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Alja [10]
2 years ago
9

What direction would the north pole of a bar magnet point if you were to hang the bar magnet from a thin string?.

Physics
1 answer:
inysia [295]2 years ago
8 0
The end that is marked with ’N’ will point to the Arctic. The ‘S’ end will point toward the Antarctic. A bar magnet is ‘just’ a big heavy compass needle. The ‘N’ and ‘S’ marking are made to help travelers find their way around the globe.

Yes, it IS confusing to have the ’N’ end of a compass point toward the north axis of the earth. These conventions were well in place LONG before anyone understood that ‘Opposites Attract’. But now that we DO know that ‘Opposites Attract’ we are left with the unfortunate reality of compasses, bar magnets and the earth. There are gazillions of magnets and only one earth! So the ‘convention’ has been made that the magnets are labeled correctly! This means (unfortunately) that the Arctic area of the globe is actually the south end of the earthly magnet. The earth is a giant (but very weak) electromagnetic and the lines of the magnetic field point to the Arctic Regions which is why all out magnets point that way. The magnetic field line dive INTO the earth in the Arctic Regions and OUT of the earth in the Antarctic regions.

I haven’t used the word ‘pole’ here, because magnetic poles don’t actually exist (as far as we know). Every magnet is actually an electromagnetic with electrical charges in motion creating the magnetic field.
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3 years ago
Two long, parallel wires are separated by a distance of 2.60 cm. The force per unit length that each wire exerts on the other is
Alex777 [14]

Answer:

<em>10.75 A</em>

<em>The current is in opposite direction since it causes a repulsion force between the wires</em>

Explanation:

Force per unit length on the wires = 4.30×10^−5 N/m

distance between wires = 2.6 cm = 0.026 m

current through one wire = 0.52 A

current on the other wire = ?

Recall that the force per unit length of two wires conducting and lying parallel and close to each other is given as

F/l = \frac{u_{0}I_{1} I_{2}  }{2\pi r }

where F/l is the force per unit length on the wires

u_{0} = permeability of vacuum = 4π × 10^−7 T-m/A

I_{1} = current on the first wire = 0.520 A

I_{2} = current on the other wire = ?

r = the distance between the two wire = 0.026 m

substituting the value into the equation, we have

4.30×10^−5 = \frac{4\pi *10^{-7}*0.520*I_{2}  }{2\pi *0.026} =  \frac{ 2*10^{-7}*0.520*I_{2}  }{0.026}

4.30×10^−5 = 4 x 10^-6 I_{2}

I_{2} = (4.30×10^-5)/(4 x 10^-6) = <em>10.75 A</em>

<em>The current is in opposite direction since it causes a repulsion force between the wires.</em>

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4 years ago
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Answer:

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Explanation:

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3 years ago
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zloy xaker [14]
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