Recent questions in Magnetism

ElectromagnetismAnswered question

Jayden Mckay 2022-05-17

State Gauss’ Law for Magnetism. What significant thing does it tell us about the universe?

ElectromagnetismAnswered question

hetriamhageh6k20 2022-05-17

Gauss's Law for magnetism is

$\mathrm{\nabla}\cdot B=0$

This allows us to write the magentic field B as the curl of another field the magnetic vector potential, A.

$B=\mathrm{\nabla}\times A$

This adhers to $\mathrm{\nabla}\cdot (\mathrm{\nabla}\times A)=0$

However, if a monopole does exist then we have

$\mathrm{\nabla}\cdot B={\rho}_{m}$

Where ${\rho}_{m}$ is some magnetic charge density however with a magentic vector potential this violates the equation, $\mathrm{\nabla}\cdot (\mathrm{\nabla}\times A)\ne 0$

Does that mean if magnetic monopoles does exist, that the magnetic field can no longer be defined by a magnetic vector potential? In which case how was dirac able to still define the magnetic field by a magnetic vector potentials?

$\mathrm{\nabla}\cdot B=0$

This allows us to write the magentic field B as the curl of another field the magnetic vector potential, A.

$B=\mathrm{\nabla}\times A$

This adhers to $\mathrm{\nabla}\cdot (\mathrm{\nabla}\times A)=0$

However, if a monopole does exist then we have

$\mathrm{\nabla}\cdot B={\rho}_{m}$

Where ${\rho}_{m}$ is some magnetic charge density however with a magentic vector potential this violates the equation, $\mathrm{\nabla}\cdot (\mathrm{\nabla}\times A)\ne 0$

Does that mean if magnetic monopoles does exist, that the magnetic field can no longer be defined by a magnetic vector potential? In which case how was dirac able to still define the magnetic field by a magnetic vector potentials?

ElectromagnetismAnswered question

revistasbibxjm87 2022-05-17

A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity 5 rad/s . If the horizontal component of earth’s magnetism is $2\times {10}^{-5}T$, then voltage developed between the two ends of the conductor is ?

ElectromagnetismAnswered question

terrasson81sgt 2022-05-15

The area of a rectangular loop is $200\text{}c{m}^{2}$, and the plane of the loop makes an angle of ${41}^{\circ}$ with a 0.28-T magnetic field. What is the magnetic flux penetrating the loop?

ElectromagnetismAnswered question

indimiamimactjcf 2022-05-14

Gauss's law for magnetism is stated as followed with the beautiful closed surface double integral :

$\underset{S}{\text{\u222f}\phantom{\rule{thinmathspace}{0ex}}}\mathbf{B}\cdot \text{d}\mathbf{A}=0$

As I understand, the idea is to say that if we sum (continuous sum since integral) all the scalar products between the vector field B (i.e., magnetic field) and surface elements dA defined by their surface normals, we get 0?

$\underset{S}{\text{\u222f}\phantom{\rule{thinmathspace}{0ex}}}\mathbf{B}\cdot \text{d}\mathbf{A}=0$

As I understand, the idea is to say that if we sum (continuous sum since integral) all the scalar products between the vector field B (i.e., magnetic field) and surface elements dA defined by their surface normals, we get 0?

ElectromagnetismAnswered question

hard12bb30crg 2022-05-13

Gauss's Law of Magnetism shows us that the divergence of Magnetic field is 0, $\u25bd\cdot \overrightarrow{B}=0$

Then how do you derive that statement by showing the divergence of a magnetic field upon an axis of a current carrying coil where radius is much smaller that distance so that we can use,

${B}_{z}=\frac{{\mu}_{o}I}{2{z}^{3}}\hat{z}$

$\therefore $

$\u25bd\cdot B\equiv \frac{\mathrm{\partial}}{\mathrm{\partial}z}\cdot \frac{{u}_{o}I}{2{z}^{3}}\hat{z}\ne 0$

This doesn't equal zero? What am I missing?

Then how do you derive that statement by showing the divergence of a magnetic field upon an axis of a current carrying coil where radius is much smaller that distance so that we can use,

${B}_{z}=\frac{{\mu}_{o}I}{2{z}^{3}}\hat{z}$

$\therefore $

$\u25bd\cdot B\equiv \frac{\mathrm{\partial}}{\mathrm{\partial}z}\cdot \frac{{u}_{o}I}{2{z}^{3}}\hat{z}\ne 0$

This doesn't equal zero? What am I missing?

ElectromagnetismAnswered question

Blaine Stein 2022-05-13

If Div B = 0, where B = magnetic field intensity, then B must be a Curl of a some vector function. What is that vector function?

ElectromagnetismAnswered question

kwisangqaquqw3 2022-05-10

Solve the equations for ${v}_{x}$ and ${v}_{y}$ :

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

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:

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

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

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:

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

ElectromagnetismAnswered question

indimiamimactjcf 2022-05-10

The current in a conductor is found to be moving from (-1, 2, -3) m to (-4, 5, -6) m. Given that the force the conductor experiences is $\overrightarrow{F}=12({a}_{y}+{a}_{z}{)}^{N}$ and the magnetic field present in the region is $\overrightarrow{B}=2{a}_{x}T$, determine the current through the conductor. a.) -1 A;

b.) -2 A;

c.) This problem has no solution.

d.) 2 A;

b.) -2 A;

c.) This problem has no solution.

d.) 2 A;

ElectromagnetismAnswered question

Bernard Mora 2022-05-10

In Maxwell's equations, I understand intuitively how: $\oint B\cdot da=0$ (because there are no monopoles and so equal number of field lines going in and coming out of the surface).

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?

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?

ElectromagnetismAnswered question

vilitatelp014 2022-05-09

What I understand is that Synthetic magnetism is just a fancy name of a method to make a charge neutral particle act like it is in a magnetic field.

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.

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.

ElectromagnetismAnswered question

lifretatox8n 2022-05-09

The given equation $\int \overrightarrow{E}\cdot \overrightarrow{ds}=-\frac{d{\mathrm{\Phi}}_{m}}{dt}$ is called dt

(a) Faraday's law

(b) Ampere - Maxwell law

(c) Gauss's law in electricity

(d) Gauss's law in magnetism

(a) Faraday's law

(b) Ampere - Maxwell law

(c) Gauss's law in electricity

(d) Gauss's law in magnetism

ElectromagnetismAnswered question

Jazlyn Raymond 2022-05-09

A microwave oven operates at 2.4 GHz with an intensity inside the oven of $2700\text{}W/{m}^{2}.$

What is the amplitude of the oscillating electric field?

What is the amplitude of the oscillating magnetic field?

What is the amplitude of the oscillating electric field?

What is the amplitude of the oscillating magnetic field?

ElectromagnetismAnswered question

poklanima5lqp3 2022-05-08

$|u|=i\overrightarrow{A}$

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

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:

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

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

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:

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

ElectromagnetismAnswered question

spazzter08dyk2n 2022-05-08

Earth has a magnetic dipole moment of $8.0\text{}{10}^{22}\text{}J/T$. Find the magnitude of current in a single turn of wire extending around Earth at its geomagnetic equator to produce a dipole having the same dipole moment. Could such an arrangement be used to cancel out Earth's magnetism?

ElectromagnetismAnswered question

Jay Barrett 2022-05-07

Lets say you have an uniform prism magnet of Iron for example. How would you calculate the demagnetization field H which the bar magnet produces? As I understand, first you need the magnetization M which is the magnetic moments per volumen. But then what ? Or would you apply and external field to the bar magnet and see how it reacts, if this is the case, how does this work ?

ElectromagnetismAnswered question

Alissa Hutchinson 2022-05-07

Gauss's law states that ${\int}_{S}\overrightarrow{B}\cdot d\overrightarrow{S}=0$. But law of induction states that $\xi =-\frac{d\varphi}{dt}$, where $\varphi ={\int}_{S}\overrightarrow{B}\cdot d\overrightarrow{S}$

So if Gauss's law was to be correct there should be no induction at all, because then ϕ would be zero through every loop.

So if Gauss's law was to be correct there should be no induction at all, because then ϕ would be zero through every loop.

ElectromagnetismAnswered question

NepanitaNesg3a 2022-05-02

How can the integral form of Gauss's law for magnetism be described as a version of general Stokes' theorem? How does it follow?

ElectromagnetismAnswered question

Poemslore8ye 2022-04-30

My question is: is earth really a magnet? Doesn anyone have any proof that earth is a magnet? Is there a magnetic core at the center of the earth? Has anyone reached the core of the earth?

ElectromagnetismAnswered question

robinmarian9nhn8 2022-04-12

Consider the circuit shown in

photo

A)What are the magnitude and direction of the current in the $20\mathrm{\Omega}$

resistor in figure

Express your answer with the appropriate units. Enter positive value if the current is clockwise and negative value if the current is counterclockwise.

photo

A)What are the magnitude and direction of the current in the $20\mathrm{\Omega}$

resistor in figure

Express your answer with the appropriate units. Enter positive value if the current is clockwise and negative value if the current is counterclockwise.

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