traucaderx7
2022-08-08
Answered

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

You can still ask an expert for help

Luna Wells

Answered 2022-08-09
Author has **19** answers

Answer:

Magnetic Induction is the process of making a non-magnetic material behave as a magnet when is inside the magnetic field of a magnet. Thus, when a non-magnetized magnetic material is brought near the pole of a permanent magnet it behave as a magnet.

A magnetic domain is a region within a magnetic material in which magnetization is uniform. Thus, in side the magnetic domain a magnetic material magnet behaves as magnet.

Magnetic Induction is the process of making a non-magnetic material behave as a magnet when is inside the magnetic field of a magnet. Thus, when a non-magnetized magnetic material is brought near the pole of a permanent magnet it behave as a magnet.

A magnetic domain is a region within a magnetic material in which magnetization is uniform. Thus, in side the magnetic domain a magnetic material magnet behaves as magnet.

asked 2021-05-04

Until he was in his seventies, Henri LaMothe excited audiences by belly-flopping from a height of 13 m into 35 cm. of water. Assuming that he stops just as he reaches the bottom of the water and estimating his mass to be 72 kg, find the magnitudes of the impulse on him from the water.

asked 2021-02-06

Starting with an initial speed of 5.00 m/s at a height of 0.300 m, a 1.50 kg ball swings downward and strikes a 4.60kg ballthat is at rest, as the drawing shows. a. using the principle of conservation of mechanicalenergy,find the speed of the 1.50 kg ball just before impact b. assuming that the collision is elastic, find the velocities( magnitude and direction ) of both balls just after thecollision c. how high does each abll swing after the collision, ignoringair resistance?

asked 2021-01-04

Four very long, current-carrying wires in the same plane intersect to form a square with side lengths of 45.0 cm, as shown in the figure blelow. The currents running through the wires are 8.0 A, 20.0 A, 10.0 A, and I.

Part A

Find the magnitude of the current I that will make the magnetic field at the center of the square equal to zero.

Part B

What is the direction of the current I (up or down)?

asked 2021-01-19

In a truck-loading station at a post office, a small 0.200-kg package is released from rest a point A on a track that is one quarter of a circle with radius 1.60 m.The size of the packageis much less than 1.60m, so the package can be treated at aparticle. It slides down the track and reaches point B with a speed of 4.80 m/s. From point B, it slides on a level surface a distanceof 3.00 m to point C, where is comes to rest.

(a) What is the coefficient of kinetic friction on the horizontal surface?

(b) How much work is done on the package by friction as it slides down the circular arc from A to B?

asked 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?

asked 2022-04-12

The Maxwell equation $\oint \overrightarrow{E}\cdot \overrightarrow{dA}=-\frac{{Q}_{in}}{{\u03f5}_{0}}$ is also called

a. Faraday's Law of Electromagnetic induction

b. Ampere's Law

c. Gauss Law of Magnetism

d. Gauss Law of Electricity

a. Faraday's Law of Electromagnetic induction

b. Ampere's Law

c. Gauss Law of Magnetism

d. Gauss Law of Electricity

asked 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:

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