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EleoNora [17]
3 years ago
6

II. Magnetic fields Magnets and magnetic fields EM 115 We have observed that magnets interact even when they are not in direct c

ontact. In electrostatics we used the idea of an electric field to account for the interaction between charges that were separated from one another. For magnetic interactions, we similarly define a magnetic field. A. Obtain a compass from a tutorial instructor. I. Use the compass to explore the region around a bar magnet. Describe the behavior of the compass needle both near the poles of the magnet and in the region between the poles.
Physics
1 answer:
nasty-shy [4]3 years ago
3 0

Solution :

We all know that a bar magnet have two poles, the north pole and the south pole. These poles interacts with each other. The ends of the magnets having similar poles will push each other away while the poles with like charges will pull each others towards it.

The compass needle is also a magnet having south polarity as well as north polarity. When the compass needle is close to the bar magnet, it is opposite to the poles or along the poles. The compass needle shows the direction or is pointed towards the north. So when the compass needle is placed near the north pole of the bar magnet, the pointer of the compass needle points towards the north, i.e. it gets deflected because of he like charges. And when it is placed near the south pole of the magnet, it gets attracted towards it and is pointed towards the pole.

Now as we move the compass needle from the poles to the region that is between the poles, the compass needle pointer points towards the north direction every time. It show a deflection always. If we place the magnetic lines, we will see that the magnetic lines will exit from the north poles and enters the south pole of the bar magnet.  

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ololo11 [35]

Answer:

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Explanation:

3 0
3 years ago
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tankabanditka [31]

Answer:

The equation of current is I=0.86 cos \left (120 t + \frac{\pi}{2}  \right )

Explanation:

Resistance, R = 12 ohm

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E (t) = 12 cos (120 t)

Compare with the standard equation,

E=E_{0}cos (2\pi ft)

2\pi ft = 120 t \\\\\\w = 2\pi f = 120 rad/s

So, the inductive reactance is

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The impedance of the circuit is

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The current leads  by 90degree so the equation of current is

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4 0
3 years ago
on an unknown temperature scale, the freezing point of water is -15°U and the boiling point is +60°U. develop a linear conversio
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Answer:

Since this is a linear equation

y = m x + b     or

U = m F + b     is a linear equation

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and ΔU = (60 - (-15)) = 75

m = 75 / 180 = 2.4 if converting F to U and a = .417

U = .417 F + b

If F = 32 then U = -15 and

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b = -15 - 13.3 = -28.3 and our equation becomes

U = .417 F - 28.3

Check: let F = 212

U = .417 * 212 - 28.3 = 60          as it should

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k= \frac{F}{mg}= \frac{80.85 N}{(15 kg)(9.81 m/s^2)}=0.55
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