the particle uses it once, since $\mathbf{F} \propto \mathbf{v} \times \mathbf{B}$, and the magnetic field is given by ...


1

Intuitive Answer: $\newcommand{\curl}[1]{\nabla \times #1}\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}}$There is an emf created because of the Lenz's Law, which means an electric field is being created. The electric field is circular (thus the force, which is proportional to the electric field, is also circular) and the charged particle would spiral ...


1

If by our goal is to see the fields due to some charge or current, look at Jefimenko's equations. For electric fields electric field they can be caused by charge density, the rate of which charge density changes or by the rate that current density changes. For magnetic fields they can be caused by current or by the rate at which current changes. Since you ...



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