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3 years ago

The magnet below is cut in half. what will be the result?

Physics

2 answers:

ozzi3 years ago

3 0

<h2>Answer:</h2>

<h2>Explanation:</h2>

This is what we call a permanent magnet. By the way, the magnetic phenomena were first observed about 2500 years ago near the ancient city of Magnesia, what is today Manisa, located in western Turkey, when people saw fragments of magnetized iron. So <em>what happens if you cut a magnet in half? </em>Well, a magnet has two ends, the first one is called a north pole or N pole while the other end is a south pole or S pole, so if you break a bar magnet, each piece has a north and south pole, no matter the size of each new bar although the smaller the piece, the weaker its magnetism. This is true because unlike electric charges, you always find magnetic poles in pairs, that is, ¡they can't be isolated! The option is C. because in the great bar the north pole is to the left while the south pole is to right.

DedPeter [7]3 years ago

3 0

Answer: C

Explanation:

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3 years ago

Read 2 more answers

Two parallel wires carry equal currents in the opposite directions. Point A is midway between the wires, and B is an equal dista

zepelin [54]

Answer:

Complete Question:(missing part)

The current in the left wire having magnitude is = 210 A

The Current in the right wire having magnitude is = 10 A

Answer:

a. $B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (I_{1} +I_{2} )$

Ratio = B 1/B 2 = 210/10 = 21

$b.B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (\frac{I_{1}}{3}-I_{1}} )$

Ratio = B 1/B 2 = 210/3 X10 = 7

Explanation:

Ratio of the magnitude of the magnetic field at point A to that at point B =?

The current in the left wire having magnitude is = 210 A

The Current in the right wire having magnitude is = 10 A

distance between two wires = d

Given that both wires are carrying equal current but in opposite direction therefore

Magnetic field linked with the left wire at point A distance =d/2= according to biot savarts law

$B = \frac{\mu_{0} \times I_{1} }{2\pi\times\frac{d}{2} }$

Magnetic field linked with the right wire at point B distance =d/2

$B = \frac{\mu_{0} \times I_{2} }{2\pi\times\frac{d}{2} }$

Net magnetic field added

B net = B 1 + B 2

$B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (I_{1} +I_{2} )$

Ratio = B 1/B 2 = 210/10 = 21

Magnetic field linked to left wire at point B(d+ d/2)

$B = \frac{\mu_{0} \times I_{1} }{2\pi\times\frac{3d}{2} }$

Magnetic field linked to right wire at point B(d+ d/2)

$B = \frac{\mu_{0} \times I_{2} }{2\pi\times\frac{3d}{2} }$

Net magnetic field subtracted

B net = B 1 - B 2

$B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (\frac{I_{1} }{3}-1 } )$

Ratio = B 1/B 2 = 210/3 X10 = 7

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3 years ago

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Given

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Explanation:

See attachment below.

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