Magnetic flux of rectangular loop w/ wire

In summary, the conversation discusses finding the magnetic flux of a rectangular loop, using the Biot-Savart Law to solve for the magnetic field, and using the integral of B x dl to show that it is independent of path. The integral is evaluated over each side of the loop, and the conversation discusses the calculation for the top and bottom segments of the loop. The conversation also addresses the confusion on whether to integrate the top and bottom segments or just the sides, ultimately concluding that only the sides need to be integrated since the magnetic field is constant along the top segment.
  • #1
Vapor88
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Homework Statement


Find the magnetic flux of the rectangular loop. Show that [tex]\int B x dl[/tex] is independent of path by calculating [tex] \oint B x dl[/tex]

loop.png



Homework Equations





The Attempt at a Solution



First step was to solve for B from the wire, I did that using Biot-Savart Law for a straight wire.

[tex]B = \frac{ \mu_{o}I}{2 \pi s}[/tex]

To get flux use

[tex] \Phi = \int B \bullet da = \int ^{b}_{a} \frac{ \mu_{o} I}{2 \pi s} w ds = \frac{ \mu_{o} I w}{2 \pi} ln ( \frac{b}{a})[/tex]

So far so good?

Okay, now I can show that it's independent of path by doing a line integral.

Here's where I'm sketchy.

I need to use [tex] \oint B \! x \! dl[/tex]

So I need to evaluate an integral over each side of the loop. The value is going to be dependent on distance from the straight wire. Now, B is coming out of the board. The magnitude of this will be the same on the left as it is on the right, but because it's a cross product, they'll cancel each other out. Then, right hand rule shows that evaluating the top segment's integral yields something that points toward the wire, and the bottom one points up. We'll say up is in the [tex]\widehat{s}[/tex] direction. So they won't cancel each other out, but I'm forgetting how to do this...

[tex]B_{top} = \frac{ \mu_{o}I}{2 \pi b}[/tex]

[tex]B_{bottom} = \frac{ \mu_{o}I}{2 \pi a}[/tex]

So now I'm confused, do I have to integrate these? I would think that the only things that I had to integrate were the sides, because B changes for them as they get further from the straight wire.
 
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  • #2
Since B is constant along the top of the wire, you will just integrate along dx. Since the length is 'w', then the integral will just be 'w'.

For cross product please use the "\times" ([itex]\times[/itex]) latex command.
 
Last edited:

What is magnetic flux?

Magnetic flux refers to the measure of the total magnetic field passing through a given area. It is represented by the symbol Φ and is measured in units of webers (Wb).

How is magnetic flux calculated?

Magnetic flux can be calculated by multiplying the magnetic field strength (B) by the area (A) perpendicular to the magnetic field. The formula is Φ = B x A.

What is a rectangular loop with wire?

A rectangular loop with wire is a closed circuit with a rectangular shape that has a wire wrapped around it. This setup allows for the creation of a magnetic field when an electrical current passes through the wire.

What factors affect the magnetic flux of a rectangular loop with wire?

The magnetic flux of a rectangular loop with wire is affected by the strength of the magnetic field, the area of the loop, and the angle between the magnetic field and the loop.

What are some practical applications of magnetic flux of a rectangular loop with wire?

Some practical applications of magnetic flux of a rectangular loop with wire include generators, electric motors, and transformers. These devices use the magnetic field created by the loop to produce and control electrical energy.

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