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3 years ago

What factors that affect magnetic force

Physics

2 answers:

Reil [10]3 years ago

7 0

Answer:

Strong electrical currents in close proximity to the magnet.

Other magnets in close proximity to the magnet.

Neo magnets will corrode in high humidity environments unless they have a protective coating.

Explanation: Heat radiation

barxatty [35]3 years ago

6 0

Answer: 1 Heat.

2 Radiation.

3 Strong electrical currents in close proximity to the magnet.

4 Other magnets in close proximity to the magnet.

5 Neo magnets will corrode in high humidity environments unless they have a protective coating. Hope this helps. Can I give me brainliest

Explanation:

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30 points - how do I do 2 b and c?

Delvig [45]

1. its must be B and 2. must be C

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3 years ago

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Answer:

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3 years ago

Four long wires are each carrying 6.0 A. The wires are located

Firdavs [7]

Answer:

$B_T=2.0*10^-5[-\hat{i}+\hat{j}]T$

Explanation:

To find the magnitude of the magnetic field, you use the following formula for the calculation of the magnetic field generated by a current in a wire:

$B=\frac{\mu_oI}{2\pi r}$

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

I: current = 6.0 A

r: distance to the wire in which magnetic field is measured

In this case, you have four wires at corners of a square of length 9.0cm = 0.09m

You calculate the magnetic field in one corner. Then, you have to sum the contribution of all magnetic field generated by the other three wires, in the other corners. Furthermore, you have to take into account the direction of such magnetic fields. The direction of the magnetic field is given by the right-hand side rule.

If you assume that the magnetic field is measured in the up-right corner of the square, the wire to the left generates a magnetic field (in the corner in which you measure B) with direction upward (+ j), the wire down (down-right) generates a magnetic field with direction to the left (- i) and the third wire generates a magnetic field with a direction that is 45° over the horizontal in the left direction (you can notice that in the image attached below). The total magnetic field will be:

$B_T=B_1+B_2+B_3\\\\B_{T}=\frac{\mu_o I_1}{2\pi r_1}\hat{j}-\frac{\mu_o I_2}{2\pi r_2}\hat{i}+\frac{\mu_o I_3}{2\pi r_3}[-cos45\hat{i}+sin45\hat{j}]$

I1 = I2 = I3 = 6.0A

r1 = 0.09m

r2 = 0.09m

$r_3=\sqrt{(0.09)^2+(0.09)^2}m=0.127m$

Then you have:

$B_T=\frac{\mu_o I}{2\pi}[(-\frac{1}{r_2}-\frac{cos45}{r_3})\hat{i}+(\frac{1}{r_1}+\frac{sin45}{r_3})\hat{j}}]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[(-\frac{1}{0.09m}-\frac{cos45}{0.127m})\hat{i}+(\frac{1}{0.09m}+\frac{sin45}{0.127m})]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[-16.67\hat{i}+16.67\hat{j}]\\\\B_T=2.0*10^-5[-\hat{i}+\hat{j}]T$

5 0

3 years ago

The best time to get my RESTING HEART RATE is right after I run.<br><br>true or false

Savatey [412]

Answer:

Your answer is: False

It is called a resting heart rate for a reason : )

Explanation:

Hope this helped : )

8 0

3 years ago

Read 2 more answers

If it has enough kinetic energy, a molecule at the surface of the Earth can "escape the Earth's gravitation", in the sense that

user100 [1]

Answer:

$K_E=mgr_E$

Explanation:

By conservation of energy, the sum of the kinetic and gravitational potential energies at the surface of the Earth must be equal than their sum at infinity, so we have:

$K_E+U_E=K_\infty+U_\infty$

$\frac{mv_E^2}{2}-\frac{GM_Em}{r_E}=\frac{mv_\infty^2}{2}-\frac{GM_Em}{r_\infty}$

Where $G=6.67\times10^{-11}Nm^2/kg^2$ is the gravitational constant, $M_E=5.97\times10^{24}kg$ and $r_E=6371000m$ are the mass and radius of the Earth, <em>m </em>is the mass of the particle, $v_E$ its velocity at the surface of the Earth (which would be its escape velocity) and $v_\infty$ and $r_\infty$ are the velocities and distance at infinity, which would be null and infinity respectively, so the right hand side of our equation is 0J, which leaves us with:

$\frac{GM_Em}{r_E}=\frac{mv_E^2}{2}=K_E$

Also, since the force the molecule experiments is the force of gravity (disregarding drag), we can write its weight in terms of Newton's Law of Gravitation:

$F=mg=\frac{GM_Em}{r_E^2}$

Which means that:

$\frac{GM_Em}{r_E}=mgr_E$

So finally putting all together we can write:

$K_E=\frac{GM_Em}{r_E}=mgr_E$

4 0

3 years ago