Recent questions in Electromagnetism

Magnetic force
Answered

hyprkathknmk
2022-05-13

I confuse the four kinds of fundamental interactions, so I think the electric force and magnetic force should not be classified as a big class called electromagnetism.

Here is my evidence:

1.The Gauss law of electric force is related to the surface integration but the Ampere's law corresponds with path integration.

2.The electric field can be caused by a single static charge while the magnetic force is caused by a moving charge or two moving infinitesimal current.

3.The electric field line is never closed, but the magnetic field line (except those to infinity) is a closed curve.

Magnetism
Answered

Blaine Stein
2022-05-13

Magnetism
Answered

kwisangqaquqw3
2022-05-10

$m\frac{d({v}_{x})}{dt}=q{v}_{y}B\phantom{\rule{2em}{0ex}}m\frac{d({v}_{y})}{dt}=-q{v}_{x}B$

by differentiating them with respect to time to obtain two equations of the form:

$\frac{{d}^{2}u}{d{t}^{2}}+{\alpha}^{2}u=0$

where $u={v}_{x}$ or ${v}_{y}$ and ${\alpha}^{2}=qB/m$. Then show that $u=C\mathrm{cos}\alpha t$ and $u=D\mathrm{sin}\alpha t$, where C and D are constants, satisfy this equation

Whenever I differentiate the first equation with respect to time, I get a resulting equation with the form:

$\frac{{d}^{2}u}{d{t}^{2}}+{\alpha}^{2}\frac{du}{dt}=0$

Magnetism
Answered

indimiamimactjcf
2022-05-10

b.) -2 A;

c.) This problem has no solution.

d.) 2 A;

Magnetic force
Answered

poklanima5lqp3
2022-05-10

Given two infinite parallel charged rods with equal charge density $\lambda $. They are moving with same constant velocity $\overrightarrow{v}$ parallel to the rods. Find the speed $v$ for which the magnetic attraction is equal to the electrostatic repulsion.

Well, I know how to solve this problem: we first find out the magnetic field created by one rod on the other using Biot's and Savart's law, then we use the definition of $\overrightarrow{B}$ ( $d\overrightarrow{F}=\overrightarrow{v}dq\times \overrightarrow{B}$ ) to find the magnetic force, then equate magnetic and electrostatic forces to find $v$, which will be greater than or equal to $c$, thus conclude it is impossible for the forces to be equal.

However, one can argue as the following:

We all know that "same laws of physics apply in all inertial frames". With a constant velocity $\overrightarrow{v}$,the rest frame of the rods is an inertial frame. Therefore, if Biot-Savart law applies in our frame, it has to apply in the rest frame. If so, none of the rods will feel a magnetic field from the other one because their relative speed is zero, and there will be no magnetic force between the rods.

I've seen this question several times before in references, exams, exercise sheets,and in many different forms (parallel planes, beam of electrons ...),but no one ever used this argument.What is the problem in it ? Is it something related to Maxwell's equations or special relativity ? Or what else ?

I know a similar question was asked before, but the answers weren't satisfying. Please provide your answers with necessary mathematics.

Magnetic force
Answered

Yasmine Larson
2022-05-10

If the magnetic force does no work on a particle with electric charge, then: How can you influence the motion of the particle? Is there perhaps another example of the work force but do not have a significant effect on the motion of the particle?

Magnetic force
Answered

Damion Hardin
2022-05-10

One can no nothing about the magnetic force and yet arrive at it by taking the relativistic effects of a current and a moving charge system into account. I ask whether there exists such an inherent force in case of gravity.\

Magnetism
Answered

Bernard Mora
2022-05-10

And then using the divergence theorem:

${\int}_{V}(\mathrm{\nabla}\cdot B)d\tau ={\oint}_{S}B\cdot da$

Then ${\int}_{V}(\mathrm{\nabla}\cdot B)d\tau $ must be = 0.

But then I'm not sure why I can say: $\mathrm{\nabla}\cdot B=0$ and forget about the integral.

Does it just mean that $\mathrm{\nabla}\cdot B$ must be zero everywhere?

Ferromagnetism
Answered

Lexi Chandler
2022-05-10

where M is the ferromagnetic order parameter and $\varphi $ is the auxiliary field from the Hubbard-Stratonovich transformation. The book argues that because the above equation is correct, the mean field theory which is derived from the Hartree-Fock approach is equivalent to the the saddle point approximation formalism for H-S transformation auxiliary field Lagrangian. But I can not understand the equation.

Ferromagnetism
Answered

garcialdaria2zky1
2022-05-09

Magnetic force
Answered

Marco Villanueva
2022-05-09

I'm here asking this as a student. Magnetic force can be effected by distance, the longer the distance, the weaker, the shorter the stronger. But does mass also effect magnetic force? Does a heavier magnet produce stronger magnetic force, compare to lighter one. when they're measured in the same distance? If so, is there an equation or formula to calculate it, the strength of the magnetic force? I've been looking for calculation formulas about magnetic forces for magnets on google, but I can't find anything about how mass affects magnetic force for a magnet. (Maybe I'm just using the wrong key word.)

Ps: This has nothing to do about electromagnetism, just normal magnets.

Magnetism
Answered

vilitatelp014
2022-05-09

A charged particle in a magnetic field acquires a geometric phase, so a neutral particle if by any method is able to acquire this geometric phase, then that method is said to create a synthetic magnetic field.

Magnetic force
Answered

Fescoisyncsibgyp8b
2022-05-09

Assume the particles generates the magnetic field with their own spin. No motions except spin.(they have their intrinsic angular momentum.) Then which one generates stronger magnetic force?

I've seen the formula which represents the magnetic moment of the 1/2 spin particles with charge $q$, mass $m$, $\overrightarrow{\mu}=\frac{{g}_{s}q}{2m}\overrightarrow{S}$ where ${g}_{s}$ called g-factor.

But I can't understand the formula.

Let me ask a question, which magnetic force is stronger between electron and proton?

Ferromagnetism
Answered

Kevin Snyder
2022-05-09

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

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?

Magnetism
Answered

lifretatox8n
2022-05-09

(a) Faraday's law

(b) Ampere - Maxwell law

(c) Gauss's law in electricity

(d) Gauss's law in magnetism

Magnetism
Answered

Jazlyn Raymond
2022-05-09

What is the amplitude of the oscillating electric field?

What is the amplitude of the oscillating magnetic field?

Magnetic force
Answered

britesoulusjhq
2022-05-09

I'm misunderstanding something very important regarding magnetic force and its relation to velocity.

According to the Lorentz force, the magnetic force ${\mathbf{F}}_{\mathbf{B}}=q\mathbf{v}\times \mathbf{B}$. Assuming the charge is not moving, then $\mathbf{v}=\left(\begin{array}{ccc}0& 0& 0\end{array}\right)$. Therefore, ${\mathbf{F}}_{\mathbf{B}}=0$

So why when I hold a magnet close to a piece of metal, I can feel the magnet applying a force on the piece of metal? Since the piece of metal and the magnet are not moving, shouldn't the net magnetic force be 0? I assume the force I am feeling is from the magnetic field from the magnet and the charges in the piece of metal.

Magnetism
Answered

poklanima5lqp3
2022-05-08

$\overrightarrow{B}=\left(2|u|\frac{\mu \mathrm{cos}\theta}{4\pi |\overrightarrow{R}{|}^{3}}\right)\hat{r}+\left(|u|\frac{\mu \mathrm{sin}\theta}{4\pi |\overrightarrow{R}{|}^{3}}\right)\hat{\theta}+0\hat{\varphi}\phantom{\rule{thinmathspace}{0ex}}.$

Gauss's Law for Magnetism stats that $\u25bd\cdot \overrightarrow{B}=0$. However, this:

$(\hat{r}\frac{\mathrm{\partial}}{\mathrm{\partial}r}+\frac{\hat{\theta}}{r}\frac{\mathrm{\partial}}{\mathrm{\partial}\theta}+\frac{\hat{\varphi}}{r\mathrm{sin}\theta}\frac{\mathrm{\partial}}{\mathrm{\partial}\varphi})\cdot [\left(2|u|\frac{\mu \mathrm{cos}\theta}{4\pi |\overrightarrow{R}{|}^{3}}\right)\hat{r}+\left(|u|\frac{\mu \mathrm{sin}\theta}{4\pi |\overrightarrow{R}{|}^{3}}\right)\hat{\theta}+0\hat{\varphi}]\ne 0\phantom{\rule{thinmathspace}{0ex}}.$

All terms should cancel, but they do not. Where am I going wrong?

What I got is:

$\u25bd\cdot \overrightarrow{B}=-\frac{6|u|\mu \cdot \hat{r}}{4\pi |R{|}^{4}}+\frac{|u|\mu \cdot \hat{r}}{4\pi |R{|}^{4}}+0\therefore \ne 0\phantom{\rule{thinmathspace}{0ex}}.$

Magnetic force
Answered

Adelyn Rodriguez
2022-05-08

A charged particle near a current-carrying wire does not experience a magnetic force when its velocity is equal to $0$. So why does a compass needle kept near a current carrying wire experience a force when the compass needle is at rest?

Magnetism
Answered

spazzter08dyk2n
2022-05-08

Electromagnetism Physics belongs to those aspects of the science that can be applied to anything from engineering to design and construction of automation for our daily life. Since we are dealing with the relations between electricity and magnetism, finding answers and questions that we present will help you to see practical examples like radio, gamma rays, or microwave emissions. Take at least one electromagnetism example and you will be able to repeat the steps or receive relevant help. Make sure to focus on the experiments and the order of the steps to gain even more clarity as you study electromagnetism.