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Anestetic [448]
4 years ago
5

Which of the following statements is true about earth's magnetic poles

Physics
1 answer:
Dahasolnce [82]4 years ago
4 0

Answer: b) they are the areas where Earth's magnetic field is weakest

Explanation:

According to classical physics, a magnetic field always has two associated magnetic poles (north and south), the same happens with magnets. This is because for <em>classical physics</em>, naturally, magnetic monopoles can not exist.  

In this context, Earth is similar to a magnetic bar with a north pole and a south pole. This means, the axis that crosses the Earth from pole to pole is like a big magnet.  

Now, by convention, on all magnets the north pole is where the magnetic lines of force leave the magnet and the south pole is where the magnetic lines of force enter the magnet.  Then, for the case of the Earth, the north pole of the magnet is located towards the geographic south pole and the south pole of the magnet is near the geographic north pole.  

Being the magnetic poles the places where the Earth's magnetic field is weakest. And it is for this reason, moreover, that the magnetic field lines enter the Earth through its magnetic south pole (which is the geographic north pole).

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The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m
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Answer:

  • The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

F = k \frac{q_1 q_2}{d^2}

where k is Coulomb's constant, q_1 and q_2 are the charges and d is the distance between the charges.

Working a little the equation, we can take:

d^2 = k \frac{q_1 q_2}{F}

d = \sqrt{ k \frac{q_1 q_2}{F}}

And this equation will give us the distance between the charges. Taking the values of the problem

k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00

(the force has a minus sign, as its attractive)

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 28,462,500 \ m^2}}

d = 5,335.026 m

And this is the distance between the charges.

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