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Ksenya-84 [330]
3 years ago
6

Jadeen says that you can increase the resistance of a copper wire by hammering the wire to make it narrower and longer. Arnell s

ays that you can increase its resistance by heating the wire. Which one, if either, is correct, and why?
1) Arnell, because the conductivity of the wire increases when it is heated.
2) Arnell, because the conductivity of the wire decreases when it is heated.
3) Jadeen, because the conductivity of a wire is directly proportional to its area and inversely proportional to its length.
4) Jadeen, because the conductivity of a copper wire does not increase and might decrease when it is hammered.
5) Both are correct because (b) and (d) are both correct.
Physics
1 answer:
katrin [286]3 years ago
5 0

Answer:

The answer is "Option 5".

Explanation:

Jadeen claims that only by hitting the wire to make it thinner and wider, it can improve a copper wire's strength, which is why a copper wire's permeability doesn't quite improve and can reduce once it is pounded. Arnell says so by heating its wire, it can improve its strength, and when it is heated, the wire's permeability reduces.

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a)

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Explanation:

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The electric force exerted on a charged particle is given by

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In this problem:

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F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N

towards positive x-direction.

The magnetic force instead is given by

F=qvB sin \theta

where

q is the charge

v is the velocity of the charge

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\theta is the angle between the directions of v and B

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b)

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

F_{E_x}=1.19\cdot 10^{-3} N (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- B_x: this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, \theta=0^{\circ}, so the force due to this field is zero.

- B_y: this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, \theta=90^{\circ}. Therefore, \theta=90^{\circ}, so the force due to this field is:

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F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N

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