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Aleks04 [339]
3 years ago
15

A long solenoid that has 810 turns uniformly distributed over a length of 0.380 m produces a magnetic field of magnitude 1.00 10

-4 T at its center. What current is required in the windings for that to occur
Physics
1 answer:
IRINA_888 [86]3 years ago
5 0

Answer:

0.037 A

Explanation:

Magnetic field = B = 1.00 e-4 T

Length = L = 0.380 m

Number of turns = 810

B = μ₀ N I / L

⇒ Current = I = B L / μ₀ N = ( 1 e-4) ( 0.380) / (4π × 10⁻⁷)(810)

                                            = 0.037 A = 37.3 mA

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The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>
  • Let's draw the free body diagram of the system using the given data.
  • From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
  • For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

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  • We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

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  • To find Ny, we need to find the tension T.
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                           F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N

  • Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,

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  • Thus, the magnitude of the force that the beam exerts on the hi.nge will be,

                 N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.

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