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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?
Physics
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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Svetach [21]

Answer:

the impulse experienced by the passenger is 630.47 kg

Explanation:

Given;

initial velocity of the car, u = 0

final velocity of the car, v = 9.41 m/s

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The impulse experienced by the passenger is calculated as;

J = ΔP = mv - mu = m(v - u)

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           = 67 x 9.41

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Therefore, the impulse experienced by the passenger is 630.47 kg

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