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Zigmanuir [339]
4 years ago
15

A straight, horizontal length of copper wire has a current i=28A through it. What are the magnitude and direction of the minimum

magnetic field B needed to suspend the wire(to balance the gravitational force on it) if the linear density of the wire is 46.6g/m?
Physics
1 answer:
seropon [69]4 years ago
6 0

Answer:

0.01631 T

Explanation:

current, i = 28 A

mass per unit length, m/l = 46.6 g/m = 0.0466 kg/m

Let the magnetic field is B.

the weight of the wire is balanced by the magnetic force .

mg = i l B

B = mg / i l

B = (m/l) x g/i

B = 0.0466 x 9.8 / 28

B = 0.01631 T

Thus, the magnetic field is 0.01631 T.

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Answer:

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Explanation:

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Hence, the correct option is (d).

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A wire carries a current in the x-direction. a positively charged particle moves in the –x-direction near the current-carrying w
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Fleming's left-hand rule:

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Learn more about the magnetic fields here:

brainly.com/question/14848188

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