Recent questions in Magnetism

ElectromagnetismAnswered question

zapri4j 2022-09-27

Hard magnets are materials, in which the process of

a. magnetization is done easily, but demagnetization is difficult

b. demagnetization is done easily, but magnetization is difficult

c. magnetization and demagnetization is done easily

d. magnetization and demagnetization is difficult

a. magnetization is done easily, but demagnetization is difficult

b. demagnetization is done easily, but magnetization is difficult

c. magnetization and demagnetization is done easily

d. magnetization and demagnetization is difficult

ElectromagnetismAnswered question

Haven Kerr 2022-09-23

Explain the concept of Ampere’s Law and cite at least two (2) importance

ElectromagnetismAnswered question

Jase Rocha 2022-09-21

If magnetic monopoles existed, how would the Gauss’s law of magnetism be modified?

ElectromagnetismAnswered question

Orion Cervantes 2022-09-15

Show that the mass required is given by $m=\frac{{\mu}_{0}L{I}^{2}}{2\pi rg}$ where g is the acceleration due to gravity on the surface of Earth.

ElectromagnetismOpen question

Nashorn0o 2022-08-19

The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth’s field in past?

ElectromagnetismOpen question

brasocas6 2022-08-18

Earth has a magnetic dipole moment of $8.0\cdot {10}^{22}\text{}J/T$.(a) What current would have to be produced in a single turn of wire extending around Earth at its geomagnetic equator if we wished to set up such a dipole? Could such an arrangement be used to cancel out Earth’s magnetism (b) at points in space wellabove Earth’s surface or (c) on Earth’s surface?

ElectromagnetismOpen question

sublimnes9 2022-08-18

True or false: Ampere's law (with displacement current) is one of Maxwell's Equations:

a. True

b. False, even the names are different

a. True

b. False, even the names are different

ElectromagnetismOpen question

assuolareuz 2022-08-13

According to Gauss' law for magnetism, the second equation of Maxwell's Equations shown below, magnetic field lines

$\oint \overrightarrow{E}\cdot \overrightarrow{d}\overrightarrow{A}=q/\u03f5$

$\oint \overrightarrow{B}\cdot d\overrightarrow{A}=0$

$\oint \overrightarrow{B}\cdot d\overrightarrow{A}=0\phantom{\rule{0ex}{0ex}}\oint \overrightarrow{E}\cdot d\overrightarrow{s}=-d{\mathrm{\Phi}}_{B}/dt\phantom{\rule{0ex}{0ex}}\oint \overrightarrow{B}\cdot d\overrightarrow{s}={\mu}_{0}i+{\mu}_{0}{\u03f5}_{0}d{\mathrm{\Phi}}_{E}/dt$

$\oint \overrightarrow{E}\cdot \overrightarrow{d}\overrightarrow{A}=q/\u03f5$

$\oint \overrightarrow{B}\cdot d\overrightarrow{A}=0$

$\oint \overrightarrow{B}\cdot d\overrightarrow{A}=0\phantom{\rule{0ex}{0ex}}\oint \overrightarrow{E}\cdot d\overrightarrow{s}=-d{\mathrm{\Phi}}_{B}/dt\phantom{\rule{0ex}{0ex}}\oint \overrightarrow{B}\cdot d\overrightarrow{s}={\mu}_{0}i+{\mu}_{0}{\u03f5}_{0}d{\mathrm{\Phi}}_{E}/dt$

ElectromagnetismAnswered question

proximumha 2022-08-12

What are the effects of height of wire to magnetism

ElectromagnetismAnswered question

traucaderx7 2022-08-08

How does induced magnetism work with magnetic induction and magnetic domain?

ElectromagnetismAnswered question

Francisco Proctor 2022-07-22

Description of the steps/processes involved based on electromagnetism in television (any concepts of magnetism)

ElectromagnetismAnswered question

Lilliana Livingston 2022-07-20

The Maxwell equation derived from Gauss's Law in magnetism is

a) $\oint \overrightarrow{B}\overrightarrow{ds}={\mu}_{0}I+{\u03f5}_{0}{\mu}_{0}\frac{d{\mathrm{\Phi}}_{E}}{dt}$

b) $\oint \overrightarrow{B}\overrightarrow{dA}=0$

c) $\oint \overrightarrow{E}\overrightarrow{dA}=\frac{{Q}_{in}}{{\u03f5}_{0}}$

d) $\oint \overrightarrow{E}\overrightarrow{ds}=-\frac{d{\mathrm{\Phi}}_{m}}{dt}$

a) $\oint \overrightarrow{B}\overrightarrow{ds}={\mu}_{0}I+{\u03f5}_{0}{\mu}_{0}\frac{d{\mathrm{\Phi}}_{E}}{dt}$

b) $\oint \overrightarrow{B}\overrightarrow{dA}=0$

c) $\oint \overrightarrow{E}\overrightarrow{dA}=\frac{{Q}_{in}}{{\u03f5}_{0}}$

d) $\oint \overrightarrow{E}\overrightarrow{ds}=-\frac{d{\mathrm{\Phi}}_{m}}{dt}$

ElectromagnetismAnswered question

Makena Preston 2022-07-17

An isolated conducting Sphere of radius R has a charge Q. What is the total stored energy? What is the radius r within which half the stored energy is contained?

ElectromagnetismAnswered question

Ellie Benjamin 2022-07-15

a) Faradays formulate two basic laws of electromagnetic induction based on Faradays experiments.

Explain Faradays experiment and at the end depending on the experiment, state Faradays laws.

b) Explain how one can increase the emf induced in the coil by change in magnetic field.

c) State Lenz law.

Explain Faradays experiment and at the end depending on the experiment, state Faradays laws.

b) Explain how one can increase the emf induced in the coil by change in magnetic field.

c) State Lenz law.

ElectromagnetismAnswered question

Jamison Rios 2022-07-15

I want to derive Gauss's law for magnetism,

$\mathrm{\nabla}\cdot \overrightarrow{B}=0\phantom{\rule{thinmathspace}{0ex}}.$

The derivation in Griffiths Introduction to elecrodynamics uses

$\mathrm{\nabla}\cdot \overrightarrow{B}\text{}=\text{}\frac{{\mu}_{0}}{4\pi}\int \mathrm{\nabla}\cdot (\overrightarrow{J}\times \frac{\overrightarrow{r}}{{r}^{3}})\phantom{\rule{thinmathspace}{0ex}},$

but to use this, I would need $\overrightarrow{J}\times \frac{\overrightarrow{r}}{{r}^{3}}$ and $\mathrm{\nabla}\cdot (\overrightarrow{J}\times \frac{\overrightarrow{r}}{{r}^{3}})$ to be continuous.

Since $\overrightarrow{J}=\rho \overrightarrow{v}$ , then if $\rho $ (charge density) has a continuous derivative, I'm done; but, I'm not sure why can I affirm this because clearly, I can construct charge distributions with jump discontinuities.

$\mathrm{\nabla}\cdot \overrightarrow{B}=0\phantom{\rule{thinmathspace}{0ex}}.$

The derivation in Griffiths Introduction to elecrodynamics uses

$\mathrm{\nabla}\cdot \overrightarrow{B}\text{}=\text{}\frac{{\mu}_{0}}{4\pi}\int \mathrm{\nabla}\cdot (\overrightarrow{J}\times \frac{\overrightarrow{r}}{{r}^{3}})\phantom{\rule{thinmathspace}{0ex}},$

but to use this, I would need $\overrightarrow{J}\times \frac{\overrightarrow{r}}{{r}^{3}}$ and $\mathrm{\nabla}\cdot (\overrightarrow{J}\times \frac{\overrightarrow{r}}{{r}^{3}})$ to be continuous.

Since $\overrightarrow{J}=\rho \overrightarrow{v}$ , then if $\rho $ (charge density) has a continuous derivative, I'm done; but, I'm not sure why can I affirm this because clearly, I can construct charge distributions with jump discontinuities.

ElectromagnetismAnswered question

garcialdaria2zky1 2022-05-20

If the magnitude of the magnetic induction is less than that in a vaccum within a solid, so, this material called

a) Ferrimagnetic

b) Ferromagnetic

c) Diamagnetic

d) Paramagnetic

a) Ferrimagnetic

b) Ferromagnetic

c) Diamagnetic

d) Paramagnetic

ElectromagnetismAnswered question

Angelique Horne 2022-05-20

How many neutral points can be obtained in a given plane perpendicular to the length of the two parallel wires conducting current in the same direction? Explain, Neglect earth's magnetic field of a bar magnet NS

ElectromagnetismAnswered question

adocidasiaqxm 2022-05-19

I know that there is some relationship between electric current and magnetism, but I am having trouble pinning down the exact relationship. Is electric current a necessary condition for the existence of magnetic field? And is electric current a sufficient condition for the existence of magnetic field?

In other words, is it true that "there is electric current if and only if there is a magnetic field"?

In other words, is it true that "there is electric current if and only if there is a magnetic field"?

ElectromagnetismAnswered question

Raphael Mccullough 2022-05-18

The inverse square law for an electric field is:

$E=\frac{Q}{4\pi {\epsilon}_{0}{r}^{2}}$

Here:

$\frac{Q}{{\epsilon}_{0}}$

is the source strength of the charge. It is the point charge divided by the vacuum permittivity or electric constant, I would like very much to know what is meant by source strength as I can't find it anywhere on the internet. Coming to the point an electric field is also described as:

$Ed=\frac{Fd}{Q}=\mathrm{\Delta}V$

This would mean that an electric field can act only over a certain distance. But according to the Inverse Square Law, the denominator is the surface area of a sphere and we can extend this radius to infinity and still have a value for the electric field. Does this mean that any electric field extends to infinity but its intensity diminishes with increasing length? If that is so, then an electric field is capable of applying infinite energy on any charged particle since from the above mentioned equation, if the distance over which the electric field acts is infinite, then the work done on any charged particle by the field is infinite, therefore the energy supplied by an electric field is infinite. This clashes directly with energy-mass conservation laws. Maybe I don't understand this concept properly, I was hoping someone would help me understand this better.

$E=\frac{Q}{4\pi {\epsilon}_{0}{r}^{2}}$

Here:

$\frac{Q}{{\epsilon}_{0}}$

is the source strength of the charge. It is the point charge divided by the vacuum permittivity or electric constant, I would like very much to know what is meant by source strength as I can't find it anywhere on the internet. Coming to the point an electric field is also described as:

$Ed=\frac{Fd}{Q}=\mathrm{\Delta}V$

This would mean that an electric field can act only over a certain distance. But according to the Inverse Square Law, the denominator is the surface area of a sphere and we can extend this radius to infinity and still have a value for the electric field. Does this mean that any electric field extends to infinity but its intensity diminishes with increasing length? If that is so, then an electric field is capable of applying infinite energy on any charged particle since from the above mentioned equation, if the distance over which the electric field acts is infinite, then the work done on any charged particle by the field is infinite, therefore the energy supplied by an electric field is infinite. This clashes directly with energy-mass conservation laws. Maybe I don't understand this concept properly, I was hoping someone would help me understand this better.

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