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ioda
3 years ago
11

The earth has a magnetic field much like a bar magnet. Explain this statement. what does a bar magnet look like? what does its f

ield look like? where are the poles? how does it relate to the earth?
Physics
2 answers:
OlgaM077 [116]3 years ago
8 0

Explanation:

The earth has a magnetic field. It is much like a bar magnet. Imagine a gigantic bar magnet inside the Earth. But there is no giant magnet inside it.

To have a pretty good idea what earth's magnetic field is shaped like we imagine a bar magnet inside the earth.

The magnetic field is made by the motion of molten iron in earth's outer core.  The swirling motion of molten iron changes all the time. Therefore, the magnetic fields will also get change. Then, the magnet poles also move.

The North pole and the south pole are two geographic poles of earth.  These poles are the places on the earth's surface that earth's imaginary spin axis passes through.

There are two magnetic poles of the earth: North magnetic pole and South magnetic pole.

Earth's magnetic field is tilted a little bit. If we imagine that earth's magnetic field is made by a giant bar magnet. Then, the bar magnet would make an with earth's spin axis.

The geographic poles and the magnetic poles are not in the same place.

If we are standing at one magnetic poles then the magnetic field lines would be straight up and down.

amm18123 years ago
7 0
The Earth magnetic fields helps balance out the planet  by the core from the north pole to the south pole in order to keep earth gravity at stand even during revolving around the solar system
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Answer:

6

Explanation:

Half the wave = 3

Wavelength = 3 x 2 = 6

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A 50 Ohm resistance causes a current of 5 milliamps to flow through a circuit connected to a battery. What is the power in the c
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Explanation:

resistance=50 ohms

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8 0
3 years ago
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Which of the following statements describes an electric generator?
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3 years ago
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When serving a tennis ball, a player hits the ball when its velocity is zero (at the highest point of a vertical toss). The racq
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8 0
3 years ago
1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x
allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}}  }

With (\epsilon_{0}) been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} }    }

Explanation:

(A) Considering a uniform linear density λ_{0} on the ring, then:

dQ=\lambda dl (1)⇒Q=\lambda_{0} 2\pi R(2)⇒\lambda_{0}=\frac{Q}{2\pi R}(3)

Applying the technique of charge integration for finite charges:

V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r'  }} \, dQ(4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

r'=\sqrt{R^{2} +Z^{2}}(5)

Using the expressions (1),(4) and (5) you obtain:

V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}}  }} \, d\phi

Integrating results:

V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}}  }   (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

|E| =-\nabla V (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

|E| =-\frac{dV(z)}{dz} (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} }    } (S_b)

4 0
3 years ago
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