Alonzo Odom
2022-07-22
Answered

Metal object attached to an electromagnet that would otherwise fall to Earth. Does the magnetic field do work in resisting the fall? If so what are the force carriers? Indeed what are the force carriers in the (relativistically related) phenomena of electric attraction and repulsion?

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Alden Holder

Answered 2022-07-23
Author has **15** answers

No displacement - no work.

As far as force carriers are concerned... Do you actually need this? The question is perfectly answerable within classical mechanics/electrodynamics, and the answer is easy. Force is 'carried' by the electromagnetic field. If you start diving into quantum field theory the force will still be carried by the electromagnetic field, but now you will need to quantize the field (and you will get photons).

As far as force carriers are concerned... Do you actually need this? The question is perfectly answerable within classical mechanics/electrodynamics, and the answer is easy. Force is 'carried' by the electromagnetic field. If you start diving into quantum field theory the force will still be carried by the electromagnetic field, but now you will need to quantize the field (and you will get photons).

asked 2022-05-18

Why is magnetic field vector perpendicular to magnetic force vector?

So recently in physics class, we learned about the magnetism right hand rules.

One of them states that the index finger points in the direction of the velocity of a particle, the middle finger points in the direction of the magnetic field, and the thumb points in the direction of the magnetic force.

I'm curious why the magnetic field vector is perpendicular to the magnetic force vector.

So recently in physics class, we learned about the magnetism right hand rules.

One of them states that the index finger points in the direction of the velocity of a particle, the middle finger points in the direction of the magnetic field, and the thumb points in the direction of the magnetic force.

I'm curious why the magnetic field vector is perpendicular to the magnetic force vector.

asked 2022-05-09

In Three Lectures On Topological Phases Of Matter section 2.1 mentioned, that:

${I}^{\mathrm{\prime}}=\int dt{d}^{3}x\phantom{\rule{thickmathspace}{0ex}}(\overrightarrow{a}\overrightarrow{E}+\overrightarrow{b}\overrightarrow{B})$

correspond to ferromagnetism and ferroelectricity. And that

${I}^{\mathrm{\prime}\mathrm{\prime}}=\int dt{d}^{3}x\phantom{\rule{thickmathspace}{0ex}}({a}_{ij}{E}^{i}{E}^{j}+{b}_{ij}{B}^{i}{B}^{j})$

correspondence to electric and magnetic susceptibility.

Could somebody clarify, why?

${I}^{\mathrm{\prime}}=\int dt{d}^{3}x\phantom{\rule{thickmathspace}{0ex}}(\overrightarrow{a}\overrightarrow{E}+\overrightarrow{b}\overrightarrow{B})$

correspond to ferromagnetism and ferroelectricity. And that

${I}^{\mathrm{\prime}\mathrm{\prime}}=\int dt{d}^{3}x\phantom{\rule{thickmathspace}{0ex}}({a}_{ij}{E}^{i}{E}^{j}+{b}_{ij}{B}^{i}{B}^{j})$

correspondence to electric and magnetic susceptibility.

Could somebody clarify, why?

asked 2022-01-11

Along coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b and outer radius c. The outer cylinder is mounted on insulating supports and has no net charge. The inner cylinder has a uniform positive charge per unit length $\lambda$. Calculate the electric field

a) at any point between the cylinders a distance r from the axis and

b) at any point outside the outer cylinder.

c) Graph the magnitude of the electric field as a function of the distance r from the axis of the cable, from r= 0 to r= 2c.

d) Find the charge per unit length on the inner surface and on the outer surface of the outer cylinder.

asked 2022-07-14

It is well established that the photon is the propagator for the electromagnetic force. How are the set of photons used in the H-field behaving differently than those in the E-field? Is it different when dealing with an electromagnet versus an intrinsic magnet? In what way do the photons need to be emitted by a bar magnet to result in a magnetic interaction? How about a charged object to interact with a charged object?

asked 2022-07-25

A wrench 30 cm long lies along the positive y-axis and grips a bolt at the origin. A force is applied in the direction <0, 3, -4>at the end of the wrench. Find the magnitude of the force needed to supply 100 N-m of torque on the bolt.

asked 2022-05-17

I was wondering what is the difference between a spin-fluid Heisenberg Magnet and spin-glass Heisenberg Magnet. As far as I understand in a spin-glass the spins are randomly oriented compared to a ferromagnet material but I don't quite understand what we mean by spin-fluid phase. And also in the article I am reading its mentioned that spin-fluid phase is found to be generically "gapless", but I am not sure what we mean by phase being gapless and how is it related to the spin phase of the material?

asked 2022-05-08

Magnetic Force Confusing Paradox

Actually I had posted a very similar question, but I wasn't quite satisfied with the answer so I am posting a new variety again.

Imagine an infinitely long wire carrying current ${I}_{1}$ from West to East. At a small distance $d$ above the wire there is another small current carrying wire of length $l$ carrying current ${I}_{2}$ from East to West (opposite to the direction of current in the below placed wire). Obviously they are magnetically repelling. And if the second wire (carrying current from East to West) is at rest the magnetic force must be equal to $mg$

The magnetic force is upward and $mg$ is down. The magnetic force can be written as

$\frac{Uo{I}_{1}{I}_{2}}{2\pi d}l=mg$

Now, if the magnetic force is greater than $mg$, the wire moves up. Now magnetic force is up and displacement is up too which means that work done by magnetic force should be positive. How is that possible when we know that work done by a magnetic force is always zero? Is it that an e.m.f. is induced which opposes the change?

Actually I had posted a very similar question, but I wasn't quite satisfied with the answer so I am posting a new variety again.

Imagine an infinitely long wire carrying current ${I}_{1}$ from West to East. At a small distance $d$ above the wire there is another small current carrying wire of length $l$ carrying current ${I}_{2}$ from East to West (opposite to the direction of current in the below placed wire). Obviously they are magnetically repelling. And if the second wire (carrying current from East to West) is at rest the magnetic force must be equal to $mg$

The magnetic force is upward and $mg$ is down. The magnetic force can be written as

$\frac{Uo{I}_{1}{I}_{2}}{2\pi d}l=mg$

Now, if the magnetic force is greater than $mg$, the wire moves up. Now magnetic force is up and displacement is up too which means that work done by magnetic force should be positive. How is that possible when we know that work done by a magnetic force is always zero? Is it that an e.m.f. is induced which opposes the change?