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Gre4nikov [31]
3 years ago
13

An extremely long wire laying parallel to the x -axis and passing through the origin carries a current of 250A running in the po

sitive x -direction. Another extremely long wire laying parallel to the x -axis and passing through the y -axis at ry= 1.8m carries a current of 50A running in the negative x -direction. What is the magnitude of the net magnetic field that the wires generate on the y -axis at ry= −3.510m ?
Physics
1 answer:
viktelen [127]3 years ago
6 0

Answer:

1.232\times 10^{-5}\ T.

Explanation:

<u>Given:</u>

  • Current through the wire, passing through the origin, I_1 = 250\ A.
  • Current through the wire, passing through the y axis, r_y=1.8\ m., I_2 = 50\ A.

According to Ampere's circuital law, the line integral of magnetic field over a closed loop, called Amperian loop, is equal to \mu_o times the net current threading the loop.

\oint \vec B \cdot d\vec l=\mu_o I.

In case of a circular loop, the directions of magnetic field and the line element d\vec l, both are along the tangent of the loop at that point, therefore, \vec B\cdot d\vec l = B\ dl.

\oint \vec B \cdot d\vec l = \oint B\ dl = B\oint dl.\\

\oint dl is the circumference of the Amperian loop = 2\pi r

Therefore,

B\ 2\pi r=\mu_o I\\B=\dfrac{\mu_o I}{2\pi r}.

It is the magnetic field due to a current carrying wire at a distance r from it.

For the first wire, passing through the origin:

Consider an Amperian loop of radius 3.510 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, r_1 = 3.510\ m.

The magnetic field at the given point due to this wire is given by:

B_1 = \dfrac{\mu_o I_1}{2\pi r_1}\\=\dfrac{4\pi \times 10^{-7}\times 250}{2\pi \times 3.510}=1.42\times 10^{-5}\ T.

For the first wire, passing through the y-axis:

Consider an Amperian loop of radius (3.510+1.8) m = 5.310 m,  concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, r_2 = 5.310\ m.

The magnetic field at the given point due to this wire is given by:

B_2 = \dfrac{\mu_o I_2}{2\pi r_2}\\=\dfrac{4\pi \times 10^{-7}\times 50}{2\pi \times 5.310}=1.88\times 10^{-6}\ T.

The directions of current in both the wires are opposite therefore, the directions of the magnetic field due to both the wires are also opposite to that of each other.

Thus, the net magnetic field at r_r=-3.510\ m is given by

B=B_1-B_2 = 1.42\times 10^{-5}-1.88\times 10^{-6}=1.232\times 10^{-5}\ T.

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A car travels 10.0 m/s. What is its velocity in km/h?
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A truck is carrying a steel beam of length 13.0 m on a freeway. An accident causes the beam to be dumped off the truck and slide
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Answer:

emf= 9.88 \times 10^{-3} T

Explanation:

Given:

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According to the Faraday's law the emf induced in a rod passing transversely through a magnetic field is given as:

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3 years ago
The level of mercury falls in a barometer<br>while taking it to a mountain.​
Alexxandr [17]

Answer:

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2 years ago
The diagram shown represents a block-and-tackle pulley system on which an effort of W Newtons supports a load of 120.0N. If the
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Answer:

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Efficiency is the ratio of work out of a system to the work in to the system.

e = Wout / Win

Work is force times distance, so:

e = (Fout × Dout) / (Fin × Din)

Rearranging:

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Fin = (Fout / e) × (Dout / Din)

Fin = (Fout / e) / (Din / Dout)

We know that e = 0.40, and Fout = 120 N.  Since there are 6 pulleys, we also know that Din/Dout = 6.

F = (120 N / 0.4) / 6

F = 50 N

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