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Juliette [100K]

3 years ago

15

4.A 31-cm long conducting wire of 9-g carrying 7- A current is placed in a uniform magnetic field. What are the strength and dir

ection of the magnetic field needed to levitate the wire?

1 answer:

Anika [276]3 years ago

7 0

Answer

Given,

Length of the wire,L = 31 cm = 0.31 m

mass of the wire, m = 9 g = 0.009 Kg

Current in the wire,I = 7 A

Magnetic field strength, B= ?

Equating magnetic force to the weight of the wire.

$BIL = m g$

$B=\dfrac{m g}{IL}$

$B=\dfrac{0.009\times 9.81}{7\times 0.31}$

B = 0.0407 T

For Force to be upward magnetic field direction should be outward of the plane of paper.

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T = 4.25 ms = 4 x 10⁻³ s, the time for rebound
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The change in velocity is
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With a small magnet with a generator it will be taken up quickly because how small it is while with a big generator it would take more force for it for the generator to attach because the larger the magnet that heavier it will be because it is attached to the North Pole magnet

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The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.

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Where,

L = Vibrating length string

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At the same time we have the expression for the number of beats described as

$n = |f_1-f_2|$

Where

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From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:

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gtnhenbr [62]

Answer:

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Answer:

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Explanation:

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