Nelson Jennings
2022-07-20
Answered

A small bar magnet is hidden in a fixed position inside a tennis ball. Describe an experiment that you could do to find the location of the north pole and the south pole of the magnet.

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Cael Cox

Answered 2022-07-21
Author has **11** answers

We need to think of a way to determine a north and south pole of a magnet by using a tennis ball with a bar magnet fixed inside.

If we understand the interaction between Earth's magnetic field and a compass, we can see that we can create a compass out of that tennis ball. By simply putting it in the water, the ball will orientate in such a way to fit the effects of the Earth's magnetic field, thus becoming a compass.

Since opposite poles attract, the south pole of the magnet bar will point towards Earth's north and magnet's north will point into Earth's south.

If we understand the interaction between Earth's magnetic field and a compass, we can see that we can create a compass out of that tennis ball. By simply putting it in the water, the ball will orientate in such a way to fit the effects of the Earth's magnetic field, thus becoming a compass.

Since opposite poles attract, the south pole of the magnet bar will point towards Earth's north and magnet's north will point into Earth's south.

enmobladatn

Answered 2022-07-22
Author has **6** answers

The magnetic poles of this hidden magnet can be found with a compass. If we assume that the magnet is strong enough that in its vicinity of the compass the Earth's magnetic field is going to be neglectable, the needle of the compas once we put it on the ball will point towards the south magnetic pole of the bar (not north which is a common misconception, the Earth's north geographic pole is its south magnetic pole and vice versa). Now we will glide along the ball's surface until the compass starts rotating freely. That means that we have located the south magnetic pole and the north magnetic pole is on the opposite side.

Result:

For such an experiment we can use the compass.

Result:

For such an experiment we can use the compass.

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An open plastic soda bottle with an opening diameter of 2.5cm is placed on a table A uniform 1.75-7 magnetic field directed upward and oriented 25 degrees from vertical encompasses the bottle, what is the total magnetic flux through the plastic of the soda bottle

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In the figure a worker lifts a weight $\omega$ by pulling down on a rope with a force $\overrightarrow{F}$. The upper pulley is attached to the ceiling by a chain,and the lower pulley is attached to the weight by another chain.The weight is lifted at constant speed. Assume that the rope,pulleys, and chains all have negligible weights.

A) In terms of $\omega$,find the tension in the lower chain.

B) In terms of $\omega$,find the tension in upper chain.

C) In terms of $\omega$,find the magnitude of the force $\overrightarrow{F}$ if the weight is lifted at constant speed.

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Hypothetically, if I were to move at the speed of light, would I see an oscillating electromagnetic wave or photons oscillating or what?

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Which of the following group of elements are diamagnetic ?

A) Argon, copper, silver

B) Hydrogen, argon, copper

C) Oxygen, copper, silver

D) Hydrogen, oxygen, Argon

A) Argon, copper, silver

B) Hydrogen, argon, copper

C) Oxygen, copper, silver

D) Hydrogen, oxygen, Argon

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:

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

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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:

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

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I'm trying to learn the connection between special relativity and magnetism. I know that if I place a positive charge, at rest, next to wire with current, I should not observe any force on it because there is no electric field and there is no magnetic force as my charge is at rest.

But here is what confuses me - the wire contains moving electrons and according to what I learned, the stationary charge should observe a length contraction of those electrons and so the density of them will increase and a negative electric field should be observed.

This is definitely not the case and I wonder if someone can explain to me what is wrong in my analysts.

But here is what confuses me - the wire contains moving electrons and according to what I learned, the stationary charge should observe a length contraction of those electrons and so the density of them will increase and a negative electric field should be observed.

This is definitely not the case and I wonder if someone can explain to me what is wrong in my analysts.